Two-part epoxy glue comes as two tubes — resin in one, hardener in the other. The tubes can sit side by side in a drawer for years without a hint of change. Squeeze the two pastes out and stir them together, and within minutes the mixture starts to harden. What does stirring make possible?
You have seen reactions written as equations — reactants turning into products. This lesson looks at where a reaction actually happens, one pair of particles at a time.
The idea
Reactant particles are always moving, and moving particles run into each other.
The moment two particles hit each other is a 'collision'.
The way the particles are lined up when they hit is their 'orientation'.
Particles can only react when they collide with enough energy and in the right orientation.
A collision without enough energy, or with the wrong orientation, leaves both particles unchanged.
In a mixture of hydrogen gas and iodine gas, new molecules form only where an H₂ molecule and an I₂ molecule collide.
That is why the epoxy hardens only after stirring: mixing finally lets the resin particles and the hardener particles collide.
A mixture of H₂ and I₂ molecules. New molecules form only where an H₂ molecule and an I₂ molecule collide.
Worked examples
Worked example 1. A sealed flask holds a mixture of hydrogen gas and chlorine gas, which react to form hydrogen chloride. Where in the flask can a new HCl molecule form?
Step 1
Answer: only where an H₂ molecule and a Cl₂ molecule collide with enough energy and in the right orientation.
Worked example 2. Two reactant particles collide, but the collision does not have enough energy. What happens to the two particles?
Step 1
Answer: both particles stay unchanged — they simply bounce apart.
You can now state that particles can only react when they collide with enough energy and in the right orientation.
Check your understanding
What must happen for two reactant particles to react with each other?
AThey must collide with enough energy and in the right orientation.correct
BThey only need to be in the same container at the same time.
This option is wrong — you stopped at nearness — particles react only when they actually collide, with enough energy and in the right orientation.
CThey must collide — any collision between them leads to a reaction.
This option is wrong — you treated every collision as a reacting collision — a collision without enough energy, or with the wrong orientation, leaves both particles unchanged.
DThey must exchange energy from a distance, without touching.
This option is wrong — you let the particles react without contact — a reaction happens only where particles collide.
Particles can only react when they collide with enough energy and in the right orientation. Being in the same container is not enough, and not every collision counts. A collision missing enough energy or the right orientation leaves both particles unchanged.
Check your understanding
A flask contains carbon monoxide gas mixed with oxygen gas, which react to form carbon dioxide. Where in the flask can a new CO₂ molecule form?
AOnly where a CO molecule and an O₂ molecule collide.correct
BAnywhere in the flask, even where the two gases never meet.
This option is wrong — you skipped the collision — new molecules form only where the two kinds of molecule actually collide.
COnly along the glass walls of the flask.
This option is wrong — you moved the reaction to the container — the walls play no part; the reaction happens wherever a CO molecule and an O₂ molecule collide.
DWherever a CO molecule absorbs enough heat, even with no O₂ molecule nearby.
This option is wrong — you let a particle react alone — a new CO₂ molecule needs a collision between a CO molecule and an O₂ molecule.
Particles can only react when they collide with enough energy and in the right orientation. A new CO₂ molecule therefore needs a CO molecule and an O₂ molecule to hit each other. It can form anywhere in the flask — but only at a collision between the two.
Check your understanding
In a mixture of hydrogen gas and bromine gas, an H₂ molecule and a Br₂ molecule collide, but the collision is too gentle to carry enough energy. What happens?
ABoth molecules bounce apart unchanged.correct
BThe molecules react partly, forming one new molecule instead of two.
This option is wrong — you gave a weak collision a partial reaction — a collision without enough energy leaves both particles completely unchanged.
CThe molecules stick together and finish reacting over the next few minutes.
This option is wrong — you let the reaction happen anyway, just slower — without enough energy the collision produces no change at all.
DThe molecules break apart into single atoms.
This option is wrong — you turned a gentle collision into a violent one — a collision without enough energy leaves the molecules whole and unchanged.
Particles can only react when they collide with enough energy and in the right orientation. This collision lacks enough energy. So both molecules stay unchanged — they simply bounce apart.
Lesson 2 of 40 · KEQ-002
Successful and unsuccessful collisions
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You have seen that particles can only react when they collide with enough energy and in the right orientation. This lesson names the two outcomes a collision can have, and shows you how to tell them apart.
The idea
Most collisions between reactant particles do not lead to a reaction.
A collision that leads to a reaction is called a 'successful collision'.
A collision that leaves both particles unchanged is called an 'unsuccessful collision'.
To classify a collision, check two things: does it carry enough energy, and are the particles in the right orientation?
An H₂ molecule and an I₂ molecule that collide fast, lined up side by side, react — a successful collision.
The same two molecules colliding slowly, still lined up side by side, stay unchanged — unsuccessful, because the collision lacks enough energy.
Three collisions between an H₂ molecule and an I₂ molecule. Only the fast, correctly lined-up collision is successful.
The same two molecules colliding fast but end-to-end stay unchanged — unsuccessful, because the orientation is wrong.
A collision missing either check — enough energy or the right orientation — is unsuccessful.
Worked examples
Worked example 1. In a hot mixture of carbon monoxide and nitrogen dioxide, a CO molecule collides at high speed with an NO₂ molecule, lined up so one of NO₂'s oxygen atoms meets the carbon — the orientation that lets an oxygen atom transfer. Classify the collision.
Step 1
Check the energy: the collision is at high speed, so it carries enough energy.
Step 2
Check the orientation: the molecules are lined up for the oxygen atom to transfer, so the orientation is right.
Step 3
Both checks pass.
Step 4
The collision is successful.
Worked example 2. In the same mixture, a CO molecule drifts slowly into an NO₂ molecule, head-on and correctly lined up. Classify the collision.
Step 1
Check the energy: the collision is slow, so it lacks enough energy.
Step 2
One failed check is enough to settle the classification.
Step 3
The collision is unsuccessful, because it lacks enough energy.
You can now classify a collision between reactant particles as successful or unsuccessful from whether the particles collide with enough energy and in the right orientation.
Check your understanding
High in the atmosphere, a nitrogen monoxide molecule collides at high speed with an ozone molecule, lined up correctly for reacting. Classify the collision.
ASuccessful — it carries enough energy and has the right orientation.correct
BUnsuccessful — it lacks enough energy.
This option is wrong — you missed the speed given — the collision is at high speed, so it carries enough energy, and the orientation is right too.
CUnsuccessful — the orientation is wrong.
This option is wrong — you missed the line-up given — the molecules are correctly lined up, and the collision carries enough energy too.
DSuccessful — any collision between two reactant molecules is successful.
This option is wrong — you skipped both checks — a collision is successful only when it carries enough energy AND has the right orientation.
Check the energy: high speed means enough energy. Check the orientation: the molecules are correctly lined up. Both checks pass, so the collision is successful.
Check your understanding
The figure shows two reactant molecules about to collide, with their speeds and line-up marked. For this reaction, side-by-side is the line-up that can react. Classify the collision.
AUnsuccessful — it lacks enough energy.correct
BSuccessful — it carries enough energy and has the right orientation.
This option is wrong — you read the short 'slow' arrows as fast — a slow collision lacks enough energy, whatever its orientation.
CUnsuccessful — the orientation is wrong.
This option is wrong — you faulted the line-up — the molecules are correctly lined up; the failed check is the slow speed.
DSuccessful — a correctly lined-up collision is successful at any speed.
This option is wrong — you let orientation carry the whole test — a successful collision needs enough energy as well.
Check the energy: the arrows are short and labeled slow, so the collision lacks enough energy. One failed check settles it. The collision is unsuccessful, because it lacks enough energy.
Check your understanding
A nitrogen monoxide molecule collides with an ozone molecule at high speed, but the molecules hit lined up the wrong way. Classify the collision.
AUnsuccessful — the orientation is wrong.correct
BUnsuccessful — it lacks enough energy.
This option is wrong — you faulted the energy — the collision is at high speed; the failed check is the wrong orientation.
CSuccessful — it carries enough energy, and energy is all that matters.
This option is wrong — you let energy carry the whole test — a successful collision also needs the right orientation.
DSuccessful — the wrong orientation only makes the reaction happen more gently.
This option is wrong — you turned a failed check into a softer reaction — a collision with the wrong orientation produces no reaction at all.
Check the energy: high speed means enough energy — this check passes. Check the orientation: the molecules hit lined up the wrong way — this check fails. A collision missing either check is unsuccessful.
Lesson 3 of 40 · KEQ-003
Activation energy
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Have You Ever Wondered?
Wonder this:
A kitchen with a gas leak can fill with methane mixed through the air, yet sit unchanged for hours. Methane and oxygen molecules are colliding in that room the whole time — trillions of collisions every second — and almost none of them react. What are nearly all of those collisions missing?
You have seen that a collision without enough energy leaves both particles unchanged. 'Enough energy' has an exact meaning for every reaction, and this lesson names it.
The idea
For every reaction, there is a minimum energy that colliding particles need for their collision to lead to a reaction.
That minimum energy is called the 'activation energy'.
A collision carrying less than the activation energy leaves both particles unchanged.
A collision carrying the activation energy or more can lead to a reaction.
At room temperature, almost no collision between a methane molecule and an oxygen molecule carries the activation energy — so the gas-filled kitchen sits unchanged.
A spark gives the particles near it a burst of extra energy.
With that burst, collisions near the spark carry the activation energy, and the reaction begins.
Worked examples
Worked example 1. What is the activation energy of a reaction?
Step 1
Answer: the minimum energy colliding particles need for their collision to lead to a reaction.
Worked example 2. In a sealed tank, a mixture of hydrogen gas and oxygen gas stays unchanged for years, even though its molecules collide constantly. What do nearly all of those collisions lack?
Step 1
Answer: the activation energy — almost none of the collisions carry the minimum energy the reaction needs.
You can now state that activation energy is the minimum energy colliding particles need for their collision to lead to a reaction.
Check your understanding
What is the activation energy of a reaction?
AThe minimum energy colliding particles need for their collision to lead to a reaction.correct
BThe energy released by the reaction as it turns reactants into products.
This option is wrong — you named the energy that comes OUT — activation energy is the minimum energy a collision must carry going in.
CThe average energy of all the particles in the reaction mixture.
This option is wrong — you averaged the mixture — activation energy is a threshold a single collision must reach, not an average over all particles.
DThe speed at which reactant particles move through the mixture.
This option is wrong — you swapped energy for speed — activation energy is the minimum collision energy, not a particle speed.
Every reaction has a minimum energy that colliding particles need. That minimum is the activation energy. A collision carrying less than the activation energy leaves both particles unchanged.
Check your understanding
A collision between two reactant particles carries less than the reaction's activation energy. What happens?
ABoth particles stay unchanged.correct
BThe particles react, but more slowly than usual.
This option is wrong — you slowed the reaction instead of stopping it — below the activation energy, a collision produces no reaction at all.
CThe particles react, but the reaction releases less energy than usual.
This option is wrong — you shrank the reaction's output — below the activation energy there is no reaction, so there is no output to shrink.
DThe collision's energy is stored up and added to the next collision.
This option is wrong — you banked the energy — a collision below the activation energy simply leaves both particles unchanged.
The activation energy is the minimum energy a collision needs to lead to a reaction. This collision carries less than that minimum. So both particles stay unchanged.
Check your understanding
Propane gas escaping from a barbecue tank mixes with the air, and the mixture sits unchanged. Propane and oxygen molecules in it are colliding constantly. What are nearly all of those collisions missing?
AThe activation energy — the minimum energy the reaction needs.correct
BContact — the propane and oxygen molecules never actually meet.
This option is wrong — you removed the collisions — the molecules collide constantly; nearly all of the collisions lack the activation energy.
COxygen — the air holds far too little oxygen for any reaction to happen.
This option is wrong — you blamed the mixture — air carries plenty of oxygen; what nearly every collision lacks is the activation energy.
DTime — the molecules have simply not finished reacting yet.
This option is wrong — you made the reaction slow instead of unstarted — collisions below the activation energy leave the molecules unchanged, however long they keep colliding.
The molecules are colliding constantly, so contact is not the problem. A collision leads to a reaction only when it carries the activation energy. At everyday temperatures, almost no propane–oxygen collision carries that minimum, so the mixture sits unchanged.
Lesson 4 of 40 · KEQ-004
Activation energy on an energy diagram
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Did You Know?
You have seen that every reaction has an activation energy, and in Unit 12 you read energy diagrams with a reactants' level and a products' level. The activation energy appears on those diagrams too — as a climb.
The idea
An energy diagram's curve does not run straight from the reactants' level to the products' level — it rises over a hump in between.
The hump is the 'energy barrier' between reactants and products.
Left: the activation energy is the climb from the reactants' level to the top of the barrier. Right: three intervals marked on an endothermic diagram for the worked examples.
The activation energy is the energy rise from the reactants' level up to the top of the barrier.
Read it as a climb: it starts at the reactants' level and ends at the very top of the hump.
The figure's left panel shows the energy diagram for methane burning in oxygen, with the activation energy labeled.
The drop from the top of the barrier down to the products' level is not the activation energy.
The gap between the reactants' level and the products' level shows the energy the reaction releases or absorbs — not the activation energy either.
Worked examples
Worked example 1. The figure's right panel shows the energy diagram for the decomposition of calcium carbonate, an endothermic reaction, with three marked intervals: interval 1 climbs from the reactants' level to the top of the barrier, interval 2 drops from the top of the barrier to the products' level, and interval 3 spans from the reactants' level to the products' level. Which interval shows the activation energy?
Step 1
The activation energy is the energy rise from the reactants' level up to the top of the barrier.
Step 2
Interval 1 starts at the reactants' level and ends at the top of the barrier.
Step 3
Interval 1 shows the activation energy.
Worked example 2. On the same diagram, a student points at interval 3 — from the reactants' level to the products' level — and calls it the activation energy. What does interval 3 actually show?
Step 1
Interval 3 spans from the reactants' level to the products' level.
Step 2
That gap is the energy the reaction absorbs, because calcium carbonate's decomposition is endothermic.
Step 3
Interval 3 shows the energy absorbed by the reaction, not the activation energy.
You can now identify the activation energy on an energy diagram as the energy rise from the reactants' level to the top of the barrier between reactants and products.
Check your understanding
The figure shows the energy diagram for the decomposition of hydrogen peroxide, an exothermic reaction, with three marked intervals. Which interval shows the activation energy?
AInterval Pcorrect
BInterval Q
This option is wrong — you took the drop from the top of the barrier down to the products' level — the activation energy is the climb that starts at the reactants' level.
CInterval R
This option is wrong — you took the gap between the reactants' and products' levels — that shows the energy released, not the activation energy.
DNone of the intervals
This option is wrong — you looked past the climb — the interval from the reactants' level to the top of the barrier is exactly the activation energy.
The activation energy is the energy rise from the reactants' level up to the top of the barrier. Interval P starts at the reactants' level and ends at the top of the barrier. So interval P shows the activation energy.
Check your understanding
The figure shows the energy diagram for nitrogen and oxygen reacting to form nitrogen monoxide, an endothermic reaction, with three marked intervals. Which interval shows the activation energy?
AInterval Ycorrect
BInterval X
This option is wrong — you took the gap between the reactants' and products' levels — that shows the energy absorbed, not the activation energy.
CInterval Z
This option is wrong — you took the drop from the top of the barrier down to the products' level — the activation energy is the climb that starts at the reactants' level.
DNone of the intervals
This option is wrong — you looked past the climb — the interval from the reactants' level to the top of the barrier is exactly the activation energy.
The activation energy is the energy rise from the reactants' level up to the top of the barrier. On an endothermic diagram the rule is unchanged — start at the reactants' level, end at the top of the barrier. Interval Y makes that climb, so it shows the activation energy.
Check your understanding
The figure shows the energy diagram for propane burning in oxygen, an exothermic reaction, with three marked intervals. Which interval shows the activation energy?
AInterval 3correct
BInterval 1
This option is wrong — you took the drop from the top of the barrier down to the products' level — the activation energy is the climb that starts at the reactants' level.
CInterval 2
This option is wrong — you took the gap between the reactants' and products' levels — that shows the energy released, not the activation energy.
DNone of the intervals
This option is wrong — you looked past the climb — the interval from the reactants' level to the top of the barrier is exactly the activation energy.
The activation energy is the energy rise from the reactants' level up to the top of the barrier. Interval 3 starts at the reactants' level and ends at the top of the barrier. So interval 3 shows the activation energy.
Lesson 5 of 40 · KEQ-005
Sketching the activation energy barrier
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Did You Know?
You have seen where the activation energy sits on an energy diagram. This lesson has you draw it in yourself — and a drawn arrow is only correct if it starts and ends in the right places.
The idea
Start the activation-energy arrow at the height of the reactants' level.
End the arrow at the very top of the barrier.
The activation-energy arrow always runs from the reactants' level up to the top of the barrier — on exothermic and endothermic diagrams alike.
Draw the arrow vertical, pointing upward, and label it 'activation energy'.
For an endothermic reaction, nothing changes: the arrow still runs from the reactants' level up to the top of the barrier.
On an endothermic diagram, that makes the arrow taller than the gap between the two levels — the barrier top sits above the products' level.
Three drawing mistakes to avoid: starting the arrow at the products' level, starting it at the bottom axis, and stretching it from the reactants' level to the products' level.
Worked examples
Worked example 1. Sketch and label the activation energy on the energy diagram for propane burning in oxygen, an exothermic reaction whose products' level sits below the reactants' level.
Step 1
Find the reactants' level — the height where the curve starts.
Step 2
Draw a vertical arrow from that height up to the top of the barrier.
Step 3
Label the arrow 'activation energy'.
Step 4
An upward arrow from the reactants' level to the top of the barrier, labeled 'activation energy'.
Worked example 2. Sketch and label the activation energy on the energy diagram for the decomposition of sodium hydrogen carbonate (baking soda), an endothermic reaction.
Step 1
The reaction is endothermic, so the products' level sits above the reactants' level.
Step 2
The arrow still starts at the reactants' level and still ends at the top of the barrier.
Step 3
That makes the arrow taller than the gap between the two levels.
Step 4
An upward arrow from the reactants' level past the products' level to the top of the barrier, labeled 'activation energy'.
You can now sketch and label the activation energy on a given energy diagram, drawing the interval from the reactants' energy level up to the top of the barrier.
Your turn
On paper, copy the energy diagram shown for the decomposition of hydrogen peroxide, an exothermic reaction. Then draw and label the activation energy. When your drawing is finished, reveal the model answer and check it against the checklist.
Model answer. The supplied diagram with one added vertical arrow that starts at the height of the reactants' level, ends exactly at the top of the barrier, points upward, and is labeled 'activation energy'.
The arrow starts at the height of the reactants' level.
The arrow ends exactly at the top of the barrier.
The arrow is vertical and points upward.
The arrow is labeled 'activation energy'.
The arrow does not stretch down to the products' level or up from the bottom axis.
Your turn
On paper, copy the energy diagram shown for the decomposition of calcium carbonate, an endothermic reaction. Then draw and label the activation energy. When your drawing is finished, reveal the model answer and check it against the checklist.
Model answer. The supplied diagram with one added vertical arrow from the reactants' level up to the top of the barrier, labeled 'activation energy'. Because the reaction is endothermic, the arrow rises past the products' level to reach the barrier top.
The arrow starts at the height of the reactants' level — the LOWER of the two levels here.
The arrow ends exactly at the top of the barrier, above the products' level.
The arrow is vertical and points upward.
The arrow is labeled 'activation energy'.
The arrow is taller than the gap between the reactants' and products' levels.
Check your understanding
The figure shows the energy diagram for zinc reacting with hydrochloric acid, an exothermic reaction, with no activation energy marked. Which drawing adds the activation-energy arrow correctly?
Acorrect
B
This option is wrong — you started the arrow at the products' level — the activation energy is climbed from the REACTANTS' level.
C
This option is wrong — you drew the energy released by the reaction — the activation-energy arrow ends at the top of the barrier, not at the products' level.
D
This option is wrong — you measured from the bottom axis — the arrow starts at the reactants' level, not at the axis.
The activation-energy arrow runs from the height of the reactants' level up to the very top of the barrier. Starting at the products' level, stopping at the products' level, or starting at the bottom axis all mark some other interval.
Lesson 6 of 40 · KEQ-006
What reaction rate means
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Did You Know?
Wonder this:
Drop an effervescent tablet into a glass of water and it is gone in a couple of minutes. Leave a bicycle out in the rain and its iron parts take months to rust. Both are chemical reactions. Chemists need one word for the difference between them.
You have seen reactions and their equations. An equation says what a reaction makes — it says nothing about how quickly.
The idea
The 'rate' of a reaction means how quickly a reactant is used up or a product forms.
In a fast reaction the rate is high: a reactant is used up quickly, or a product forms quickly.
In a slow reaction the rate is low: the same changes take a long time.
A reaction producing gas bubbles rapidly has a higher rate than the same reaction producing bubbles slowly.
Rate is about time, not amount — how quickly product forms, not how much of it forms.
Worked examples
Worked example 1. Two strips of zinc sit in two acid solutions. One strip gives a steady stream of bubbles; the other gives a bubble only now and then. Which reaction has the higher rate?
Step 1
Answer: the steady-stream reaction — its product is forming more quickly.
Worked example 2. What does the rate of a reaction mean?
Step 1
Answer: how quickly a reactant is used up or a product forms.
You can now state that the rate of a reaction means how quickly a reactant is used up or a product forms.
Check your understanding
A copper roof turns green over about twenty years. A sparkler burns out in about a minute. Which reaction has the higher rate?
AThe sparkler's — its reactants are used up far more quickly.correct
BThe roof's — its reaction keeps running for far longer.
This option is wrong — you measured how LONG the reaction runs — rate is how quickly change happens, and the sparkler's change is far quicker.
CBoth have the same rate, because both reactions eventually finish.
This option is wrong — you judged by completion — rate compares how quickly the changes happen, not whether they finish.
DThe two cannot be compared, because the reactions make different products.
This option is wrong — you let different products block the comparison — rate compares how quickly ANY reactant is used up or product forms.
The rate of a reaction means how quickly a reactant is used up or a product forms. The sparkler's reactants are used up in a minute; the roof's take twenty years. So the sparkler's reaction has the higher rate.
Check your understanding
What does the rate of a reaction mean?
AHow quickly a reactant is used up or a product forms.correct
BHow much product a reaction finally makes.
This option is wrong — you swapped quickness for amount — rate is about time, not about how much product forms.
CHow much energy a reaction releases.
This option is wrong — you swapped quickness for energy — a reaction can release a lot of energy quickly or slowly; rate is the quickly-or-slowly part.
DHow many different products a reaction makes.
This option is wrong — you counted the products — rate is how quickly a reactant is used up or a product forms.
The rate of a reaction means how quickly a reactant is used up or a product forms. A high rate means fast; a low rate means slow.
Check your understanding
A bottle of hydrogen peroxide breaks down gradually over many months. Which statement describes this reaction's rate?
AThe rate is low — the reactant is used up slowly.correct
BThe rate is high — the reaction runs for a very long time.
This option is wrong — you read a long run as a high rate — rate is how quickly the reactant is used up, and months of gradual change means slowly.
CThe rate is zero — nothing is happening inside the bottle.
This option is wrong — you rounded slow down to stopped — the peroxide IS breaking down, just slowly, so the rate is low rather than zero.
DThe reaction has no rate, because no gas bubbles are visible.
This option is wrong — you tied rate to bubbles — rate is about any reactant being used up or product forming, visible bubbles or not.
The rate of a reaction means how quickly a reactant is used up or a product forms. The peroxide is used up gradually over months — slowly. So the rate is low.
Lesson 7 of 40 · KEQ-007
Reactant and product curves
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Did You Know?
You have seen what rate means. From Unit 10, concentration says how much dissolved or gaseous substance is present per volume. Chemists watch a reaction's rate by graphing concentration against time.
The idea
Chemists follow a reaction by measuring each substance's concentration again and again as time passes.
Plotting those measurements gives a 'concentration-time graph': time along the bottom, concentration up the side.
A reactant is used up as the reaction runs, so a reactant's curve falls.
Left: the reactant's curve (H₂O₂) falls while the product's curve (O₂) rises. Right: the worked examples' graph, curves labeled A and B.
A product forms as the reaction runs, so a product's curve rises.
The figure's left panel shows both curves for hydrogen peroxide breaking down into water and oxygen: the H₂O₂ curve falls while the O₂ curve rises.
To identify an unlabeled curve, read its direction: falling means reactant, rising means product.
Worked examples
Worked example 1. The figure's right panel shows the concentration-time graph for dinitrogen pentoxide (N₂O₅) breaking down into nitrogen dioxide and oxygen, with two curves labeled A and B. Curve A rises from zero; curve B falls from a high start. Which curve shows the reactant, N₂O₅?
Step 1
A reactant is used up as the reaction runs, so its curve falls.
Step 2
Curve B is the falling curve.
Step 3
Curve B shows the reactant, N₂O₅.
Worked example 2. On the same graph, which curve shows a product, and why?
Step 1
A product forms as the reaction runs, so its curve rises.
Step 2
Curve A rises from zero.
Step 3
Curve A shows a product.
You can now identify which curve on a concentration-time graph shows a reactant and which shows a product, using that reactant concentration falls while product concentration rises.
Check your understanding
The figure shows the concentration-time graph for hydrogen iodide breaking down into hydrogen and iodine, with two curves labeled X and Y. Which statement matches the graph?
ACurve X shows a reactant and curve Y shows a product.correct
BCurve X shows a product and curve Y shows a reactant.
This option is wrong — you matched the reactant to the rising curve — a reactant is used up, so its concentration falls.
CCurves X and Y both show reactants.
This option is wrong — you read the rising curve as a reactant — a substance whose concentration rises is being made, so it is a product.
DCurves X and Y both show products.
This option is wrong — you read the falling curve as a product — a substance whose concentration falls is being used up, so it is a reactant.
Read each curve's direction. Curve X falls — a reactant is used up, so the falling curve is the reactant. Curve Y rises — a product forms, so the rising curve is the product.
Check your understanding
The figure shows the concentration-time graph for ozone in a lab flask breaking down into oxygen, with two curves labeled P and Q. Which curve shows the product, O₂?
ACurve Pcorrect
BCurve Q
This option is wrong — you picked the falling curve — a product is being made, so its concentration rises, and curve Q falls.
CBoth curves
This option is wrong — you gave the product both directions — only the rising curve shows a substance being made.
DNeither curve
This option is wrong — you looked past the rising curve — a product's concentration rises from zero, which is exactly what curve P does.
A product forms as the reaction runs, so its curve rises. Curve P rises from zero. So curve P shows the product, O₂.
Check your understanding
Bromine reacts with methanoic acid in solution, producing carbon dioxide. The figure shows the concentration-time graph with two curves labeled M and N. Which curve shows the bromine?
ACurve Mcorrect
BCurve N
This option is wrong — you picked the rising curve — bromine is a reactant here, and a reactant is used up, so its curve falls.
CBoth curves
This option is wrong — you gave one substance two curves — each curve tracks one substance, and bromine's is the falling one.
DNeither curve — a reactant's concentration stays constant.
This option is wrong — you held the reactant steady — a reactant is used up as the reaction runs, so its concentration falls.
Bromine is a reactant in this mixture — it is used up. A reactant's curve falls. Curve M is the falling curve, so it shows the bromine.
Lesson 8 of 40 · KEQ-008
Steeper means faster
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Did You Know?
You have seen which curve is which on a concentration-time graph. Plotting two runs of the same reaction on one set of axes shows which run is faster.
The idea
On a concentration-time graph, a steep curve shows concentration changing quickly, and a shallow curve shows it changing slowly.
The rate of a reaction is how quickly a reactant is used up or a product forms — and the curve's steepness shows exactly that.
Left: Run 1's product curve climbs more steeply — Run 1 is faster. Right: the worked examples' reactant curves, Run B falling more steeply.
So the steeper curve belongs to the faster run.
The rule works for product curves and reactant curves alike: rising more steeply or falling more steeply both mean faster.
In the figure's left panel, Run 1's product curve climbs more steeply than Run 2's — Run 1 is the faster run.
Worked examples
Worked example 1. The figure's right panel shows the reactant's curve for two runs of the decomposition of hydrogen peroxide, labeled Run A and Run B. Run B's curve falls more steeply. Which run is faster?
Step 1
A reactant's curve falls as the reactant is used up.
Step 2
Run B's curve falls more steeply, so its reactant is being used up more quickly.
Step 3
Run B is the faster run.
Worked example 2. In the same panel, what does the shallower fall of Run A's curve say about Run A?
Step 1
A shallow fall means the concentration is changing slowly.
Step 2
Run A is the slower run — its reactant is being used up slowly.
You can now compare the rates of two reaction runs from their concentration-time graphs, using that a steeper curve means a faster reaction.
Check your understanding
Zinc reacts with sulfuric acid, producing hydrogen. The figure shows the hydrogen curves for two runs, labeled Run 1 and Run 2. Which run is faster?
ARun 2correct
BRun 1
This option is wrong — you picked the shallower curve — steeper means concentration changing more quickly, and Run 2's curve is the steeper one.
CBoth runs have the same rate.
This option is wrong — you looked past the steepness difference — the two curves climb at clearly different steepnesses, so the rates differ.
DThe graph cannot show which run is faster.
This option is wrong — you looked past the steepness — the steeper curve IS the faster run.
The steeper curve shows concentration changing more quickly. Run 2's product curve climbs more steeply. So Run 2 is the faster run.
Check your understanding
Bromine reacts away in two runs of the same solution reaction. The figure shows the bromine curves, labeled Run P and Run Q. Which run is faster?
ARun Pcorrect
BRun Q
This option is wrong — you picked the shallower fall — a reactant curve falling more steeply means the reactant is used up more quickly, and Run P's falls more steeply.
CBoth runs have the same rate.
This option is wrong — you looked past the steepness difference — the curves fall at clearly different steepnesses, so the rates differ.
DFalling curves cannot show rate.
This option is wrong — you limited the rule to rising curves — falling more steeply means faster, just as rising more steeply does.
The rule covers falling curves too: falling more steeply means faster. Run P's bromine curve falls more steeply. So Run P is the faster run.
Check your understanding
The figure shows the product curves for two runs of the same reaction, labeled Run 1 and Run 2. Run 2's curve ends at a greater height; Run 1's climbs more steeply. Which run is faster?
ARun 1correct
BRun 2
This option is wrong — you judged by the final height — rate is in the steepness, and Run 1's curve climbs more steeply.
CBoth runs have the same rate.
This option is wrong — you canceled height against steepness — only steepness shows rate, and the steepnesses differ.
DThe graph cannot show which run is faster.
This option is wrong — you gave up on the comparison — the steeper climb settles it, whatever the final heights.
Height shows how much product has formed; steepness shows how quickly it is forming. Run 1's curve climbs more steeply. So Run 1 is the faster run, even though Run 2's curve ends higher.
Lesson 9 of 40 · KEQ-009
How rate changes during a reaction
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You have seen that steeper means faster when comparing two runs. Steepness also changes within a single run — one curve tells the story of a whole reaction.
The idea
A reaction is fastest at the very start — its curve is steepest there.
As the reaction proceeds, it slows — the curve becomes shallower.
One run, three parts: steep start (fastest), shallowing middle (slowing), flat end (stopped). Left: a reactant curve. Right: the worked examples' product curve.
When the curve turns flat, the concentration has stopped changing — the reaction has stopped.
Read a single run in three parts: steep start, shallowing middle, flat end.
The figure's left panel shows this on the acid's curve for magnesium dissolving in dilute sulfuric acid, with the three time regions marked.
Worked examples
Worked example 1. The figure's right panel shows the product's curve for zinc reacting with acid in a sealed flask, with time regions J, K, and L marked. In region L the curve is flat. What is the reaction doing during region L?
Step 1
A flat curve means the concentration is no longer changing.
Step 2
No product forming means the reaction has stopped.
Step 3
During region L the reaction has stopped.
Worked example 2. On the same curve, during which region is the reaction fastest?
Step 1
The curve is steepest in region J, at the start of the run.
Step 2
Steepest means the product is forming most quickly.
Step 3
The reaction is fastest during region J, at the start.
You can now describe how the rate of a reaction changes over its course from a concentration-time graph — fastest at the start, slowing as the reaction proceeds, and zero once the curve is flat.
Check your understanding
The figure shows the reactant's curve for a run of the thiosulfate–acid reaction, with time regions 1, 2, and 3 marked. During which region has the reaction stopped?
ARegion 3correct
BRegion 1
This option is wrong — you picked the steep start — steep means fastest, and flat means stopped.
CRegion 2
This option is wrong — you picked the shallowing middle — the reaction is still running there, just slowing.
DThe reaction never stops on this graph.
This option is wrong — you missed the flat end — a flat curve means the concentration no longer changes, so the reaction has stopped.
A flat curve means the concentration has stopped changing. The curve is flat in region 3. So the reaction has stopped during region 3.
Check your understanding
The figure shows the product's curve for a run of marble chips reacting with acid. When is the reaction fastest?
AAt the start of the run.correct
BIn the middle of the run.
This option is wrong — you picked the shallowing section — the curve is steepest at the start, and steepest means fastest.
CNear the end of the run.
This option is wrong — you picked the flattening section — the curve is shallowest there, so the reaction is at its slowest.
DIt runs at the same rate throughout.
This option is wrong — you flattened the whole story — the curve visibly changes from steep to flat, so the rate changes too.
The curve is steepest at the very start. Steepest means the product is forming most quickly. So the reaction is fastest at the start.
Check your understanding
The figure shows a reactant's curve for one run of a solution reaction. Which description matches the curve?
ASteep at first, shallower later, flat at the end.correct
BEqually steep the whole way down.
This option is wrong — you straightened the curve — the fall visibly starts steep and shallows before flattening.
CShallow at first, steepest at the end.
This option is wrong — you ran the story backwards — the reaction is fastest at the start, so the curve is steepest there.
DFlat at first, then steadily falling.
This option is wrong — you delayed the start — the reaction begins at its fastest, so the curve falls steeply from the first moment.
Read the steepness along the curve. It falls steeply at first, shallows through the middle, and turns flat at the end. That is the standard shape of a single run: fastest at the start, slowing, then stopped.
Lesson 10 of 40 · KEQ-010
Concentration and rate
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Did You Know?
You have seen how to read rates from graphs. The next lessons change one condition at a time and predict what the rate does. The first condition is concentration.
The idea
Increasing the concentration of a reactant makes a reaction faster, when other conditions stay unchanged.
Decreasing the concentration of a reactant makes the reaction slower.
Air is about one-fifth oxygen; pure oxygen is a much higher concentration of O₂.
So steel wool burns faster in pure oxygen than in air — the same steel wool, a higher O₂ concentration, a faster reaction.
Worked examples
Worked example 1. Marble chips react with hydrochloric acid, giving off carbon dioxide. Identical chips are dropped into concentrated acid and into dilute acid. Which reaction is faster?
Step 1
The concentrated acid has the higher concentration of the reacting acid.
Step 2
Increasing the concentration of a reactant makes a reaction faster, other conditions unchanged.
Step 3
The chips in the concentrated acid react faster.
Worked example 2. Sodium thiosulfate solution reacts with acid, slowly turning the mixture cloudy. The same acid is added to a concentrated thiosulfate solution and to a dilute one. Which mixture turns cloudy sooner?
Step 1
The concentrated thiosulfate solution has the higher concentration of that reactant.
Step 2
Higher reactant concentration means a faster reaction, other conditions unchanged.
Step 3
The concentrated mixture turns cloudy sooner — its reaction is faster.
You can now predict that increasing the concentration of a reactant makes a reaction faster, other conditions unchanged.
Check your understanding
Zinc granules react with dilute sulfuric acid. The acid is swapped for a more concentrated sulfuric acid, with everything else unchanged. What happens to the rate?
AThe reaction speeds up.correct
BThe reaction slows down.
This option is wrong — you flipped the effect — increasing a reactant's concentration makes the reaction faster.
CThe rate stays the same.
This option is wrong — you separated concentration from rate — a higher concentration of the reacting acid means a faster reaction.
DThe reaction speeds up at first, then drops back to its original rate.
This option is wrong — you made the boost temporary — while the concentration stays higher, the reaction stays faster.
One condition changed: the acid's concentration went up. Increasing the concentration of a reactant makes a reaction faster, other conditions unchanged. So the zinc reacts faster in the more concentrated acid.
Check your understanding
Bleach reacts with a colored stain. The bleach is diluted with water before use, with everything else unchanged. What happens to the rate of the stain-removing reaction?
AThe reaction slows down.correct
BThe reaction speeds up.
This option is wrong — you flipped the effect — diluting lowers the bleach's concentration, and a lower reactant concentration means a slower reaction.
CThe rate stays the same.
This option is wrong — you separated concentration from rate — with the bleach diluted, the reaction runs slower.
DThe reaction stops completely.
This option is wrong — you rounded slower down to stopped — the diluted bleach still reacts, just more slowly.
One condition changed: dilution lowered the bleach's concentration. Decreasing the concentration of a reactant makes the reaction slower. So the stain is removed more slowly.
Check your understanding
A charcoal ember on a barbecue glows gently in air. Predict what happens when the glowing ember is lowered into a jar of pure oxygen instead.
AIt burns faster.correct
BIt burns more slowly.
This option is wrong — you flipped the effect — pure oxygen is a higher concentration of the reacting gas, so the burning is faster.
CIt burns at the same rate.
This option is wrong — you treated air and pure oxygen as the same — pure oxygen holds far more O₂ per volume, and higher reactant concentration means faster.
DIt stops burning entirely.
This option is wrong — you treated pure oxygen as smothering — oxygen is the reacting gas, and more of it per volume means faster burning.
One condition changed: the oxygen concentration went up from air to pure oxygen. Increasing the concentration of a reactant makes a reaction faster, other conditions unchanged. So the ember burns faster — a fiercer, brighter glow.
Lesson 11 of 40 · KEQ-011
Why concentration changes the rate
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You have seen that a higher reactant concentration means a faster reaction. Collisions explain why.
The idea
Raising a reactant's concentration packs more of its particles into the same space.
With more particles in the same space, collisions between reactant particles happen more often.
Each collision has the same chance of success as before — there are simply more collisions every second.
The reaction runs faster because more frequent or more energetic collisions mean more successful collisions per second.
Steel wool burns faster in pure oxygen than in air: with far more O₂ molecules in the same space, collisions with the iron happen far more often.
Same box, same iron surface at the base, different O₂ concentrations. More O₂ molecules in the same space means more frequent collisions.
Worked examples
Worked example 1. Zinc granules react faster in concentrated hydrochloric acid than in dilute hydrochloric acid. Explain why, using collisions.
Step 1
The concentrated acid packs more acid particles into the same space.
Step 2
With more acid particles in the same space, collisions between acid particles and the zinc happen more often.
Step 3
More frequent or more energetic collisions mean more successful collisions per second.
Step 4
More successful collisions per second — the zinc reacts faster in the concentrated acid.
Worked example 2. The cloudiness-making reaction between sodium thiosulfate solution and acid runs faster when the thiosulfate solution is more concentrated. Explain why.
Step 1
The concentrated solution holds more thiosulfate particles in the same space.
Step 2
Collisions between thiosulfate particles and acid particles therefore happen more often.
Step 3
More frequent or more energetic collisions mean more successful collisions per second.
Step 4
More successful collisions per second — the cloudiness appears sooner.
You can now explain why increasing the concentration of a reactant makes a reaction faster — with more reactant particles in the same space, collisions happen more often, and because more frequent or more energetic collisions mean more successful collisions per second.
Check your understanding
Why does increasing a reactant's concentration make a reaction faster?
AMore particles in the same space collide more often, so there are more successful collisions per second.correct
BThe extra particles collide harder, so each collision carries more energy.
This option is wrong — you gave the particles extra energy — concentration changes how OFTEN collisions happen, not how hard each one is.
CEach collision becomes more likely to lead to a reaction, so the same number of collisions gives more successes.
This option is wrong — you raised the success chance of each collision — the chance per collision is unchanged; there are simply more collisions per second.
DThe added particles react with each other as well as with the other reactant.
This option is wrong — you invented a second reaction — the speed-up is more frequent collisions between the same reacting particles.
Higher concentration packs more reactant particles into the same space. With more particles in the same space, collisions happen more often. More frequent or more energetic collisions mean more successful collisions per second — so the reaction is faster.
Check your understanding
A glowing wooden splint flares up in a jar of pure oxygen but only glows in air. Why does the pure oxygen speed up the burning?
AWith more O₂ molecules in the same space, collisions with the wood happen more often, so more succeed each second.correct
BIn pure oxygen, each collision with the wood carries more energy than it does in air.
This option is wrong — you gave the collisions extra energy — pure oxygen changes how often O₂ molecules collide with the wood, not how hard each one hits.
CIn pure oxygen, every collision with the wood leads to a reaction instead of only some of them.
This option is wrong — you made every collision succeed — the chance per collision is unchanged; there are simply far more collisions per second.
DPure oxygen burns by itself, adding its own flame to the wood's.
This option is wrong — you made oxygen a fuel — oxygen does not burn; its molecules collide with the wood more often, so more collisions succeed each second.
Pure oxygen packs far more O₂ molecules into the same space than air does. Collisions between O₂ molecules and the wood therefore happen far more often. More frequent or more energetic collisions mean more successful collisions per second — the splint flares.
Check your understanding
Using a more concentrated vinegar makes its fizzing reaction with baking soda faster. Which chain of causes is correct?
AMore acid particles in the same space → collisions happen more often → more successful collisions per second → faster reaction.correct
BMore acid particles in the same space → each collision carries more energy → more successful collisions → faster reaction.
This option is wrong — you routed the speed-up through collision energy — concentration raises how often collisions happen, not the energy each one carries.
CMore acid particles in the same space → the activation energy falls → faster reaction.
This option is wrong — you moved the activation energy — concentrating the vinegar does not move the energy threshold; it changes how often collisions happen.
DMore acid particles in the same space → more product can form in total → faster reaction.
This option is wrong — you routed rate through amount — how much product could form says nothing about how quickly; the rate rises because collisions happen more often.
Higher concentration means more acid particles in the same space. More particles in the same space means collisions happen more often. More frequent or more energetic collisions mean more successful collisions per second — the fizzing is faster.
Lesson 12 of 40 · KEQ-012
Why reactions slow down as they run
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You have seen that a reaction is fastest at the start and slows until its curve turns flat. Concentration and collisions explain why.
The idea
As a reaction runs, its reactant is used up, so the reactant's concentration falls.
With fewer reactant particles in the same space, collisions happen less often.
Less frequent collisions mean fewer successful collisions per second — so the reaction slows.
That is why — as long as concentration is the only thing changing (temperature steady, nothing added mid-run) — the reaction is fastest at the very start: the reactant's concentration is highest then.
The figure shows this on the acid's curve for magnesium reacting with hydrochloric acid: crowded particles and a steep fall early, sparse particles and a shallow fall late.
Eventually the acid is used up, and no acid particles remain to collide with the magnesium.
The acid's falling curve with particle snapshots: crowded and steep early, sparse and shallow late.
With no collisions left between the reactants, the reaction stops — the curve runs flat.
Worked examples
Worked example 1. Sodium thiosulfate solution reacts with acid, and the thiosulfate's concentration-time curve is steepest in the first seconds. Explain why the reaction is fastest at the start.
Step 1
At the start, none of the thiosulfate has been used up, so its concentration is at its highest.
Step 2
With the most reactant particles in the space, collisions happen most often.
Step 3
The most collisions per second means the most successful collisions per second.
Step 4
The reaction is fastest at the start, when the reactant's concentration is highest.
Worked example 2. The same run's curve turns flat after four minutes and stays flat. Explain, with collisions, why the reaction has stopped.
Step 1
By four minutes, the thiosulfate has been used up.
Step 2
No thiosulfate particles remain to collide with the acid particles.
Step 3
With no reactant collisions left, the reaction has stopped — so the curve runs flat.
You can now explain why a reaction is fastest at the start and slows as it proceeds, using that the reactant concentration falls as reactant is used up, so collisions become less frequent.
Check your understanding
A solution reaction runs at a steady temperature, with nothing added after the start. Why is the reaction fastest at the very start of the run?
AThe reactant's concentration is highest at the start, so collisions happen most often then.correct
BThe reactant's particles carry the most energy at the start.
This option is wrong — you gave the early particles extra energy — their energy does not change through the run; it is the crowd of reactant particles that thins out.
CThe mixture holds the most product at the start, pushing the reaction along.
This option is wrong — you made product the engine — the rate follows the reactant's concentration, which is highest at the start.
DThe container is fullest at the start.
This option is wrong — you counted everything in the container — only the REACTANT's concentration sets the collision rate, and it is highest at the start.
At the start, no reactant has been used up yet, so its concentration is at its highest. With the most reactant particles in the same space, collisions happen most often. More collisions each second means more successful collisions each second — the fastest moment of the run.
Check your understanding
A glow stick's light comes from a reaction between two dissolved chemicals. It glows brightly at first and dims through the night. Why does the reaction slow down?
AThe dissolved reactants are used up, so their concentrations fall and collisions happen less often.correct
BThe stick runs out of stored light rather than reacting more slowly.
This option is wrong — you swapped the reaction for a battery — the light is made by the reaction, and it dims because the reactant concentrations fall.
CThe reaction's activation energy rises as the stick works.
This option is wrong — you moved the activation energy — using up reactant does not move the energy threshold; the reactant crowd thins, so collisions happen less often.
DThe product made early on walls off the remaining particles from each other.
This option is wrong — you built a wall of product — the slowdown is simply fewer reactant particles left to collide.
The reacting chemicals are used up as the stick glows. Their concentrations fall, so collisions between them happen less often. Fewer collisions each second means fewer successful collisions each second — the glow dims.
Check your understanding
The figure marks two times, t₁ and t₂, on a reactant's falling curve. Compare the collisions between reactant particles at the two times.
ACollisions happen more often at t₁, because the reactant's concentration is higher there.correct
BCollisions happen more often at t₂, because the particles have had longer to find each other.
This option is wrong — you let collisions build up over time — by t₂ the reactant crowd has thinned, so collisions happen LESS often.
CCollisions happen equally often at both times, because the same mixture is present.
This option is wrong — you kept the crowd constant — reactant particles are used up between t₁ and t₂, so fewer remain to collide.
DNo collisions happen at t₂.
This option is wrong — you emptied the flask early — the curve is still falling at t₂, so reactant particles remain and collisions continue, just less often.
Read the curve: the concentration at t₁ is higher than at t₂. More reactant particles in the same space means more frequent collisions. So collisions happen more often at t₁ than at t₂.
Lesson 13 of 40 · KEQ-013
Temperature and rate
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Did You Know?
You have seen what concentration does to rate. The second condition is temperature.
The idea
Raising the temperature makes a reaction faster, when other conditions stay unchanged.
Lowering the temperature makes the reaction slower.
The reactions that spoil food run faster on a warm counter than in a refrigerator — the same food spoils sooner when warm.
A refrigerator does not stop the spoiling reactions — it slows them down.
Worked examples
Worked example 1. A glow stick glows by a chemical reaction. Dropped into hot water, does it glow more brightly or more dimly than an identical stick in cool water?
Step 1
Raising the temperature makes a reaction faster, other conditions unchanged.
Step 2
A faster glowing reaction gives off its light more quickly.
Step 3
More brightly in the hot water — the reaction runs faster there.
Worked example 2. The same effervescent tablet is dropped into warm water and into ice-cold water. In which glass does the fizzing reaction finish sooner?
Step 1
The warm glass is at the higher temperature.
Step 2
Raising the temperature makes a reaction faster, other conditions unchanged.
Step 3
In the warm water — the faster reaction finishes sooner.
You can now predict that raising the temperature makes a reaction faster, other conditions unchanged.
Check your understanding
Epoxy glue hardens by a chemical reaction. The glued joint is warmed gently with a heat lamp, with everything else unchanged. What happens to the hardening reaction?
AIt speeds up.correct
BIt slows down.
This option is wrong — you flipped the effect — raising the temperature makes a reaction faster.
CIts rate stays the same.
This option is wrong — you separated temperature from rate — warming the joint makes the hardening reaction run faster.
DIt stops until the joint cools again.
This option is wrong — you turned warming into a pause — heat speeds the reaction up; it never halts it.
One condition changed: the temperature went up. Raising the temperature makes a reaction faster, other conditions unchanged. So the glue hardens sooner under the lamp.
Check your understanding
A cut apple browns by a chemical reaction. The slices are moved into a picnic cooler full of ice, with everything else unchanged. What happens to the browning reaction?
AIt slows down.correct
BIt speeds up.
This option is wrong — you flipped the effect — lowering the temperature makes a reaction slower.
CIts rate stays the same.
This option is wrong — you separated temperature from rate — in the cold the browning reaction runs slower.
DIt stops completely the moment the slices are cold.
This option is wrong — you rounded slower down to stopped — cold slows the browning; the reaction still runs.
One condition changed: the temperature went down. Lowering the temperature makes a reaction slower. So the slices brown more slowly in the cooler.
Check your understanding
Two identical thiosulfate–acid mixtures are prepared for the disappearing-cross experiment. One is warmed to 40 °C; the other stays at 20 °C. In which mixture does the cloudiness appear sooner?
AThe 40 °C mixture.correct
BThe 20 °C mixture.
This option is wrong — you flipped the effect — the warmer mixture reacts faster, so its cloudiness appears sooner.
CBoth mixtures at the same time.
This option is wrong — you erased the temperature effect — raising the temperature makes the reaction faster, so the warm mixture clouds first.
DNeither — warming changes how cloudy the mixture gets, not how soon.
This option is wrong — you swapped rate for amount — temperature changes how QUICKLY the cloudiness forms, and the warmer mixture forms it sooner.
One condition differs: temperature. Raising the temperature makes a reaction faster, other conditions unchanged. The 40 °C mixture reacts faster, so its cloudiness appears sooner.
Lesson 14 of 40 · KEQ-014
Why temperature changes the rate
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Did You Know?
You have seen that raising the temperature makes reactions faster. The explanation has two parts — and both come from collisions.
The idea
Particles move faster at higher temperature.
First part: faster-moving particles collide more often.
Same box, same number of particles — only the speeds differ. Faster particles collide more often AND hit harder.
Second part: faster-moving particles hit harder, so more collisions carry enough energy to react.
Both parts push the same way, because more frequent or more energetic collisions mean more successful collisions per second.
Food spoils faster on a warm counter than in a refrigerator: the spoiling reactions' particles collide more often, and more of those collisions carry enough energy.
Cooling runs both parts in reverse: particles move slower, collide less often, and fewer collisions carry enough energy.
Worked examples
Worked example 1. Epoxy glue hardens faster on a warm day than on a cold one. Explain why, in two parts.
Step 1
Particles move faster at higher temperature.
Step 2
First: the faster-moving particles collide more often.
Step 3
Second: the faster-moving particles hit harder, so more collisions carry enough energy to react.
Step 4
More frequent or more energetic collisions mean more successful collisions per second.
Step 5
Two parts — more collisions, and more of them energetic enough — so the glue hardens faster on the warm day.
Worked example 2. A bottle of hydrogen peroxide kept in a cool cupboard keeps its strength longer than one kept somewhere warm. Explain why the cool bottle's decomposition is slower.
Step 1
At the lower temperature, the particles move slower.
Step 2
First: the slower-moving particles collide less often.
Step 3
Second: the slower-moving particles hit more gently, so fewer collisions carry enough energy to react.
Step 4
Fewer collisions, and fewer of them energetic enough — the cool bottle's peroxide breaks down more slowly.
You can now explain why raising the temperature makes a reaction faster in two ways, because particles move faster at higher temperature — colliding more often, with more collisions carrying enough energy to react — and because more frequent or more energetic collisions mean more successful collisions per second.
Check your understanding
Why does raising the temperature make a reaction faster? Give the full two-part explanation.
AParticles move faster, so they collide more often AND more collisions carry enough energy to react.correct
BThe particles themselves expand at higher temperature, so they are bigger targets and easier to hit.
This option is wrong — you swelled the particles — particles do not grow when heated; they move faster, colliding more often and harder.
CHeat lowers the activation energy the reaction needs.
This option is wrong — you moved the threshold — heating does not change the activation energy; it means more collisions carry enough energy to clear it.
DWarming adds more reactant particles to the mixture.
This option is wrong — you turned temperature into concentration — warming speeds the particles up; it does not add particles.
Particles move faster at higher temperature. Faster particles collide more often, and more of their collisions carry enough energy to react. More frequent or more energetic collisions mean more successful collisions per second.
Check your understanding
A warmed thiosulfate–acid mixture turns cloudy sooner than a cold one. What is happening at the particle level in the warm mixture?
AIts particles move faster — colliding more often, with more collisions carrying enough energy to react.correct
BWarming packs its particles closer together, so they are more concentrated than the cold mixture's.
This option is wrong — you turned temperature into concentration — warming speeds particles up; it does not pack in more of them.
CEvery collision in the warm mixture leads to a reaction.
This option is wrong — you made warm collisions perfect — more succeed, but success still needs enough energy and the right orientation.
DThe warm acid becomes a stronger acid.
This option is wrong — you changed the acid's identity — warming leaves the acid the same and simply makes its particles move faster.
Particles move faster at higher temperature. They collide more often, and more collisions carry enough energy to react. More frequent or more energetic collisions mean more successful collisions per second — the cloudiness appears sooner.
Check your understanding
A glow stick in a freezer glows dimly but lasts all night. Explain the dim glow.
AThe cold particles move slower — fewer collisions each second, and fewer with enough energy to react.correct
BThe cold thickens the glow stick's liquid, so the light made inside cannot escape it.
This option is wrong — you dimmed the light instead of the reaction — the glow is dim because the reaction is slower: fewer, gentler collisions.
CCold raises the reaction's activation energy, so almost no collision can clear it.
This option is wrong — you moved the threshold — cooling does not change the activation energy; in the cold, fewer collisions carry enough energy to clear it.
DThe particles in the stick stop moving completely once it is below freezing.
This option is wrong — you froze the particles solid — they still move at freezer temperatures, only slower, so the reaction runs slowly rather than not at all.
At the lower temperature, the particles move slower. They collide less often, and fewer collisions carry enough energy to react. Fewer successful collisions per second means a slower reaction — a dimmer glow that lasts longer.
Lesson 15 of 40 · KEQ-015
Surface area and rate
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Have You Ever Wondered?
Wonder this:
A grain elevator stores tons of wheat. Solid grain is hard to set alight — yet grain dust floating in the elevator's air can explode. Same substance, same air. What turns a slow-burning solid into an explosive?
You have seen that a reactant's concentration and the mixture's temperature each change a reaction's rate. This lesson adds a factor that matters whenever one reactant is a solid: the size of its pieces.
The idea
Take any solid lump: only its outside is available for other substances to touch.
The total amount of outside a solid has is called its 'surface area'.
Cutting a lump into smaller pieces uncovers new faces, so the same mass of solid ends up with more surface area.
Breaking a solid reactant into smaller pieces makes its reaction faster, other conditions unchanged.
Cutting a solid into smaller pieces uncovers new faces: same mass, more surface area.
A whole stick of chalk reacts away slowly in acid; the same chalk ground to a powder reacts away much faster in the same acid.
That is the grain elevator's danger: dust is grain broken into countless tiny pieces, so its burning reaction runs explosively fast.
Worked examples
Worked example 1. A camper wants a fire going quickly. Predict which catches and burns faster: a thick log, or the same mass of wood split into thin kindling sticks.
Step 1
Splitting the log into kindling breaks the solid into smaller pieces.
Step 2
Breaking a solid reactant into smaller pieces makes its reaction faster, other conditions unchanged.
Step 3
The kindling catches and burns faster.
Worked example 2. An effervescent antacid tablet reacts with water and fizzes. Predict which finishes fizzing sooner: a whole tablet, or the same tablet crushed to powder, each dropped into identical glasses of water.
Step 1
Crushing the tablet breaks the solid into smaller pieces.
Step 2
Smaller pieces of a solid reactant react faster, so the crushed tablet's reaction runs faster.
Step 3
The crushed tablet finishes fizzing sooner.
You can now predict that breaking a solid reactant into smaller pieces, which increases its surface area, makes its reaction faster, other conditions unchanged.
Check your understanding
A student drops a lump of zinc into dilute hydrochloric acid. Into a second beaker of the same acid, at the same temperature, she drops the same mass of zinc as a fine powder. Which zinc finishes reacting first?
BThe zinc lump — one big piece reacts faster than many small ones.
This option is wrong — you reversed the direction — breaking a solid into smaller pieces makes its reaction faster, so the powder finishes first.
CBoth finish together, because the two samples have the same mass.
This option is wrong — you compared masses instead of piece sizes — equal masses react at different speeds when their pieces differ in size.
DNeither reacts, because powdering a metal stops its reactions.
This option is wrong — you treated powdering as changing the metal itself — the powder is still zinc, and smaller pieces of it react faster.
Both beakers hold the same acid at the same temperature, so only the zinc's piece size differs. Breaking a solid reactant into smaller pieces makes its reaction faster, other conditions unchanged. So the powdered zinc reacts faster and finishes first.
Check your understanding
Equal masses of limestone are added to two identical beakers of acid: one beaker gets large chips, the other gets fine powder. Which statement is correct?
AThe powder reacts faster than the chips.correct
BThe chips react faster than the powder.
This option is wrong — you reversed the direction — smaller pieces of a solid reactant react faster.
CThe two react at the same speed in the same acid.
This option is wrong — you let the acid alone set the speed — the solid's piece size changes the rate even in identical acid.
DThe powder reacts more slowly because it settles at the bottom.
This option is wrong — you reasoned from where the solid sits — position is not a rate factor here; piece size is, and smaller pieces react faster.
The beakers differ in one condition only: the limestone's piece size. Breaking a solid reactant into smaller pieces makes its reaction faster, other conditions unchanged. So the powdered limestone reacts faster.
Check your understanding
Equal masses of copper carbonate are added to three identical beakers of acid: as one pressed pellet, as coarse grains, and as fine powder. Which ranking, fastest reaction first, is correct?
APowder, then grains, then pellet.correct
BPellet, then grains, then powder.
This option is wrong — you reversed the whole ranking — the smaller the pieces, the faster the reaction, so the powder leads and the pellet trails.
CGrains, then powder, then pellet.
This option is wrong — you demoted the finest powder — piece size sets the order all the way down, so the finest pieces react fastest.
DAll three react equally fast.
This option is wrong — you compared masses instead of piece sizes — equal masses react at different speeds when their pieces differ in size.
All three beakers hold the same acid, so piece size is the only difference. Breaking a solid reactant into smaller pieces makes its reaction faster, other conditions unchanged. Smallest pieces first: powder, then grains, then the pellet.
Lesson 16 of 40 · KEQ-016
Why surface area changes the rate
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You have seen that breaking a solid reactant into smaller pieces makes its reaction faster. But why should piece size matter at all? The answer sits at the particle level.
The idea
A solid keeps most of its particles locked inside, buried beneath its surface.
The other reactant's particles can only collide with the solid particles they can reach: the ones at the surface.
A buried particle gets no collisions until the layers above it have reacted away.
Same particles, smaller pieces: far more of them sit at a surface, where the other reactant can collide with them.
Breaking the lump into smaller pieces uncovers particles that were buried.
So the reaction speeds up because breaking a solid into smaller pieces exposes more particles at the surface, where collisions with the other reactant can happen.
That is why powdered chalk reacts away faster than a whole stick of chalk in the same liquid: the powder offers the acid far more chalk particles to collide with.
Worked examples
Worked example 1. A flame sets fine steel wool glowing and burning, while an iron nail in the same flame only glows. Both are iron. Explain why the steel wool reacts with oxygen so much faster.
Step 1
Steel wool is iron drawn into very thin strands, so most of its iron particles sit at a surface.
Step 2
The steel wool burns faster because breaking a solid into smaller pieces exposes more particles at the surface, where collisions with the other reactant can happen.
Step 3
In the nail, most iron particles are buried where oxygen cannot collide with them.
Step 4
The thin strands expose far more iron particles to collisions with oxygen, so the steel wool burns while the nail only glows.
Worked example 2. A metal powder reacts quickly with acid. Pressing the same powder into one solid pellet makes the reaction slower. Explain why.
Step 1
Pressing the powder into a pellet buries particles that used to sit at a surface.
Step 2
Buried particles cannot be reached, so the acid's particles have fewer metal particles to collide with.
Step 3
Fewer reachable particles means fewer collisions that can lead to reaction.
Step 4
The pellet hides most of the metal's particles from collisions, so the reaction slows down.
You can now explain why breaking a solid reactant into smaller pieces makes its reaction faster, because breaking a solid into smaller pieces exposes more particles at the surface, where collisions with the other reactant can happen.
Check your understanding
Equal masses of magnesium are dropped into identical beakers of acid: fine filings in one, a single block in the other. The filings react away in minutes; the block takes far longer. Why do the filings react faster?
ABreaking the metal into small pieces exposes more particles at the surface, where the acid's particles can collide with them.correct
BThe filings' particles carry more energy than the block's particles, so their collisions succeed more often.
This option is wrong — you gave the small pieces extra energy — piece size changes how many particles can be reached, not how energetic they are.
CFiling the metal turns it into a more reactive substance than the block.
This option is wrong — you changed the chemistry — the filings are still magnesium; only the number of reachable particles changed.
DThe acid gathers more thickly around small pieces, so the filings meet stronger acid.
This option is wrong — you moved the change to the acid — the acid is identical in both beakers; the filings simply expose more magnesium particles to it.
In the block, most magnesium particles are buried where the acid cannot reach them. The filings react faster because breaking a solid into smaller pieces exposes more particles at the surface, where collisions with the other reactant can happen. More reachable particles means more collisions that can lead to reaction.
Check your understanding
A lump of solid reactant sits in a solution of another reactant. Which of the lump's particles can the solution's particles collide with?
AOnly the particles at the lump's surface.correct
BAll of the lump's particles, wherever they sit.
This option is wrong — you let collisions reach buried particles — a particle inside the lump is covered by other particles, so nothing can collide with it.
CNone of them, until the lump melts.
This option is wrong — you required melting — solids react right at their intact surface, without melting first.
DOnly particles that have broken free of the lump.
This option is wrong — you made particles leave the solid first — collisions happen at the intact surface itself.
A solid keeps most of its particles buried beneath its surface. The other reactant can only collide with the particles it can reach: the ones at the surface. That is why uncovering more particles — by breaking the solid into smaller pieces — speeds the reaction.
Check your understanding
A block of wood burns in a flame. When does a wood particle at the center of the block get its chance to react with oxygen?
AOnly after the layers of wood above it have reacted away.correct
BRight away — oxygen particles pass freely through solid wood to reach it.
This option is wrong — you let oxygen through the solid — buried particles are covered, and collisions only happen at the surface.
CNever — particles at the center of a solid cannot react at all.
This option is wrong — you buried the particle forever — burning eats the block from the outside in, and every layer becomes surface in its turn.
DOnly if the block first melts and frees the particle.
This option is wrong — you required melting — wood burns from its surface without melting; the center just has to wait its turn.
Collisions with oxygen happen only at the surface. A buried particle gets no collisions until the layers above it have reacted away. That waiting is exactly why one large piece reacts more slowly than many small ones.
Lesson 17 of 40 · KEQ-017
What a catalyst is
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Have You Ever Wondered?
Wonder this:
Inside every gasoline car's exhaust pipe sits a metal honeycomb coated with platinum. The platinum speeds up reactions that clean the exhaust — every drive, for the life of the car. It is never topped up and never runs low. How can a substance take part in speeding a reaction, day after day, without ever being spent?
You have seen three ways to speed up a reaction by changing the reactants' conditions: raise the concentration, raise the temperature, or break a solid into smaller pieces. This lesson adds a rate changer of a different kind: an extra substance.
The idea
Some substances speed up a reaction without being used up by it.
Such a substance is called a 'catalyst'.
A catalyst is still present, chemically unchanged, when the reaction ends.
So the mass of a catalyst is the same before and after the reaction it speeds up.
Because it is not used up, a small amount of catalyst can serve over and over.
That is the platinum's secret: it speeds up the exhaust-cleaning reactions without being used up by them, so one thin coating lasts the car's life.
Worked examples
Worked example 1. A student adds 2.0 g of manganese dioxide powder to speed up a reaction. What mass of manganese dioxide remains when the reaction ends?
Step 1
The manganese dioxide is acting as a catalyst, and a catalyst is not used up by the reaction it speeds up.
Step 2
2.0 g — the full amount remains, chemically unchanged.
Worked example 2. What is a substance called that speeds up a reaction without being used up?
Step 1
A catalyst.
You can now state that a catalyst is a substance that speeds up a reaction without being used up, so it is still present and unchanged when the reaction ends.
Check your understanding
In ammonia manufacture, iron powder speeds up the reaction between nitrogen and hydrogen. The iron is still there, chemically unchanged, when the process ends. What is the iron?
AA catalystcorrect
BA reactant
This option is wrong — you treated the iron as a substance the reaction uses up — its unchanged state at the end shows nothing consumed it.
CA product
This option is wrong — you treated the iron as something the reaction made — the iron was added at the start and only sped the reaction up.
DA solvent
This option is wrong — you called the iron the substance the reaction happens in — the iron is a solid added to speed the reaction up.
The iron speeds up the reaction. It is not used up: it is still there, chemically unchanged, at the end. A substance that speeds up a reaction without being used up is a catalyst.
Check your understanding
Catalase, a substance in potato, speeds up the breakdown of hydrogen peroxide. What happens to the amount of catalase while the reaction it speeds up runs to the end?
AIt stays the same from start to finish.correct
BIt falls steadily as the reaction uses it up.
This option is wrong — you let the catalyst be consumed like a reactant — a catalyst is not used up by the reaction it speeds up.
CIt grows as the reaction produces more of it.
This option is wrong — you treated the catalyst as a product — the reaction neither makes nor destroys its catalyst.
DIt drops to zero at the moment the reaction finishes.
This option is wrong — you spent the catalyst at the finish line — the full amount is still present, unchanged, when the reaction ends.
A catalyst speeds up a reaction without being used up. So the amount of catalase is the same before, during, and after the breakdown it speeds up.
Check your understanding
A powder is added to a slow reaction, and the reaction speeds up. When the reaction ends, the powder's mass has fallen from 3.0 g to 1.5 g. Can the powder be a catalyst?
ANo — half of it was used up, and a catalyst is not used up.correct
BYes — it sped the reaction up, and that is all a catalyst has to do.
This option is wrong — you kept only the speed half of the definition — a catalyst must also come through the reaction unchanged, with its mass the same.
CYes — a catalyst normally loses about half its mass in each reaction.
This option is wrong — you built mass loss into the definition — a catalyst's mass is the same before and after the reaction.
DNo — a true catalyst gains mass during the reaction it speeds up.
This option is wrong — you flipped the mass change — a catalyst neither gains nor loses mass; it comes through unchanged.
Check both halves of the definition: speeds the reaction up, and is not used up. This powder sped the reaction up but lost half its mass — the reaction consumed it. A substance the reaction consumes is a reactant, not a catalyst.
Lesson 18 of 40 · KEQ-018
How a catalyst speeds up a reaction
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You have seen what a catalyst is: a substance that speeds up a reaction without being used up. That raises a puzzle — how can something speed up a reaction while staying out of the accounting? The activation energy holds the answer.
The idea
In an ordinary mixture, many collisions carry less than the activation energy, so they lead to no reaction.
A catalyst gives the reacting particles a different path — a different way for the same overall reaction to happen.
The catalyzed path has a lower activation energy than the original path.
It is like crossing a hill range through a low pass instead of over the summit: the same journey, less energy needed.
With a lower energy bar to clear, collisions that were too weak to react before are now strong enough.
So the reaction runs faster because a catalyst provides a different path for the reaction with a lower activation energy, so more collisions have enough energy to react.
The particles' own energies have not changed — only the amount of energy a successful collision needs has dropped.
That is why hydrogen peroxide breaks down faster when a catalyst powder is added: on the catalyzed path, far more of the collisions have enough energy to react.
Worked examples
Worked example 1. Margarine makers bubble hydrogen through hot vegetable oil. With nickel powder present, the reaction runs quickly; without it, it barely runs at all. Explain why the nickel speeds up the reaction.
Step 1
The nickel is a catalyst for this reaction.
Step 2
The reaction runs faster because a catalyst provides a different path for the reaction with a lower activation energy, so more collisions have enough energy to react.
Step 3
On the nickel-provided path, collisions between hydrogen and oil that were too weak to react can now succeed, so the reaction speeds up.
Worked example 2. A catalyst speeds up a reaction without heating the mixture. What does the catalyst change, and what does it leave unchanged?
Step 1
Changed: the path the reaction takes, which has a lower activation energy than the original path.
Step 2
Unchanged: the particles' own energies, and the reactants and products themselves.
Step 3
The catalyst lowers the energy a successful collision needs; it gives the particles no energy of its own.
You can now explain why a catalyst makes a reaction faster, because a catalyst provides a different path for the reaction with a lower activation energy, so more collisions have enough energy to react.
Check your understanding
Platinum in a car's exhaust system speeds up the reaction that turns poisonous carbon monoxide into carbon dioxide. Why does the reaction run faster on the platinum?
AThe platinum provides a different path for the reaction with a lower activation energy, so more collisions have enough energy to react.correct
BThe platinum heats the passing exhaust gases, and hotter particles collide with each other more often and more energetically.
This option is wrong — you swapped in the temperature factor — the catalyst adds no heat; it lowers the energy a successful collision needs.
CThe platinum gives its own energy to the gas particles until nearly every collision succeeds.
This option is wrong — you had the catalyst hand out energy — the particles' energies are unchanged; the needed energy is what drops.
DThe platinum raises the activation energy, forcing the particles to collide much harder and faster.
This option is wrong — you raised the barrier — a catalyst's path has a LOWER activation energy, which is why weaker collisions can now succeed.
A catalyst changes neither the particles' energies nor the mixture's temperature. The reaction speeds up because a catalyst provides a different path for the reaction with a lower activation energy, so more collisions have enough energy to react.
Check your understanding
A reaction can run with or without a catalyst. Compare the activation energy of the catalyzed path with the activation energy of the original path.
AThe catalyzed path's activation energy is lower.correct
BThe catalyzed path's activation energy is higher.
This option is wrong — you flipped the comparison — the catalyzed path needs LESS energy, which is exactly why more collisions succeed on it.
CThe catalyzed path has no activation energy at all.
This option is wrong — you removed the barrier entirely — the catalyzed path still has an activation energy, just a lower one.
DThe two are equal, but particles collide more often on the catalyzed path.
This option is wrong — you kept the barrier and changed collision frequency — a catalyst changes the energy the path needs, not how often particles collide.
A catalyst provides a different path for the reaction with a lower activation energy. Lower — but not zero: collisions still need to carry enough energy for the catalyzed path.
Check your understanding
A catalyst is added to one of two identical reaction mixtures held at the same temperature. In which mixture do more collisions lead to reaction, and why?
AThe catalyzed one — its path needs less energy, so more of the same collisions are strong enough to react.correct
BThe catalyzed one — the catalyst makes its particles move faster than the other mixture's particles.
This option is wrong — you sped the particles up — at the same temperature the particles in both mixtures move the same way; the difference is the lower energy requirement.
CBoth equally — at the same temperature the same share of collisions must succeed.
This option is wrong — you let temperature alone set the success rate — the catalyzed path's lower activation energy lets more of the very same collisions succeed.
DThe uncatalyzed one — the catalyst's particles get in the way of the reactants.
This option is wrong — you turned the catalyst into an obstacle — a catalyst opens an easier path; it does not block the reactants.
Same temperature means the particles in both mixtures collide with the same range of energies. In the catalyzed mixture, the reaction can take a path with a lower activation energy. So more of those same collisions carry enough energy to react — the catalyzed mixture reacts faster.
Lesson 19 of 40 · KEQ-019
A catalyst on the energy diagram
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Did You Know?
You have seen energy diagrams: the reactants' level, the products' level, and the activation energy as the rise from the reactants' level to the top of the barrier. You have also seen that a catalyst provides a path with a lower activation energy. One diagram can show both paths at once.
The idea
To show a catalyst on an energy diagram, a second barrier curve is drawn between the same reactant and product levels.
The catalyzed curve rises to a lower peak than the uncatalyzed curve.
Left: two paths for the same reaction — the catalyzed curve peaks lower, but both start and end at the same levels. Right: the worked examples' diagram, the same two paths labeled only 1 and 2.
The catalyzed path's activation energy is the smaller rise: from the reactants' level up to the lower peak.
The reactants' energy level does not move when a catalyst is added.
The products' energy level does not move either.
So the energy difference between reactants and products — the energy the reaction gives out or takes in — is exactly the same with or without the catalyst.
Only the barrier changes: the catalyst lowers the hill, never the start or the end.
Worked examples
Worked example 1. The figure's right panel shows an energy diagram with two barrier curves between the same reactant and product levels, labeled only 1 and 2. Curve 1 peaks high; curve 2 peaks lower. Which curve shows the catalyzed path?
Step 1
A catalyst provides a path with a lower activation energy, so its curve rises to the lower peak.
Step 2
Curve 2 — the lower peak — shows the catalyzed path.
Worked example 2. The reaction in the figure's right panel gives out energy — its products' level sits below its reactants' level. When a catalyst is added, what happens to the amount of energy the reaction gives out, as read from the diagram?
Step 1
The energy given out is the drop from the reactants' level to the products' level.
Step 2
A catalyst moves neither level — it only lowers the barrier's peak.
Step 3
The energy given out is unchanged — the same drop, with or without the catalyst.
You can now identify the effect of a catalyst on an energy diagram — a lower barrier between reactants and products, with the reactant and product energy levels unchanged.
Check your understanding
The energy diagram shows a reaction that can run with or without a catalyst, with four labeled arrows. Which arrow shows the activation energy of the catalyzed path?
AArrow xcorrect
BArrow w
This option is wrong — you read the uncatalyzed barrier — the catalyzed path is the curve with the lower peak.
CArrow y
This option is wrong — you read the reaction's energy change — activation energy is the rise from the reactants' level to a peak, not the reactant-to-product drop.
DArrow z
This option is wrong — you measured how much the catalyst lowered the peak — each activation energy starts from the reactants' level, not from the other peak.
The catalyzed path is the curve with the lower peak. Its activation energy is the rise from the reactants' level up to that lower peak — arrow x.
Check your understanding
A catalyst is added to a reaction. Which feature of the reaction's energy diagram changes?
AThe height of the barrier's peak.correct
BThe reactants' energy level.
This option is wrong — you moved the starting level — the reactants are the same substances with the same energy, catalyst or not.
CThe products' energy level.
This option is wrong — you moved the finishing level — the products are the same substances with the same energy, catalyst or not.
DThe energy gap between reactants and products.
This option is wrong — you changed the reaction's energy change — with both levels fixed, the gap between them cannot move.
A catalyst provides a path with a lower activation energy, drawn as a curve with a lower peak. The reactant and product levels stay exactly where they were. Only the barrier changes: the hill gets lower, never the start or the end.
Check your understanding
The diagram shows a reaction's two paths, with and without a catalyst. Which statement matches the diagram?
ABoth paths start and end at the same levels; the catalyzed path clears a lower peak.correct
BThe catalyzed path ends at a lower product level, so it gives out more energy.
This option is wrong — you dropped the product level — both curves end at the same level, so both paths give out the same energy.
CThe catalyzed path starts at a higher reactant level, giving the particles a head start.
This option is wrong — you raised the starting level — both curves begin at the same reactants' level; only the peak differs.
DThe two paths differ only in how much time the reaction takes, which the horizontal axis shows.
This option is wrong — you read the horizontal axis as time — it shows the progress of the reaction, and the curves differ in barrier height, not clock time.
Check the three landmarks: reactant level, product level, and peak. The two curves share their start and end; only the peaks differ, and the catalyzed peak is lower. Same start and end means the reaction's energy change is the same on either path.
Lesson 20 of 40 · KEQ-020
Reading rate experiments
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Did You Know?
You have seen the four rate factors — concentration, temperature, surface area, and catalysts — and how to read concentration-time graphs. Rate experiments put all of that together: two runs of the same reaction, alike in every condition but one.
The idea
A fair rate experiment changes exactly one condition between runs and keeps every other condition the same.
First, compare the runs' condition lists and find the one condition that differs.
Then read the result: the steeper curve — or the sooner-finished observation — is the faster run.
When both runs start with the same amounts of reactants, their product curves level off at the same final height; only the steepness differs.
Finally, check the direction: the faster run should be the one with the higher concentration, the higher temperature, the smaller pieces, or the catalyst present.
Two runs of zinc in the same dilute acid, alike except for temperature — 20 °C and 40 °C — give two product curves: the 40 °C curve climbs more steeply to the same final height.
Left: same reaction, same amounts, different temperatures — the 40 °C curve is steeper, but both level off at the same final height. Right: the worked example's magnesium runs, labels lost, Run 1 steeper.
Worked examples
Worked example 1. The figure's right panel shows two runs of magnesium ribbon in hydrochloric acid that were identical except that one used dilute acid and the other used concentrated acid — the graph's condition labels were lost, so the curves are labeled only Run 1 and Run 2. Run 1's curve is steeper, and both level off at the same height. Which run used the concentrated acid?
Step 1
The steeper curve shows the faster run, so run 1 was faster.
Step 2
Higher concentration makes a reaction faster, other conditions unchanged.
Step 3
So the faster run is the one with the concentrated acid.
Step 4
Run 1 — its steeper curve marks the faster, concentrated-acid run.
Worked example 2. Two runs of an effervescent tablet in identical glasses of water: in run A a whole tablet fizzes for 90 seconds; in run B the same mass of tablet, crushed, fizzes for 25 seconds. Which condition changed between the runs, and which run was faster?
Step 1
The runs differ in one condition only: the tablet's piece size — its surface area.
Step 2
The sooner-finished run is the faster run, so run B was faster.
Step 3
That matches the factor's direction: smaller pieces react faster.
Step 4
The changed condition is surface area, and run B — the crushed tablet — was the faster run.
You can now interpret concentration-time graphs or observations from reaction-rate experiments that differ in one condition, identifying which condition changed and which run is faster.
Check your understanding
Two runs of the same reaction were identical, except that a spoonful of manganese dioxide catalyst was added to one of them. The graph shows the two product curves, labeled R and S. Which run contained the catalyst?
ARun R — its curve is steeper, and a catalyst makes the run faster.correct
BRun S — its curve keeps rising for longer, showing more reaction.
This option is wrong — you read a longer rise as more action — the slower run just takes longer to make the same product; steepness marks speed.
CNeither run — both curves reach the same height, so the runs were identical.
This option is wrong — you judged speed from the final height — equal heights show equal amounts of product, and the speed difference lives in the steepness.
DIt cannot be read from the graph, because a catalyst leaves no trace on a product curve.
This option is wrong — you looked past the steepness — a catalyst speeds the run up, and the steeper curve is that faster run.
One condition differs between the runs: the catalyst. The steeper curve shows the faster run, and the catalyst run is the faster one. So curve R, the steeper curve, is the run with the manganese dioxide.
Check your understanding
Two runs of iron reacting with acid are recorded with their conditions. Run 1: large lumps, 30 °C, dilute acid. Run 2: large lumps, 50 °C, dilute acid. Which condition changed between the runs?
AThe temperaturecorrect
BThe iron's piece size
This option is wrong — you picked a condition the lists show unchanged — both runs used large lumps.
CThe acid's concentration
This option is wrong — you picked a condition the lists show unchanged — both runs used dilute acid.
DThe presence of a catalyst
This option is wrong — you added a condition the lists never mention — the only listed difference is 30 °C against 50 °C.
Compare the two condition lists entry by entry. Piece size matches, acid matches; only 30 °C against 50 °C differs. A fair rate experiment changes exactly one condition — here, the temperature.
Check your understanding
Runs P and Q of the same reaction differ in one condition only: run P used acid stored in a warm cupboard, run Q used acid from a refrigerator. The graph shows their product curves. Which claim do the graph and conditions support?
AThe warmer run was faster — the steeper curve matches the higher temperature.correct
BThe colder run was faster — cold acid is fresher and reacts more keenly.
This option is wrong — you gave cold the speed advantage — lower temperature slows a reaction, and the cold run's curve is the shallow one.
CThe two runs went at the same rate, because the same acid was used in both.
This option is wrong — you let the acid's identity settle it — the same acid at different temperatures reacts at different rates, as the two steepnesses show.
DThe warmer run made more product than the colder run.
This option is wrong — you turned faster into more — both curves level off at the same height, so both runs made the same amount, just at different speeds.
One condition differs: the acid's temperature. Curve P is steeper, so run P was faster — and the warm-cupboard acid is the warmer one. Faster is not more: both curves flatten at the same final height.
Summary video — Reaction rates — collisions, the energy barrier, and the factors that change them
Heat blue copper sulfate crystals gently and they turn into a white powder as water is driven out. Add a few drops of water back to the white powder and it turns blue again — the products have turned back into the starting substance. What kind of reaction can be run backward like that?
You have seen reactions written with a single arrow, running from reactants to products. Some reactions refuse to stay one-way.
The idea
In some reactions, the products can react to re-form the reactants.
A reaction whose products can re-form its reactants is called a 'reversible reaction'.
Chemists mark a reversible reaction by writing the symbol ⇌ between reactants and products, in place of the single arrow — you met this double arrow with weak acids, where it showed the change also runs backward.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) reads: nitrogen and hydrogen react to form ammonia, AND ammonia reacts to re-form nitrogen and hydrogen.
A one-way reaction keeps the ordinary single arrow →.
The copper sulfate change is reversible: heating drives the reaction one way, and adding water drives it back.
Worked examples
Worked example 1. The equation 2NO₂(g) ⇌ N₂O₄(g) carries the symbol ⇌. What does that symbol tell you about the reaction?
Step 1
The reaction is reversible — it runs in both directions, so N₂O₄ can react to re-form NO₂.
Worked example 2. In the reversible reaction H₂(g) + I₂(g) ⇌ 2HI(g), what can the product HI do?
Step 1
React to re-form the reactants, H₂ and I₂.
You can now state that in a reversible reaction the products can react to re-form the reactants, and that the symbol ⇌ between reactants and products marks a reaction as reversible.
Check your understanding
Sulfur dioxide reacts with oxygen: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). What does the symbol ⇌ in this equation mean?
AThe reaction is reversible — it runs in both directions.correct
BThe reaction is balanced, with equal atom counts on both sides.
This option is wrong — you read ⇌ as a balancing mark — balancing is checked from the coefficients, and ⇌ says the reaction runs both ways.
CThe reaction runs only from right to left as written.
This option is wrong — you kept one direction and flipped it — ⇌ means BOTH directions run, not the reverse one alone.
DThe reaction has stopped and nothing is changing.
This option is wrong — you read the symbol as a stop sign — ⇌ marks a reaction that runs both ways, not one that has ended.
The symbol ⇌ replaces the single arrow in a reversible reaction. It reads: SO₂ and O₂ react to form SO₃, and SO₃ reacts to re-form SO₂ and O₂.
Check your understanding
In the reversible reaction PCl₃(g) + Cl₂(g) ⇌ PCl₅(g), what can the product PCl₅ do?
AReact to re-form PCl₃ and Cl₂.correct
BNothing further — a product is the reaction's final, unchangeable result.
This option is wrong — you treated the product as an endpoint — in a reversible reaction the products can react to re-form the reactants.
CReact only with more Cl₂ to form new products.
This option is wrong — you sent the product forward again — the ⇌ says PCl₅ can run the reaction backward, re-forming PCl₃ and Cl₂.
DSplit into phosphorus atoms and chlorine atoms.
This option is wrong — you broke the compound into bare atoms — the reverse reaction re-forms the reactants PCl₃ and Cl₂, not separate atoms.
The ⇌ marks this reaction as reversible. So the product, PCl₅, can react to re-form the reactants, PCl₃ and Cl₂.
Check your understanding
A chemist finds that carbon monoxide and hydrogen react to form methanol, and that methanol can react to re-form carbon monoxide and hydrogen. Which equation writes this correctly?
ACO(g) + 2H₂(g) ⇌ CH₃OH(g)correct
BCO(g) + 2H₂(g) → CH₃OH(g)
This option is wrong — you used the one-way arrow — a reaction whose products re-form the reactants needs the two-way symbol ⇌.
CCO(g) + 2H₂(g) = CH₃OH(g)
This option is wrong — you used an equals sign — chemical equations use arrows, and a reversible reaction takes the two-way symbol ⇌.
DCH₃OH(g) → CO(g) + 2H₂(g)
This option is wrong — you wrote only the reverse direction — one single arrow, either way, cannot show a reaction that runs both ways.
Both directions run: forward to methanol, and back to carbon monoxide and hydrogen. A reaction that runs in both directions is written with ⇌ between reactants and products.
Lesson 22 of 40 · KEQ-022
Forward and reverse run together
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You have seen that a reversible reaction runs in both directions. That raises a scheduling question: does the mixture run the reaction forward first, and only later run it backward?
The idea
The left-to-right direction of a reversible reaction is called the 'forward reaction'.
The right-to-left direction is called the 'reverse reaction'.
Both directions run at the same time in the same mixture.
In a reversible reaction mixture, the forward and reverse reactions happen at the same time.
While reactant particles are forming products, product particles elsewhere in the same mixture are re-forming reactants.
In a sealed container of N₂, H₂, and NH₃, ammonia is being made and being broken apart at the same time.
There is no taking turns: both directions run together, in the same mixture, at every moment.
Worked examples
Worked example 1. For the reversible reaction 2NO₂(g) ⇌ N₂O₄(g), name the forward reaction and the reverse reaction.
Step 1
The forward reaction is the left-to-right direction: NO₂ molecules combining to form N₂O₄.
Step 2
The reverse reaction is the right-to-left direction: N₂O₄ breaking apart to re-form NO₂.
Worked example 2. A sealed flask holds a mixture of H₂, I₂, and HI, where H₂(g) + I₂(g) ⇌ 2HI(g). While HI is being formed, what else is happening in the same flask?
Step 1
In a reversible reaction mixture, the forward and reverse reactions happen at the same time.
Step 2
Elsewhere in the flask, HI molecules are breaking apart to re-form H₂ and I₂ — at the same time.
You can now state that in a reversible reaction mixture the forward reaction and the reverse reaction happen at the same time, with reactants forming products while products re-form reactants.
Check your understanding
For the reversible reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), which change is the forward reaction?
ASO₂ and O₂ reacting to form SO₃.correct
BSO₃ breaking apart to re-form SO₂ and O₂.
This option is wrong — you named the reverse reaction — the forward reaction is the left-to-right direction as written.
CSO₂ reacting with SO₃ to form O₂.
This option is wrong — you mixed substances from both sides into one reaction — the forward reaction takes the left side to the right side as written.
DO₂ molecules splitting into oxygen atoms.
This option is wrong — you invented a side reaction — the forward reaction is the written left-to-right change, SO₂ and O₂ forming SO₃.
The forward reaction is the left-to-right direction of the equation as written. Left side: SO₂ and O₂. Right side: SO₃. So the forward reaction is SO₂ and O₂ reacting to form SO₃.
Check your understanding
A sealed container holds a reacting mixture of PCl₃, Cl₂, and PCl₅, where PCl₃(g) + Cl₂(g) ⇌ PCl₅(g). When does the reverse reaction happen?
AAt the same time as the forward reaction.correct
BOnly after the forward reaction has completely finished.
This option is wrong — you put the directions in sequence — in a reversible mixture both directions run together, at every moment.
COnly when the container is heated to restart the mixture.
This option is wrong — you made the reverse reaction wait for a push — it runs alongside the forward reaction without any trigger.
DNever — the reverse reaction is only a possibility, not something that actually runs.
This option is wrong — you read ⇌ as a mere possibility — in the mixture, product particles really are re-forming reactants as reactant particles form products.
In a reversible reaction mixture, the forward and reverse reactions happen at the same time. While PCl₃ and Cl₂ are forming PCl₅, other PCl₅ molecules are breaking apart again. There is no taking turns.
Check your understanding
In a sealed flask, carbon monoxide and hydrogen react reversibly: CO(g) + 2H₂(g) ⇌ CH₃OH(g). Which pair of events is happening at the same moment in the flask?
ACO and H₂ forming CH₃OH, while CH₃OH re-forms CO and H₂.correct
BCO and H₂ forming CH₃OH, while all the CH₃OH waits unchanged.
This option is wrong — you froze the product — the product is reacting too, re-forming CO and H₂ at the same time.
CAll the CO and H₂ reacting away first, then the CH₃OH starting to break down.
This option is wrong — you put the directions in sequence — both directions run together in the same mixture, not in phases.
DCH₃OH re-forming CO and H₂, while the forward reaction pauses to let it happen.
This option is wrong — you made the directions take turns — neither direction pauses; both run at every moment.
Both directions of a reversible reaction run at the same time in the same mixture. Forward: CO and H₂ form CH₃OH. Reverse: CH₃OH re-forms CO and H₂. Both are happening at this moment, side by side.
Lesson 23 of 40 · KEQ-023
Dynamic equilibrium
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Have You Ever Wondered?
Wonder this:
Seal brown NO₂ gas in a glass tube and watch: the brown fades for a while as the gas reacts, then the color settles and stops changing entirely. The tube looks finished. But is anything still happening inside?
You have seen that in a reversible reaction mixture, the forward and reverse reactions run at the same time. Each direction also has a rate — and the two rates need not stay different forever.
The idea
As a reversible reaction runs in a sealed container, the two rates drift toward each other.
Eventually the rate of the forward reaction equals the rate of the reverse reaction.
From that point on, the mixture is at 'dynamic equilibrium'.
At dynamic equilibrium, both reactions are still running — neither has stopped.
At dynamic equilibrium the forward and reverse reactions run at the same rate.
The word 'dynamic' records exactly that: the mixture is balanced, but never still.
Equilibrium is reached in a closed container, where nothing escapes; if a substance leaks away, the balance is never struck.
A sealed mixture of N₂, H₂, and NH₃ is at equilibrium when ammonia is formed exactly as fast as it breaks apart.
That is the settled tube: the color stops changing because the two reactions have matched rates — not because they have stopped.
Worked examples
Worked example 1. In a sealed flask, H₂(g) + I₂(g) ⇌ 2HI(g). Measurements show that HI is being formed exactly as fast as it is breaking apart. What state has the mixture reached?
Step 1
HI formed exactly as fast as it breaks apart means the forward rate equals the reverse rate.
Step 2
The mixture is at dynamic equilibrium.
Worked example 2. A sealed reversible reaction mixture is at dynamic equilibrium. Has the reaction stopped?
Step 1
At dynamic equilibrium the forward and reverse reactions are still running.
Step 2
What is equal is their rates — not zero, just matched.
Step 3
No — both reactions continue, at equal rates.
You can now state that a reversible reaction in a sealed container reaches dynamic equilibrium when the rate of the forward reaction equals the rate of the reverse reaction.
Check your understanding
A sealed container holds the reversible reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). What condition puts this mixture at dynamic equilibrium?
AThe forward reaction and the reverse reaction run at the same rate.correct
BBoth the forward and the reverse reaction have completely stopped.
This option is wrong — you stopped the reactions — at dynamic equilibrium both keep running; it is their rates that match.
CThe container holds the same amount of SO₃ as of SO₂.
This option is wrong — you matched amounts instead of rates — equilibrium is about the two REACTION RATES being equal.
DAll of the SO₂ and O₂ has been converted into SO₃.
This option is wrong — you ran the reaction to completion — a reversible reaction settles with both reactants and products still present.
Dynamic equilibrium is defined by the rates. The mixture is at dynamic equilibrium when the forward rate equals the reverse rate. Both reactions are still running — just as fast as each other.
Check your understanding
A sealed flask of the reversible reaction PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) has reached dynamic equilibrium. What are the two reactions doing?
ABoth are still running, at equal rates.correct
BBoth have stopped running entirely.
This option is wrong — you read equilibrium as a standstill — 'dynamic' records that both reactions keep going.
COnly the forward reaction is still running.
This option is wrong — you switched the reverse reaction off — at equilibrium the reverse runs exactly as fast as the forward.
DThey are taking turns, each running while the other rests.
This option is wrong — you made the directions alternate — both run at every moment, at matched rates.
'Dynamic' means the mixture is balanced but never still. At dynamic equilibrium the forward and reverse reactions both continue, at the same rate.
Check your understanding
Carbon monoxide and hydrogen react reversibly toward methanol: CO(g) + 2H₂(g) ⇌ CH₃OH(g). In which set-up can this mixture reach dynamic equilibrium?
AIn a sealed container, where nothing escapes.correct
BIn an open flask, where the CH₃OH vapor can drift away.
This option is wrong — you let a substance escape — with product leaving, the reverse reaction is starved and the two rates can never match.
COnly in a container from which all the product is steadily removed.
This option is wrong — you removed the very substance the reverse reaction needs — equilibrium needs everything to stay in the container.
DIn any container at all — containment makes no difference to equilibrium.
This option is wrong — you dropped the closed-container condition — if substances leak away, the balance of rates is never struck.
Equilibrium is reached in a closed container, where nothing escapes. The reverse reaction needs the products to still be there. Sealed in, the two rates can drift together and match.
Lesson 24 of 40 · KEQ-024
Concentrations hold constant at equilibrium
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Did You Know?
You have seen that a sealed reversible reaction mixture reaches dynamic equilibrium, with both reactions still running at equal rates. Something measurable follows from that balance.
The idea
At dynamic equilibrium, the concentration of every reactant and every product stays constant over time.
Constant means unchanging: measure any substance now and again later, and the value is the same.
The amounts stop changing even though the reaction has not stopped.
In a sealed N₂, H₂, and NH₃ mixture at equilibrium, the amount of NH₃ stops changing — while ammonia is still being made and broken apart.
From equilibrium on, the NH₃ concentration holds constant — though the reactions keep running.
Before equilibrium, concentrations are still changing; from equilibrium on, every one of them holds steady.
Worked examples
Worked example 1. A sealed flask of H₂(g) + I₂(g) ⇌ 2HI(g) is at dynamic equilibrium. What happens to the concentration of HI over the next hour, if nothing disturbs the flask?
Step 1
At dynamic equilibrium the concentrations of all reactants and products stay constant over time.
Step 2
It stays constant — the same value all hour.
Worked example 2. In that same flask, the concentration of H₂ is also holding steady. Does that mean the H₂ has stopped reacting?
Step 1
The amounts stop changing even though the reaction has not stopped.
Step 2
Both reactions are still running — the steadiness is a feature of equilibrium, not of stillness.
Step 3
No — H₂ is still reacting; its concentration holds steady anyway.
You can now state that at dynamic equilibrium the concentrations of all reactants and products stay constant over time.
Check your understanding
A sealed tube of the reversible reaction N₂O₄(g) ⇌ 2NO₂(g) has reached dynamic equilibrium. What happens to the concentration of N₂O₄ from now on, if nothing disturbs the tube?
AIt stays constant.correct
BIt falls slowly to zero as the forward reaction keeps splitting it.
This option is wrong — you let one direction win — at equilibrium every concentration holds steady, N₂O₄ included.
CIt rises steadily as NO₂ keeps recombining.
This option is wrong — you let the other direction win — at equilibrium every concentration holds steady, neither rising nor falling.
DIt swings up and down around an average value.
This option is wrong — you set the amounts oscillating — at equilibrium each concentration simply holds its value.
At dynamic equilibrium the concentrations of all reactants and products stay constant over time. Measure the N₂O₄ now and an hour from now: the same value.
Check your understanding
A sealed flask of 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is at dynamic equilibrium. Which substances have constant concentrations?
AAll of them — SO₂, O₂, and SO₃ alike.correct
BOnly the product, SO₃.
This option is wrong — you steadied the product alone — at equilibrium the reactants' concentrations hold constant too.
COnly the reactants, SO₂ and O₂.
This option is wrong — you steadied the reactants alone — at equilibrium the product's concentration holds constant too.
DNone of them — in a running reaction, every concentration keeps changing.
This option is wrong — you tied change to running — at equilibrium the reactions keep running while every concentration holds steady.
The rule covers every substance in the mixture. At dynamic equilibrium the concentrations of ALL reactants and products stay constant over time.
Check your understanding
A chemist measures the concentration of CH₃OH in a sealed CO(g) + 2H₂(g) ⇌ CH₃OH(g) mixture at noon, and again at 3 p.m., finding exactly the same value, with the flask undisturbed between measurements. The mixture was at dynamic equilibrium all afternoon. Was the CH₃OH still reacting between the measurements?
AYes — the reactions kept running; at equilibrium the concentrations hold steady anyway.correct
BNo — an unchanged value proves that no reaction touched the CH₃OH.
This option is wrong — you read steady as stopped — at equilibrium the amounts stop changing even though the reactions have not stopped.
CNo — the CH₃OH finished reacting at noon, which is why the readings match.
This option is wrong — you ended the reaction at the first measurement — the steadiness comes from equilibrium, not from the chemistry finishing.
DYes — but only in short bursts between the two measurements.
This option is wrong — you made the reactions intermittent — both directions run continuously at equilibrium, with the amounts steady throughout.
At dynamic equilibrium the amounts stop changing even though the reaction has not stopped. The CH₃OH was being made and reacting away all afternoon — and its concentration held the same value throughout.
Lesson 25 of 40 · KEQ-025
Why equilibrium concentrations stay constant
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Did You Know?
You have seen that at dynamic equilibrium every concentration holds constant — while both reactions keep running. Those two facts look like they should clash: how can the amounts stand still while the chemistry never does?
The idea
Pick any substance in the mixture: one direction of the reaction makes it, and the other direction uses it up.
The concentrations stay constant because the forward and reverse reactions keep happening at the same rate, so each substance is made exactly as fast as it is used up.
Making and using cancel exactly, so the measured amount never moves.
A sink with the tap running and the drain open behaves the same way: when water pours in exactly as fast as it drains out, the water level holds constant while water never stops flowing.
In a sealed N₂, H₂, and NH₃ mixture at equilibrium, the amount of NH₃ holds steady while ammonia is still being made and broken apart — every molecule lost is replaced at the same rate.
Made exactly as fast as it is used up: the amount holds constant while both flows keep running.
Worked examples
Worked example 1. A sealed flask of H₂(g) + I₂(g) ⇌ 2HI(g) is at dynamic equilibrium. Explain why the concentration of HI stays constant even though both reactions are still running.
Step 1
The forward reaction makes HI; the reverse reaction uses HI up.
Step 2
The concentration stays constant because the forward and reverse reactions keep happening at the same rate, so each substance is made exactly as fast as it is used up.
Step 3
HI is made exactly as fast as it is used up, so its concentration never moves.
Worked example 2. In a sealed reversible reaction mixture, the forward reaction briefly runs faster than the reverse reaction. Can the concentrations stay constant while that lasts?
Step 1
With the forward reaction faster, products are made faster than they are used up.
Step 2
Making and using no longer cancel, so the product concentrations rise and the reactant concentrations fall.
Step 3
No — constant concentrations need the two rates equal; unequal rates mean changing amounts.
You can now explain why concentrations stay constant at dynamic equilibrium even though both reactions are still running, because the forward and reverse reactions keep happening at the same rate, so each substance is made exactly as fast as it is used up.
Check your understanding
A sealed flask of 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is at dynamic equilibrium. Why does the amount of SO₃ stay steady even though both reactions are still running?
AThe forward and reverse reactions keep happening at the same rate, so the SO₃ is made exactly as fast as it is used up.correct
BThe SO₃ molecules have become too stable to react any further.
This option is wrong — you retired the product — SO₃ keeps reacting; it is replaced at exactly the rate it is used.
CThe forward reaction stopped running at the very moment that the amount of SO₃ first became steady.
This option is wrong — you read steady as stopped — both reactions keep running, and their equal rates are what hold the amount steady.
DThe sealed flask physically prevents the amounts inside it from changing in any way.
This option is wrong — you credited the container — sealing keeps substances in, but it is the matched rates that hold each amount constant.
One direction makes SO₃; the other uses it up. The amount stays steady because the forward and reverse reactions keep happening at the same rate, so each substance is made exactly as fast as it is used up.
Check your understanding
At dynamic equilibrium, how fast is each substance in the mixture being made, compared with how fast it is being used up?
AExactly as fast — the making and the using cancel.correct
BSlightly faster — a small surplus is what keeps the mixture topped up.
This option is wrong — you left a surplus — any surplus would make the amount grow, and at equilibrium every amount holds constant.
CSlightly slower — the amounts shrink too slowly to measure.
This option is wrong — you left a deficit — any deficit would drain the amount, and at equilibrium every amount holds constant.
DNeither — at equilibrium nothing is being made or used at all.
This option is wrong — you stopped the chemistry — both reactions keep running; their rates are equal, not zero.
The forward and reverse reactions keep happening at the same rate. So each substance is made exactly as fast as it is used up — and its amount never moves.
Check your understanding
In a sealed flask of N₂O₄(g) ⇌ 2NO₂(g), suppose the reverse reaction briefly ran faster than the forward reaction. What would happen to the amount of NO₂ while that lasted?
AIt would fall — NO₂ would be used up faster than it is made.correct
BIt would stay constant — amounts in a sealed flask cannot change.
This option is wrong — you credited the seal — constancy comes from EQUAL rates, and unequal rates mean the amounts change.
CIt would rise — a faster reverse reaction makes extra NO₂.
This option is wrong — you ran the reverse reaction the wrong way — the reverse of this equation USES NO₂, re-forming N₂O₄.
DIt would fall to zero instantly.
This option is wrong — you emptied the flask in one step — a rate difference changes amounts gradually, not all at once.
The reverse reaction of N₂O₄ ⇌ 2NO₂ uses NO₂ to re-form N₂O₄. If using outpaces making, the NO₂ amount falls. Constant amounts need the two rates equal — that is the whole mechanism.
Lesson 26 of 40 · KEQ-026
Constant is not the same as equal
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Did You Know?
You have seen that at dynamic equilibrium every concentration holds constant. The word 'constant' sits one letter-slip away from a different claim — that the amounts are all EQUAL — and the two claims could hardly be more different.
The idea
Constant describes one substance across time: its concentration stops changing.
Equal would compare different substances at one moment: the same concentration of reactant as of product.
Equilibrium promises constant, never equal.
At equilibrium the reactant and product amounts are usually different — one side is often far larger than the other.
An equilibrium mixture can be mostly product, mostly reactant, or anywhere in between.
Constant, not equal: each bar keeps its own height over time — and the two heights differ.
A sealed N₂, H₂, and NH₃ mixture at equilibrium might hold much more NH₃ than N₂ — or much less; either way, each amount holds its own steady value.
Worked examples
Worked example 1. A sealed tube of N₂O₄(g) ⇌ 2NO₂(g) is at equilibrium. A student claims: 'There must be exactly as much NO₂ in the tube as N₂O₄.' Judge the claim.
Step 1
Equilibrium means each concentration is constant across time — it compares nothing between substances.
Step 2
The two amounts are usually different at equilibrium.
Step 3
The claim is incorrect — the tube's NO₂ and N₂O₄ amounts are each constant, but they need not be equal.
Worked example 2. A sealed equilibrium mixture of H₂(g) + I₂(g) ⇌ 2HI(g) is found to be mostly HI, with only small amounts of H₂ and I₂. Can this mixture really be at equilibrium?
Step 1
An equilibrium mixture can be mostly product, mostly reactant, or anywhere in between.
Step 2
What equilibrium requires is that each amount holds constant — which lopsided amounts can do perfectly well.
Step 3
Yes — a mostly-product mixture is a perfectly good equilibrium mixture, as long as every amount holds steady.
You can now distinguish the constant concentrations of an equilibrium mixture from equal concentrations, recognizing that reactant and product amounts at equilibrium are usually different.
Check your understanding
A sealed flask of PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) at equilibrium holds much more PCl₅ than PCl₃, and every amount has held steady for an hour. Is the flask at equilibrium?
AYes — equilibrium needs constant amounts, and lopsided amounts can be constant.correct
BNo — at equilibrium the amounts of reactant and product must be the same.
This option is wrong — you swapped constant for equal — equilibrium fixes each amount across time and says nothing about amounts matching each other.
CNo — a mixture holding mostly product must still be reacting toward balance.
This option is wrong — you treated even amounts as the destination — the destination is steady rates, and a mostly-product mixture can sit there indefinitely.
DYes — but only because PCl₅ stopped reacting once it became the majority.
This option is wrong — you froze the majority substance — every substance keeps reacting at equilibrium; the steady amounts come from matched rates.
Equilibrium promises constant, never equal. Each amount holding its own steady value — however lopsided — is exactly what equilibrium looks like.
Check your understanding
What does 'constant' mean, said of the concentrations in an equilibrium mixture?
AEach substance's concentration stops changing over time.correct
BEvery substance in the mixture has the same concentration.
This option is wrong — you read constant as equal — constant compares one substance with itself across time, not different substances with each other.
CThe concentrations change at a constant, steady speed.
This option is wrong — you kept the amounts moving — at equilibrium the concentrations stop changing altogether.
DThe concentrations are fixed at the values the reaction started with.
This option is wrong — you rewound to the start — the amounts changed during the approach and then held at their equilibrium values, not their starting ones.
Constant describes one substance across time: measure it now and later, same value. It makes no comparison between substances — reactant and product amounts usually differ.
Check your understanding
A student reasons: 'The amounts in our sealed 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) mixture stopped changing, so there must now be as much SO₃ as SO₂.' Which correction is right?
AAmounts that stop changing need not match — each holds its own steady value, and the values usually differ.correct
BNo correction is needed — amounts that stop changing have necessarily become equal.
This option is wrong — you accepted the slip — stopping is about time, matching is about substances, and equilibrium only promises the first.
CThe amounts cannot really have stopped changing, because unequal amounts keep reacting toward equality.
This option is wrong — you made equality the destination — unequal amounts hold perfectly steady at equilibrium; nothing pushes them toward matching.
DThe student should have compared SO₃ with O₂ instead, since those two must be equal.
This option is wrong — you moved the equality to another pair — no pair of substances is required to match at equilibrium.
Constant is a claim about time; equal is a claim between substances. Equilibrium delivers the first and promises nothing about the second. The steady SO₃ and SO₂ amounts can differ — and usually do.
Lesson 27 of 40 · KEQ-027
Reading an approach to equilibrium
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Did You Know?
You have seen how a reversible mixture reaches dynamic equilibrium: rates match, every concentration holds constant, and the steady amounts are usually unequal. A concentration-time graph shows that whole story in one picture.
The idea
On a reversible reaction's concentration-time graph, the reactant curve falls while the product curve rises.
At first, both curves change steeply: the mixture is still approaching equilibrium.
Left: equilibrium is the time from which both curves stay flat — at their own, usually different, heights. Right: the worked examples' H₂/HI graph, flat from the 6-minute mark, with no marker drawn.
Then each curve bends and levels off.
The time from which BOTH curves stay flat marks the arrival at equilibrium.
The flat curves usually sit at different heights — constant, not equal.
From the flat region on, both reactions are still running, at equal rates; the graph goes quiet, the chemistry does not.
For a sealed N₂, H₂, and NH₃ mixture, the moment the N₂ and NH₃ curves both run flat is the moment the mixture reached equilibrium.
Worked examples
Worked example 1. The figure's right panel shows a sealed H₂(g) + I₂(g) ⇌ 2HI(g) run: the H₂ curve falls, the HI curve rises, and both curves are flat from the 6-minute mark onward. When did the mixture reach equilibrium?
Step 1
Equilibrium is marked by the time from which BOTH curves stay flat.
Step 2
Both curves here are flat from 6 minutes onward.
Step 3
At 6 minutes — from then on the mixture is at equilibrium.
Worked example 2. On that same right-panel graph, a student points at the 10-minute mark, where both curves are flat, and says the reaction stopped 4 minutes ago. Judge the claim.
Step 1
Flat curves mean constant concentrations, and constant concentrations at equilibrium come with both reactions still running at equal rates.
Step 2
The graph goes quiet; the chemistry does not.
Step 3
The claim is incorrect — at 10 minutes both reactions are still running, at equal rates; only the amounts have stopped changing.
You can now interpret a concentration-time graph of a reversible reaction, identifying when the system reaches equilibrium and judging claims about the rates and concentrations from that point on.
Check your understanding
The graph shows a sealed run of N₂O₄(g) ⇌ 2NO₂(g): the N₂O₄ curve falls, the NO₂ curve rises, and a time axis runs from 0 to 12 minutes. From the graph, when does the mixture reach equilibrium?
AAt 8 minutes — the time from which both curves stay flat.correct
BAt 0 minutes — the reaction is in balance from the very start.
This option is wrong — you called the start equilibrium — at the start the curves are at their steepest, and the mixture is still far from balanced.
CAt 12 minutes — equilibrium is wherever the graph ends.
This option is wrong — you read the graph's edge as the event — equilibrium arrived when the curves flattened, minutes before the graph stops.
DAt 4 minutes — where the curves are changing fastest.
This option is wrong — you picked the steep region — steep curves mean the amounts are still changing, the opposite of equilibrium.
Find where each curve stops changing. Both curves run flat from 8 minutes onward. The time from which both curves stay flat marks the arrival at equilibrium.
Check your understanding
The graph shows a sealed run of 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) whose curves have been flat since the 5-minute mark. Which claim about the mixture at 9 minutes is correct?
ABoth reactions are still running, at equal rates.correct
BBoth reactions stopped at the 5-minute mark.
This option is wrong — you read flat as stopped — flat curves mean constant amounts, held by two reactions still running at matched rates.
COnly the forward reaction is still running at 9 minutes.
This option is wrong — you switched off the reverse reaction — from equilibrium on, both directions run, exactly as fast as each other.
DThe rates of the two reactions are both zero.
This option is wrong — you set the matched rates to zero — the rates are equal AND nonzero; the chemistry never pauses.
Flat from 5 minutes means the mixture is at equilibrium from then on. At equilibrium the forward and reverse reactions both keep running, at the same rate. The graph goes quiet; the chemistry does not.
Check your understanding
The graph shows a sealed run of PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) with both curves flat after 7 minutes — the PCl₅ curve well above the PCl₃ curve. A student makes four claims about the mixture at 10 minutes. Which claim is correct?
AThe concentrations are constant, and the mixture holds more PCl₅ than PCl₃.correct
BThe concentrations are constant, so the amounts of PCl₅ and PCl₃ must be the same.
This option is wrong — you slid from constant to equal — the flat curves sit at different heights, and equilibrium never levels them.
CThe PCl₅ curve will eventually fall back down to meet the PCl₃ curve.
This option is wrong — you sent the curves toward each other — from equilibrium on, each curve simply keeps its own height.
DThe mixture stopped reacting at 7 minutes, which is why the curves are flat.
This option is wrong — you read flat as stopped — the amounts hold steady while both reactions keep running at equal rates.
From 7 minutes the mixture is at equilibrium: every concentration constant. The heights differ — constant, not equal — so there is more PCl₅ than PCl₃. And beneath the flat curves, both reactions are still running at matched rates.
Summary video — Reversible reactions and dynamic equilibrium
You have seen that a sealed mixture at dynamic equilibrium holds steady because its two rates are matched. Matched rates can be unmatched: if something briefly makes one direction faster, the mixture moves — and chemists have a name for where it ends up.
The idea
Suppose a change makes the forward reaction temporarily run faster than the reverse.
While the forward direction is ahead, product is made faster than it is used, so product builds up.
The mixture keeps changing until the two rates are equal again, settling into a new equilibrium that holds more product than before.
Chemists say the equilibrium has 'shifted toward the products'.
In the opposite case — the reverse reaction temporarily faster — the new equilibrium holds more reactant, and the equilibrium has 'shifted toward the reactants'.
A shift toward the products: the new equilibrium mixture holds more product than the old one did.
Some books say the equilibrium 'shifts right' for toward the products and 'shifts left' for toward the reactants, since products sit on the right of the equation — same meaning, different words.
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), a shift toward the products leaves more NH₃ in the mixture than before.
Worked examples
Worked example 1. A sealed 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) mixture is disturbed, and when it settles into its new equilibrium it holds more SO₃ than before. Which way did the equilibrium shift?
Step 1
SO₃ is the product, and the new mixture holds more of it.
Step 2
The equilibrium shifted toward the products.
Worked example 2. In a sealed H₂(g) + I₂(g) ⇌ 2HI(g) mixture, a change makes the reverse reaction temporarily faster than the forward reaction. Which way does the equilibrium shift, and what happens to the amount of HI?
Step 1
With the reverse direction ahead, HI is used up faster than it is made.
Step 2
The mixture keeps changing until the rates are equal again, settling with more reactant and less HI.
Step 3
The equilibrium shifts toward the reactants, and the amount of HI falls.
You can now state that an equilibrium mixture shifts toward the products when the forward reaction temporarily runs faster than the reverse, leaving more product in the new mixture, and shifts toward the reactants in the opposite case.
Check your understanding
A sealed PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) equilibrium is disturbed. When the mixture settles into its new equilibrium, it holds more PCl₅ than before. Which way did the equilibrium shift?
AToward the products.correct
BToward the reactants.
This option is wrong — you named the wrong side — PCl₅ is the product, and a mixture that ends holding more product has shifted toward the products.
CIt did not shift — equilibrium mixtures cannot end up with different amounts.
This option is wrong — you froze the equilibrium — a disturbed mixture settles into a NEW equilibrium, and its amounts can differ from the old ones.
DIt shifted toward both sides at once.
This option is wrong — you split the shift — a shift has one direction, named for the side the new mixture holds more of.
Find which side the new mixture holds more of. More PCl₅ means more product. A new equilibrium holding more product marks a shift toward the products.
Check your understanding
In a sealed CO(g) + 2H₂(g) ⇌ CH₃OH(g) mixture at equilibrium, a change makes the reverse reaction temporarily faster than the forward reaction. Which way does the equilibrium shift?
AToward the reactants.correct
BToward the products.
This option is wrong — you followed the wrong direction — a temporarily faster REVERSE reaction unmakes product, settling the mixture with more reactant.
CNowhere — a temporary rate difference cannot move an equilibrium.
This option is wrong — you dismissed the imbalance — while one direction is ahead, the amounts change, and the mixture settles into a new equilibrium.
DToward whichever side had less to begin with.
This option is wrong — you steered the shift by the starting amounts — the shift follows the temporarily faster direction, wherever the amounts began.
The reverse direction runs from products back to reactants. While it is ahead, reactants are made faster than they are used. The mixture settles with more reactant: a shift toward the reactants.
Check your understanding
A sealed N₂O₄(g) ⇌ 2NO₂(g) equilibrium shifts toward the reactants after a disturbance. Compare the new equilibrium mixture with the old one.
AIt holds more N₂O₄ and less NO₂ than before.correct
BIt holds more NO₂ and less N₂O₄ than before.
This option is wrong — you swapped the sides — N₂O₄ is the reactant here, and a shift toward the reactants leaves more of it.
CIt holds the same amounts as before, once the rates are equal again.
This option is wrong — you returned the mixture to its old amounts — the rates re-match, but at NEW amounts, with more reactant than before.
DIt holds only N₂O₄, the NO₂ having been used up completely.
This option is wrong — you ran the shift to completion — a shift moves the balance point; it never empties one side.
In this equation N₂O₄ is the reactant and NO₂ is the product. A shift toward the reactants settles the mixture with more reactant and less product. More N₂O₄, less NO₂ — with both still present.
Lesson 29 of 40 · KEQ-029
Le Chatelier's principle
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Have You Ever Wondered?
Wonder this:
An ammonia factory's reactor never finishes the job: at equilibrium the mixture still holds plenty of unreacted N₂ and H₂ alongside the NH₃. The plant's profit depends on coaxing more ammonia out of that settled mixture. What could make a settled equilibrium move?
You have seen that an equilibrium mixture shifts toward the products when the forward reaction temporarily outruns the reverse, and toward the reactants in the opposite case. One rule predicts which way any change pushes a settled equilibrium.
The idea
A mixture at dynamic equilibrium keeps its constant concentrations only while its conditions stay the same.
Change a condition — how much of a substance is present, the temperature, or the pressure — and the mixture usually responds by shifting.
One rule predicts the direction of the response.
When a change is made to a system at dynamic equilibrium, the equilibrium shifts in the direction that counteracts the change.
A shift that counteracts a change works against that change, partly undoing it.
The shift never undoes the change completely, and it never reinforces the change.
This rule is called **'Le Chatelier's principle'**, after the French chemist Henri Le Chatelier.
If more N₂ is added to the N₂(g) + 3H₂(g) ⇌ 2NH₃(g) equilibrium, Le Chatelier's principle says the mixture shifts in whichever direction counteracts that addition — later lessons work out each direction, change by change — including the few changes that turn out to shift nothing.
Worked examples
Worked example 1. A change is made to a system at dynamic equilibrium. According to Le Chatelier's principle, in which direction does the equilibrium shift?
Step 1
Answer: in the direction that counteracts the change.
Worked example 2. The pressure on the 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) equilibrium is changed. Which principle predicts how the mixture responds?
Step 1
Answer: Le Chatelier's principle.
You can now state Le Chatelier's principle — when a change is made to a system at dynamic equilibrium, the equilibrium shifts in the direction that counteracts the change.
Check your understanding
Which statement is Le Chatelier's principle?
AWhen a change is made to a system at dynamic equilibrium, the equilibrium shifts in the direction that counteracts the change.correct
BWhen a change is made to a system at dynamic equilibrium, the equilibrium shifts in the direction that reinforces the change.
This option is wrong — you flipped the response — the shift works AGAINST the change, partly undoing it, never strengthening it.
CWhen a change is made to a system at dynamic equilibrium, the equilibrium always shifts toward the products, no matter what the change is.
This option is wrong — you fixed the direction in advance — the direction depends on the change made; the shift is whichever way counteracts it.
DWhen a change is made to a system at dynamic equilibrium, the forward and reverse reactions both stop.
This option is wrong — you stopped the reactions — both keep running; the mixture shifts until the two rates are equal again.
When a change is made to a system at dynamic equilibrium, the equilibrium shifts in the direction that counteracts the change. Counteracting means working against the change, partly undoing it. The shift never reinforces the change, and the reactions never stop.
Check your understanding
A student adds more Cl₂ to the PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) equilibrium and wants to predict how the mixture responds. Which rule makes that prediction?
ALe Chatelier's principlecorrect
BThe law of conservation of mass
This option is wrong — you reached for the mass law — conservation says atoms are neither created nor destroyed, but it cannot say which direction an equilibrium moves.
CThe definition of dynamic equilibrium
This option is wrong — you named the settled state itself — dynamic equilibrium describes equal forward and reverse rates, but predicting the response to a change takes Le Chatelier's principle.
DThe law of definite proportions
This option is wrong — you named a rule about a compound's fixed makeup — it says nothing about how an equilibrium mixture responds to a change.
The rule that predicts an equilibrium's response to any change is Le Chatelier's principle. When a change is made to a system at dynamic equilibrium, the equilibrium shifts in the direction that counteracts the change.
Check your understanding
More O₂ is added to the 2NO(g) + O₂(g) ⇌ 2NO₂(g) equilibrium. What kind of response does Le Chatelier's principle predict?
AA shift in the direction that counteracts the addition.correct
BA shift in the direction that reinforces the addition.
This option is wrong — you ran the principle backwards — the shift works against the change, partly undoing it.
CNo response of any kind, because a settled equilibrium mixture cannot change.
This option is wrong — you froze the mixture — an equilibrium holds only while conditions stay the same, and a change triggers a counteracting shift.
DA complete stop of both the forward and the reverse reactions.
This option is wrong — you stopped the reactions — both keep running; the mixture shifts until the two rates are equal again.
When a change is made to a system at dynamic equilibrium, the equilibrium shifts in the direction that counteracts the change. Adding O₂ is a change, so the mixture shifts in whichever direction counteracts that addition. Which direction that is — and what it does to each amount — is the work of the next lessons.
Lesson 30 of 40 · KEQ-030
Adding a substance
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You have seen Le Chatelier's principle: when a change is made to a system at dynamic equilibrium, the equilibrium shifts in the direction that counteracts the change. The first change to master is adding more of one substance.
The idea
Add more of one substance to an equilibrium mixture, and the equilibrium shifts in the direction that uses up the added substance.
Using the added substance up is how the shift counteracts the addition.
Adding a reactant therefore shifts the equilibrium toward the products, because the forward direction uses reactants up.
Adding a product therefore shifts the equilibrium toward the reactants, because the reverse direction uses products up.
Add N₂ to the N₂(g) + 3H₂(g) ⇌ 2NH₃(g) equilibrium: the shift toward the products uses up added N₂, and the new mixture holds more NH₃.
Add NH₃ to the same equilibrium instead, and the shift toward the reactants uses up added NH₃.
The shift only partly undoes the addition — some of the added substance is used up, never all of it.
Worked examples
Worked example 1. The water-gas shift equilibrium CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) is running in a sealed reactor when more steam (H₂O) is added. Which way does the equilibrium shift?
Step 1
H₂O is a reactant, and the equilibrium shifts in the direction that uses up the added substance.
Step 2
The direction that uses H₂O up is the forward direction.
Step 3
The equilibrium shifts toward the products.
Worked example 2. More H₂ is added to the same sealed equilibrium, CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g). Which way does the equilibrium shift?
Step 1
H₂ is a product, and the equilibrium shifts in the direction that uses up the added substance.
Step 2
The direction that uses H₂ up is the reverse direction.
Step 3
The equilibrium shifts toward the reactants.
You can now predict the direction an equilibrium shifts when more of one substance is added, using that the equilibrium shifts in the direction that uses up the added substance.
Check your understanding
More Cl₂ is added to the PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) equilibrium. Which way does the equilibrium shift?
AToward the products.correct
BToward the reactants.
This option is wrong — you shifted toward the added substance — the equilibrium moves the OTHER way, in the direction that uses the added Cl₂ up.
CIt does not shift.
This option is wrong — you treated the addition as no change — adding a substance disturbs the settled mixture, and it shifts until it settles again.
DThe forward and reverse reactions both stop.
This option is wrong — you stopped the reactions — both keep running; the mixture shifts toward the products and settles at a new equilibrium.
The equilibrium shifts in the direction that uses up the added substance. Cl₂ is a reactant, and the forward direction uses reactants up. So the equilibrium shifts toward the products.
Check your understanding
More NO₂ is added to the N₂O₄(g) ⇌ 2NO₂(g) equilibrium. Which way does the equilibrium shift?
AToward the reactants.correct
BToward the products.
This option is wrong — you shifted toward the added substance's own side — the mixture uses the added NO₂ up by shifting the other way.
CIt does not shift, because NO₂ was already in the mixture.
This option is wrong — you assumed only new substances disturb an equilibrium — adding more of a substance already present still counts as a change and triggers a shift.
DIt shifts, but the direction cannot be predicted from the equation.
This option is wrong — you missed that the added substance fixes the direction — the equilibrium always shifts in the direction that uses up what was added.
The equilibrium shifts in the direction that uses up the added substance. NO₂ is a product, and the reverse direction uses products up. So the equilibrium shifts toward the reactants.
Check your understanding
The deep-red ion FeSCN²⁺ forms in the equilibrium Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq). More SCN⁻ is added to the solution. What happens to the amount of FeSCN²⁺?
AIt increases.correct
BIt decreases.
This option is wrong — you shifted the wrong way — SCN⁻ is a reactant, so the equilibrium shifts toward the products, making more FeSCN²⁺.
CIt stays exactly the same.
This option is wrong — you treated the settled amounts as fixed — adding a reactant makes the mixture shift toward the products until it settles at new amounts.
DIt first increases and then falls back to its old value.
This option is wrong — you let the shift bounce back — the mixture settles at a NEW equilibrium, holding more FeSCN²⁺ than before.
The equilibrium shifts in the direction that uses up the added substance. SCN⁻ is a reactant, so the shift runs toward the products. More FeSCN²⁺ forms, and the solution turns a deeper red.
Lesson 31 of 40 · KEQ-031
Why adding a reactant shifts the balance
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Did You Know?
You have seen that adding a reactant shifts an equilibrium toward the products, and that the shift counteracts the addition. This lesson opens the machinery: what the added particles actually do to the two rates.
The idea
At equilibrium, the forward and reverse reactions run at the same rate.
Add more of a reactant, and that reactant's concentration rises.
With more reactant particles in the same space, collisions between reactant particles happen more often.
More frequent or more energetic collisions mean more successful collisions per second, so the forward reaction speeds up.
The reverse reaction has not been touched, so for a while the forward reaction outruns the reverse.
While the forward reaction runs faster, reactants are used up and products build up.
Adding a reactant makes the forward rate jump; the gap between the rates closes until the two are equal again, at a new equilibrium.
The falling reactant concentration slows the forward reaction, and the rising product concentration speeds up the reverse.
The shift ends when the two rates meet again — a new equilibrium, holding more product than before.
One sentence carries the whole story: the equilibrium shifts because the change makes one direction temporarily faster than the other, and the mixture keeps changing until the two rates are equal again.
Adding N₂ to the N₂(g) + 3H₂(g) ⇌ 2NH₃(g) equilibrium runs exactly this script: the forward rate jumps first, and the mixture settles with more NH₃ than before.
Worked examples
Worked example 1. More O₂ is added to the 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) equilibrium. Explain why the mixture ends up holding more SO₃.
Step 1
The added O₂ raises a reactant concentration, so collisions between reactant particles happen more often and the forward reaction speeds up.
Step 2
The reverse reaction is unchanged at first, so the forward reaction temporarily outruns it and SO₃ builds up.
Step 3
The mixture keeps changing until the two rates are equal again — a new equilibrium with more SO₃.
Step 4
More SO₃, because the change makes one direction temporarily faster than the other, and the mixture keeps changing until the two rates are equal again.
Worked example 2. More H₂ is added to the C₂H₄(g) + H₂(g) ⇌ C₂H₆(g) equilibrium. Which rate changes first, and what happens to it?
Step 1
The added H₂ raises a reactant concentration.
Step 2
More reactant particles in the same space collide more often, so the forward rate rises — the reverse rate has not changed yet.
Step 3
The forward rate rises first; the reverse rate rises only later, as product builds up.
You can now explain why adding a reactant shifts an equilibrium toward the products, because the change makes one direction temporarily faster than the other, and the mixture keeps changing until the two rates are equal again.
Check your understanding
More Cl₂ is added to the PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) equilibrium. What happens to the two rates at the moment of the addition?
AThe forward rate rises, and the reverse rate is unchanged at first.correct
BThe reverse rate rises, and the forward rate is unchanged at first.
This option is wrong — you attached the extra collisions to the wrong direction — the added Cl₂ is a reactant, so it feeds the FORWARD reaction.
CBoth rates rise instantly by exactly the same amount, so the mixture stays settled.
This option is wrong — you let the addition touch both directions at once — only the forward direction uses Cl₂, so only the forward rate jumps; the reverse rate rises later, as product builds up.
DBoth rates fall, because the mixture has been disturbed.
This option is wrong — you treated any disturbance as a brake — extra reactant particles mean MORE collisions per second, not fewer.
The added Cl₂ raises a reactant concentration, so collisions between reactant particles happen more often. More frequent or more energetic collisions mean more successful collisions per second — the forward rate rises. The reverse rate changes only later, as product builds up.
Check your understanding
More CO is added to the CO(g) + 2H₂(g) ⇌ CH₃OH(g) equilibrium. Why does the mixture end up with more CH₃OH?
ABecause the change makes one direction temporarily faster than the other, and the mixture keeps changing until the two rates are equal again.correct
BBecause the added CO is converted completely into CH₃OH, so none of the addition remains in the final mixture.
This option is wrong — you used up ALL of the added substance — the shift only partly undoes the addition, and some extra CO remains at the new equilibrium.
CBecause adding any substance always shifts an equilibrium toward the products, whatever the substance is.
This option is wrong — you fixed the direction in advance — adding a PRODUCT shifts the other way; here the added CO is a reactant, so the forward direction speeds up.
DBecause the reverse reaction stops completely the moment the extra CO enters, and it never starts again until all of the extra CO has been removed.
This option is wrong — you switched the reverse reaction off — it keeps running throughout, and it even speeds up as product builds.
The added CO raises the forward rate first, so the forward direction temporarily outruns the reverse. Product builds up while the gap between the rates closes. The equilibrium shifts because the change makes one direction temporarily faster than the other, and the mixture keeps changing until the two rates are equal again.
Check your understanding
More I₂ is added to the H₂(g) + I₂(g) ⇌ 2HI(g) equilibrium, and the mixture starts shifting toward the products. What brings the shift to an end?
AReactants are used up and products build up, until the forward and reverse rates are equal again.correct
BThe added I₂ runs out completely, which stops the forward reaction.
This option is wrong — you emptied the addition — the shift uses up only part of the added I₂, and the forward reaction keeps running at the new equilibrium.
CThe temperature of the mixture falls back to its starting value, which slows the forward reaction down again.
This option is wrong — you brought temperature in — nothing changed the temperature; changing concentrations are what pull the two rates back together.
DThe forward reaction stops for good once enough HI has formed.
This option is wrong — you stopped the forward reaction — at the new equilibrium BOTH reactions keep running, at equal rates.
While the forward direction outruns the reverse, reactants are used up and products build up. The falling reactant concentration slows the forward rate, and the rising product concentration speeds up the reverse rate. The mixture keeps changing until the two rates are equal again — the new equilibrium.
Lesson 32 of 40 · KEQ-032
Removing a substance
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Did You Know?
You have seen that adding a substance makes an equilibrium shift in the direction that uses the addition up. Removing a substance is the same move run in reverse.
The idea
Remove some of one substance from an equilibrium mixture, and the equilibrium shifts in the direction that replaces the removed substance.
Replacing what was removed is how the shift counteracts the removal.
Removing a product therefore shifts the equilibrium toward the products, because the forward direction makes more of it.
Removing a reactant therefore shifts the equilibrium toward the reactants, because the reverse direction makes more of it.
Remove NH₃ from the N₂(g) + 3H₂(g) ⇌ 2NH₃(g) equilibrium: the shift toward the products replaces part of the removed NH₃.
The machinery is the rate story you have seen, mirrored: removing NH₃ lowers its concentration, so the reverse reaction slows while the forward reaction is untouched.
The forward direction temporarily outruns the reverse — the equilibrium shifts because the change makes one direction temporarily faster than the other, and the mixture keeps changing until the two rates are equal again.
Factories exploit this: keep drawing the product off as it forms, and the equilibrium keeps shifting toward the products, making more.
Worked examples
Worked example 1. CH₃OH is drawn off from the sealed CO(g) + 2H₂(g) ⇌ CH₃OH(g) equilibrium as it forms. Which way does the equilibrium shift?
Step 1
CH₃OH is a product, and the equilibrium shifts in the direction that replaces the removed substance.
Step 2
The direction that makes more CH₃OH is the forward direction.
Step 3
The equilibrium shifts toward the products.
Worked example 2. Some Cl₂ leaks out of the PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) equilibrium mixture. Which way does the equilibrium shift?
Step 1
Cl₂ is a reactant, and the equilibrium shifts in the direction that replaces the removed substance.
Step 2
The direction that makes more Cl₂ is the reverse direction.
Step 3
The equilibrium shifts toward the reactants.
You can now predict the direction an equilibrium shifts when one substance is removed, using that the equilibrium shifts in the direction that replaces the removed substance.
Check your understanding
SO₃ is removed from the 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) equilibrium. Which way does the equilibrium shift?
AToward the products.correct
BToward the reactants.
This option is wrong — you shifted away from the removed substance — the equilibrium moves in the direction that REPLACES what was removed, and the forward direction makes more SO₃.
CIt does not shift.
This option is wrong — you treated the removal as no disturbance — taking SO₃ out slows the reverse reaction, and the two rates fall out of step.
DThe forward and reverse reactions both stop.
This option is wrong — you stopped the reactions — both keep running; the mixture shifts toward the products and settles at a new equilibrium.
The equilibrium shifts in the direction that replaces the removed substance. SO₃ is a product, and the forward direction makes more of it. So the equilibrium shifts toward the products.
Check your understanding
The deep-red ion FeSCN²⁺ sits in the equilibrium Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq). Some SCN⁻ is removed from the solution. What happens to the amount of FeSCN²⁺?
AIt decreases.correct
BIt increases.
This option is wrong — you shifted the wrong way — SCN⁻ is a reactant, so the equilibrium shifts toward the reactants to replace it, breaking some FeSCN²⁺ apart.
CIt stays exactly the same.
This option is wrong — you assumed only the removed substance changes — the shift the removal triggers remakes the amount of every substance in the mixture.
DIt drops to zero.
This option is wrong — you emptied a substance entirely — shifts change amounts partway; no substance in an equilibrium mixture is driven to zero.
The equilibrium shifts in the direction that replaces the removed substance. SCN⁻ is a reactant, so the shift runs toward the reactants — the reverse direction breaks FeSCN²⁺ apart. The amount of FeSCN²⁺ falls, and the red color fades.
Check your understanding
N₂O₄ is removed from the N₂O₄(g) ⇌ 2NO₂(g) equilibrium. Which way does the equilibrium shift?
AToward the reactants.correct
BToward the products.
This option is wrong — you shifted away from the removed substance — the shift must REPLACE the removed N₂O₄, and the reverse direction is what makes it.
CIt does not shift, because only additions disturb an equilibrium.
This option is wrong — you limited shifts to additions — removing a substance also knocks the two rates out of step, and the mixture shifts until they match again.
DIt shifts, but the direction cannot be predicted from the equation.
This option is wrong — you missed that the removed substance fixes the direction — the equilibrium always shifts in the direction that replaces what was removed.
The equilibrium shifts in the direction that replaces the removed substance. N₂O₄ is the reactant, and the reverse direction makes more of it. So the equilibrium shifts toward the reactants.
Lesson 33 of 40 · KEQ-033
Heat in the equation
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You have seen Le Chatelier's principle handle added and removed substances, and you have seen reactions classified as exothermic or endothermic. Writing heat into the equation itself puts an equilibrium's heat where the principle can reach it.
The idea
An exothermic reaction releases heat as it runs in the forward direction.
Released heat comes out with the products, so heat is written on the product side of the equation.
The reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is exothermic, so it is written N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + heat.
An endothermic reaction absorbs heat as it runs in the forward direction.
Absorbed heat goes in with the reactants, so heat is written on the reactant side of the equation.
The reaction N₂O₄(g) ⇌ 2NO₂(g) is endothermic, so it is written heat + N₂O₄(g) ⇌ 2NO₂(g).
The same written heat reads in both directions: the reverse of an exothermic reaction absorbs that heat back in.
Where heat is written
Forward reaction
Heat is written
Example
exothermic (releases heat)
on the product side
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + heat
endothermic (absorbs heat)
on the reactant side
heat + N₂O₄(g) ⇌ 2NO₂(g)
Heat is written on the side it comes out with: with the products for an exothermic reaction, with the reactants for an endothermic one.
Heat never appears on both sides of one equation — a reversible reaction either releases heat in the forward direction or absorbs it.
Worked examples
Worked example 1. The reaction CO(g) + 2H₂(g) ⇌ CH₃OH(g) is exothermic. Write heat into the equation.
Step 1
Exothermic means heat is released in the forward direction, so heat is written on the product side.
Step 2
CO(g) + 2H₂(g) ⇌ CH₃OH(g) + heat
Worked example 2. The reaction CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g) is endothermic. Write heat into the equation.
Step 1
Endothermic means heat is absorbed in the forward direction, so heat is written on the reactant side.
Step 2
heat + CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)
You can now identify heat as a reactant in an endothermic reversible reaction and as a product in an exothermic one, writing it on the matching side of the equation.
Check your understanding
The reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is exothermic. Which equation writes heat correctly?
A2SO₂(g) + O₂(g) ⇌ 2SO₃(g) + heatcorrect
Bheat + 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
This option is wrong — you placed heat with the reactants — an EXOTHERMIC reaction releases heat in the forward direction, so heat belongs on the product side.
Cheat + 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) + heat
This option is wrong — you wrote heat on both sides — a reversible reaction either releases heat forward or absorbs it forward, never both.
D2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
This option is wrong — you left heat out — the word 'exothermic' is exactly the instruction to write heat on the product side.
Exothermic means the forward reaction releases heat. Released heat comes out with the products, so heat is written on the product side. 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) + heat
Check your understanding
The reaction N₂(g) + O₂(g) ⇌ 2NO(g) is endothermic. Which equation writes heat correctly?
Aheat + N₂(g) + O₂(g) ⇌ 2NO(g)correct
BN₂(g) + O₂(g) ⇌ 2NO(g) + heat
This option is wrong — you placed heat with the products — an ENDOTHERMIC reaction absorbs heat in the forward direction, so heat belongs on the reactant side.
Cheat + N₂(g) + O₂(g) ⇌ 2NO(g) + heat
This option is wrong — you wrote heat on both sides — a reversible reaction either releases heat forward or absorbs it forward, never both.
DN₂(g) + O₂(g) ⇌ 2NO(g)
This option is wrong — you left heat out — the word 'endothermic' is exactly the instruction to write heat on the reactant side.
Endothermic means the forward reaction absorbs heat. Absorbed heat goes in with the reactants, so heat is written on the reactant side. heat + N₂(g) + O₂(g) ⇌ 2NO(g)
Check your understanding
The equilibrium PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) + heat is shown with heat written in. What does the written heat say about the reaction?
AThe forward reaction releases heat — the reaction is exothermic.correct
BThe forward reaction absorbs heat — the reaction is endothermic.
This option is wrong — you read the label from the wrong side — heat written with the PRODUCTS is heat coming out of the forward reaction.
CThe mixture is at a high temperature.
This option is wrong — you read heat as a temperature reading — the written heat records the direction of energy flow, not how hot the mixture is.
DHeat is one of the chemical substances that the reaction makes.
This option is wrong — you promoted the label to a substance — heat is energy; it is written like a product only to show which direction releases it.
Heat on the product side means the forward reaction releases heat. A heat-releasing forward reaction is an exothermic reaction. The written heat is a label for energy flow — not a substance, and not a temperature.
Lesson 34 of 40 · KEQ-034
Raising the temperature
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You have seen heat written into a reversible equation on the side it comes out with, and you have seen that adding a substance shifts an equilibrium in the direction that uses the addition up. Raising the temperature plays the same game with heat.
The idea
Raising the temperature of an equilibrium mixture supplies extra heat — the change is 'more heat'.
The equilibrium shifts in the direction that uses up the added heat, just as it shifts to use up an added substance.
The direction that uses heat up is the direction that absorbs heat — the direction leading away from the side where heat is written.
So raising the temperature shifts an equilibrium away from the side where heat is written.
The equation N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + heat carries heat on the product side, so raising the temperature shifts this equilibrium toward the reactants.
The new mixture holds less NH₃ — for this reaction, heating costs product.
The equation heat + N₂O₄(g) ⇌ 2NO₂(g) carries heat on the reactant side, so raising the temperature shifts this equilibrium toward the products.
That shift is visible: NO₂ is brown, and a sealed tube of this mixture darkens as it warms.
Worked examples
Worked example 1. The temperature of the 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) + heat equilibrium is raised. Which way does the equilibrium shift?
Step 1
Heat is written on the product side.
Step 2
Raising the temperature shifts the equilibrium away from the side where heat is written.
Step 3
The equilibrium shifts toward the reactants.
Worked example 2. The temperature of the heat + CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g) equilibrium is raised. Which way does the equilibrium shift?
Step 1
Heat is written on the reactant side.
Step 2
Raising the temperature shifts the equilibrium away from the side where heat is written.
Step 3
The equilibrium shifts toward the products.
You can now predict that raising the temperature shifts an equilibrium away from the side where heat is written — the direction that absorbs the added heat.
Check your understanding
The equilibrium CO(g) + 2H₂(g) ⇌ CH₃OH(g) + heat is heated. Which way does the equilibrium shift?
AToward the reactants.correct
BToward the products.
This option is wrong — you shifted toward the written heat — raising the temperature ADDS heat, so the mixture moves away from the side where heat is written, using the extra heat up.
CIt does not shift.
This option is wrong — you treated temperature as a spectator — a temperature change is a change of conditions, and the written heat tells you which way the shift runs.
DBoth directions speed up by the same amount, so the amounts stay the same.
This option is wrong — you assumed the two directions gain equally — the heat-absorbing direction gains more, and the mixture shifts away from the written heat.
Raising the temperature adds heat. The equilibrium shifts away from the side where heat is written, using the extra heat up. Heat sits with the products here, so the shift runs toward the reactants.
Check your understanding
The equilibrium heat + N₂(g) + O₂(g) ⇌ 2NO(g) is heated. Which way does the equilibrium shift?
AToward the products.correct
BToward the reactants.
This option is wrong — you shifted toward the written heat — the mixture moves AWAY from the side where heat is written, and here heat sits with the reactants.
CIt does not shift.
This option is wrong — you treated temperature as a spectator — a temperature change is a change of conditions, and the written heat tells you which way the shift runs.
DThe forward and reverse reactions both stop.
This option is wrong — you stopped the reactions — both keep running; the mixture shifts toward the products and settles at a new equilibrium.
Raising the temperature adds heat. The equilibrium shifts away from the side where heat is written. Heat sits with the reactants here, so the shift runs toward the products.
Check your understanding
The equilibrium CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) + heat is heated in a sealed reactor. What happens to the amount of H₂?
AIt decreases.correct
BIt increases.
This option is wrong — you shifted toward the products under heating — heat sits on the product side, so the mixture shifts AWAY from it, toward the reactants.
CIt stays exactly the same.
This option is wrong — you treated temperature as a spectator — the written heat makes a temperature change shift the equilibrium like any other change.
DIt rises at first and then falls to zero.
This option is wrong — you emptied a substance — shifts change amounts partway; no substance in an equilibrium mixture is driven to zero.
Raising the temperature adds heat, and the equilibrium shifts away from the side where heat is written. Heat sits with the products, so the shift runs toward the reactants. H₂ is a product, so its amount falls at the new equilibrium.
Lesson 35 of 40 · KEQ-035
Lowering the temperature
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Did You Know?
You have seen that raising the temperature shifts an equilibrium away from the side where heat is written. Lowering the temperature is the exact mirror.
The idea
Lowering the temperature of an equilibrium mixture removes heat — the change is 'less heat'.
The equilibrium shifts in the direction that replaces the removed heat, just as it shifts to replace a removed substance.
The direction that replaces heat is the direction that releases heat — the direction leading toward the side where heat is written.
So lowering the temperature shifts an equilibrium toward the side where heat is written.
The equation N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + heat carries heat on the product side, so lowering the temperature shifts this equilibrium toward the products.
The new mixture holds more NH₃ — for this reaction, cooling pays product.
The equation heat + N₂O₄(g) ⇌ 2NO₂(g) carries heat on the reactant side, so lowering the temperature shifts this equilibrium toward the reactants.
That shift is visible too: the brown NO₂ fades as a sealed tube of the mixture chills in ice water.
Worked examples
Worked example 1. The temperature of the 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) + heat equilibrium is lowered. Which way does the equilibrium shift?
Step 1
Heat is written on the product side.
Step 2
Lowering the temperature shifts the equilibrium toward the side where heat is written.
Step 3
The equilibrium shifts toward the products.
Worked example 2. The temperature of the heat + CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g) equilibrium is lowered. Which way does the equilibrium shift?
Step 1
Heat is written on the reactant side.
Step 2
Lowering the temperature shifts the equilibrium toward the side where heat is written.
Step 3
The equilibrium shifts toward the reactants.
You can now predict that lowering the temperature shifts an equilibrium toward the side where heat is written — the direction that releases heat to replace what was removed.
Check your understanding
The equilibrium CO(g) + 2H₂(g) ⇌ CH₃OH(g) + heat is cooled. Which way does the equilibrium shift?
AToward the products.correct
BToward the reactants.
This option is wrong — you shifted away from the written heat — COOLING removes heat, so the mixture moves toward the side where heat is written, replacing what was lost.
CIt does not shift.
This option is wrong — you treated temperature as a spectator — a temperature change is a change of conditions, and the written heat tells you which way the shift runs.
DBoth directions slow down by the same amount, so the amounts stay the same.
This option is wrong — you assumed the two directions lose equally — the heat-releasing direction holds up better, and the mixture shifts toward the written heat.
Lowering the temperature removes heat. The equilibrium shifts toward the side where heat is written, replacing the removed heat. Heat sits with the products here, so the shift runs toward the products.
Check your understanding
The equilibrium heat + N₂(g) + O₂(g) ⇌ 2NO(g) is cooled. Which way does the equilibrium shift?
AToward the reactants.correct
BToward the products.
This option is wrong — you shifted away from the written heat — the cooled mixture moves TOWARD the side where heat is written, and here heat sits with the reactants.
CIt does not shift.
This option is wrong — you treated temperature as a spectator — a temperature change is a change of conditions, and the written heat tells you which way the shift runs.
DThe forward and reverse reactions both stop.
This option is wrong — you stopped the reactions — both keep running, more slowly; the mixture shifts toward the reactants and settles at a new equilibrium.
Lowering the temperature removes heat. The equilibrium shifts toward the side where heat is written. Heat sits with the reactants here, so the shift runs toward the reactants.
Check your understanding
The equilibrium CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) + heat is cooled in a sealed reactor. What happens to the amount of H₂?
AIt increases.correct
BIt decreases.
This option is wrong — you shifted away from the written heat — cooling moves the mixture TOWARD the side where heat is written, and H₂ sits on that side.
CIt stays exactly the same.
This option is wrong — you treated temperature as a spectator — the written heat makes a temperature change shift the equilibrium like any other change.
DIt falls at first and then rises without limit.
This option is wrong — you let the amount grow forever — the shift ends when the two rates are equal again, at a new settled amount.
Lowering the temperature removes heat, and the equilibrium shifts toward the side where heat is written. Heat sits with the products, so the shift runs toward the products. H₂ is a product, so its amount rises at the new equilibrium.
Lesson 36 of 40 · KEQ-036
Counting gas moles on each side
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You have seen that the coefficients of a balanced equation count moles, and that the state symbol (g) marks a substance as a gas. Some questions about a gas-phase equilibrium turn on one number per side — how many moles of gas each side holds. Later lessons put the count to work; this one builds it.
The idea
In a balanced equation, each substance's coefficient counts its moles.
A substance written with no coefficient counts as 1.
To count the moles of gas on one side, add the coefficients of every substance marked (g) on that side.
In N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the left side holds 1 + 3 = 4 moles of gas.
The right side holds 2 moles of gas.
So this equation holds more moles of gas on the reactant side — 4 against 2.
The two sides do not have to match: balancing makes the atoms match, not the moles of gas.
Worked examples
Worked example 1. How many moles of gas are on each side of 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), and which side holds more?
Step 1
Left side: 2 + 1 = 3 moles of gas.
Step 2
Right side: 2 moles of gas.
Step 3
3 moles of gas on the left against 2 on the right — the reactant side holds more.
Worked example 2. How many moles of gas are on each side of CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g), and which side holds more?
Step 1
Left side: 1 + 1 = 2 moles of gas.
Step 2
Right side: 1 + 3 = 4 moles of gas.
Step 3
2 moles of gas on the left against 4 on the right — the product side holds more.
You can now compare the number of moles of gas on each side of a balanced reversible equation by adding the coefficients of the gaseous substances.
Check your understanding
How many moles of gas are on the reactant side of PCl₃(g) + Cl₂(g) ⇌ PCl₅(g)?
Accepted answer: 2 moles of gas
Add the coefficients of the substances marked (g) on the reactant side. PCl₃ has no written coefficient, so it counts as 1; Cl₂ also counts as 1. 1 + 1 = 2 moles of gas.
Check your understanding
How many moles of gas are on the product side of 2NO(g) + O₂(g) ⇌ 2NO₂(g)?
Accepted answer: 2 moles of gas
Add the coefficients of the substances marked (g) on the product side. The only product is 2NO₂. The product side holds 2 moles of gas.
Check your understanding
Which side of N₂O₄(g) ⇌ 2NO₂(g) holds more moles of gas?
AThe product side.correct
BThe reactant side.
This option is wrong — you compared molecule sizes — N₂O₄ is the bigger molecule, but moles of gas come from coefficients: 1 on the left against 2 on the right.
CThe two sides hold equal moles of gas.
This option is wrong — you matched the sides because the equation is balanced — balancing matches atoms, not moles of gas; the counts here are 1 and 2.
DIt cannot be decided from the equation alone.
This option is wrong — you looked past the coefficients — the coefficients alone carry the count: 1 mole of gas on the left, 2 on the right.
Left side: N₂O₄ counts as 1 mole of gas. Right side: 2NO₂ counts as 2 moles of gas. 2 is more than 1, so the product side holds more moles of gas.
Lesson 37 of 40 · KEQ-037
Increasing the pressure
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You have seen how to count the moles of gas on each side of a reversible equation, and you have seen that gas pressure comes from particles colliding with the container walls. Together, the two ideas predict what squeezing an equilibrium does.
The idea
Increasing the pressure on a gas-phase equilibrium means squeezing the mixture into a smaller volume.
The same particles in a smaller space hit the walls more often, so the pressure rises.
The equilibrium can counteract the squeeze by cutting its number of gas particles, because fewer gas particles in the same space means lower pressure.
The direction that cuts the particle count is the direction toward the side with fewer moles of gas.
So increasing the pressure shifts a gas-phase equilibrium toward the side with fewer moles of gas.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) holds 4 moles of gas on the left and 2 on the right, so increasing the pressure shifts it toward the products.
The rule carries a validity condition: when both sides hold equal moles of gas, no shift occurs — neither direction can change the particle count.
H₂(g) + I₂(g) ⇌ 2HI(g) holds 2 moles of gas on each side, so increasing the pressure leaves it unshifted.
Worked examples
Worked example 1. The 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) equilibrium is compressed into a smaller volume. Which way does the equilibrium shift?
Step 1
Left side: 2 + 1 = 3 moles of gas; right side: 2 moles of gas.
Step 2
Increasing the pressure shifts the equilibrium toward the side with fewer moles of gas.
Step 3
The equilibrium shifts toward the products.
Worked example 2. The CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) equilibrium is compressed into a smaller volume. Which way does the equilibrium shift?
Step 1
Left side: 1 + 1 = 2 moles of gas; right side: 1 + 1 = 2 moles of gas.
Step 2
Both sides hold equal moles of gas, so neither direction can change the particle count.
Step 3
No shift — the amounts stay as they were.
You can now predict that increasing the pressure on a gas-phase equilibrium, by decreasing its volume, shifts it toward the side with fewer moles of gas, and that no shift occurs when both sides have equal moles of gas.
Check your understanding
A piston is pushed in, increasing the pressure on the N₂O₄(g) ⇌ 2NO₂(g) equilibrium. Which way does the equilibrium shift?
AToward the reactants.correct
BToward the products.
This option is wrong — you shifted toward MORE moles of gas — a squeeze is counteracted by making FEWER gas particles, and the reactant side holds 1 mole against 2.
CIt does not shift.
This option is wrong — you treated the two sides as equal — count the coefficients: 1 mole of gas on the left against 2 on the right.
DThe forward and reverse reactions both stop.
This option is wrong — you stopped the reactions — both keep running; the mixture shifts toward the reactants and settles at a new equilibrium.
Count the moles of gas: 1 on the left, 2 on the right. Increasing the pressure shifts the equilibrium toward the side with fewer moles of gas. The reactant side holds fewer, so the shift runs toward the reactants.
Check your understanding
The pressure on the PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) equilibrium is increased by shrinking its container. Which way does the equilibrium shift?
AToward the products.correct
BToward the reactants.
This option is wrong — you shifted toward MORE moles of gas — a squeeze is counteracted by making FEWER gas particles, and the product side holds 1 mole against 2.
CIt does not shift.
This option is wrong — you treated the two sides as equal — count the coefficients: 1 + 1 = 2 moles of gas on the left against 1 on the right.
DIt shifts toward the products and then back to the reactants.
This option is wrong — you let the shift bounce back — the mixture settles at a new equilibrium and stays there while the pressure holds.
Count the moles of gas: 2 on the left, 1 on the right. Increasing the pressure shifts the equilibrium toward the side with fewer moles of gas. The product side holds fewer, so the shift runs toward the products.
Check your understanding
The pressure on the N₂(g) + O₂(g) ⇌ 2NO(g) equilibrium is increased by pushing a piston in. Which way does the equilibrium shift?
AIt does not shift.correct
BToward the products.
This option is wrong — you expected every squeeze to shift something — with 2 moles of gas on each side, neither direction can cut the particle count.
CToward the reactants.
This option is wrong — you expected every squeeze to shift something — with 2 moles of gas on each side, neither direction can cut the particle count.
DIt cannot be predicted without knowing the temperature.
This option is wrong — you reached for temperature — pressure shifts are decided by the gas-mole counts alone, and equal counts mean no shift.
Count the moles of gas: 1 + 1 = 2 on the left, 2 on the right. When both sides hold equal moles of gas, no shift occurs. Neither direction can change the particle count, so the squeeze changes nothing about the amounts.
Lesson 38 of 40 · KEQ-038
Decreasing the pressure
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Did You Know?
You have seen that increasing the pressure shifts a gas-phase equilibrium toward the side with fewer moles of gas. Decreasing the pressure is the exact mirror.
The idea
Decreasing the pressure on a gas-phase equilibrium means letting the mixture expand into a larger volume.
The same particles in a larger space hit the walls less often, so the pressure falls.
The equilibrium can counteract the drop by raising its number of gas particles, because more gas particles in the same space means higher pressure.
The direction that raises the particle count is the direction toward the side with more moles of gas.
So decreasing the pressure shifts a gas-phase equilibrium toward the side with more moles of gas.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) holds 4 moles of gas on the left and 2 on the right, so decreasing the pressure shifts it toward the reactants.
The validity condition is unchanged: when both sides hold equal moles of gas, no shift occurs — neither direction can change the particle count.
H₂(g) + I₂(g) ⇌ 2HI(g) holds 2 moles of gas on each side, so decreasing the pressure leaves it unshifted.
Worked examples
Worked example 1. The CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g) equilibrium is allowed to expand into a larger volume. Which way does the equilibrium shift?
Step 1
Left side: 1 + 1 = 2 moles of gas; right side: 1 + 3 = 4 moles of gas.
Step 2
Decreasing the pressure shifts the equilibrium toward the side with more moles of gas.
Step 3
The equilibrium shifts toward the products.
Worked example 2. The N₂(g) + O₂(g) ⇌ 2NO(g) equilibrium is allowed to expand into a larger volume. Which way does the equilibrium shift?
Step 1
Left side: 1 + 1 = 2 moles of gas; right side: 2 moles of gas.
Step 2
Both sides hold equal moles of gas, so neither direction can change the particle count.
Step 3
No shift — the amounts stay as they were.
You can now predict that decreasing the pressure on a gas-phase equilibrium, by increasing its volume, shifts it toward the side with more moles of gas, and that no shift occurs when both sides have equal moles of gas.
Check your understanding
A piston is pulled out, decreasing the pressure on the N₂O₄(g) ⇌ 2NO₂(g) equilibrium. Which way does the equilibrium shift?
AToward the products.correct
BToward the reactants.
This option is wrong — you shifted toward FEWER moles of gas — a pressure drop is counteracted by making MORE gas particles, and the product side holds 2 moles against 1.
CIt does not shift.
This option is wrong — you treated the two sides as equal — count the coefficients: 1 mole of gas on the left against 2 on the right.
DThe forward and reverse reactions both stop.
This option is wrong — you stopped the reactions — both keep running; the mixture shifts toward the products and settles at a new equilibrium.
Count the moles of gas: 1 on the left, 2 on the right. Decreasing the pressure shifts the equilibrium toward the side with more moles of gas. The product side holds more, so the shift runs toward the products.
Check your understanding
The pressure on the CO(g) + 2H₂(g) ⇌ CH₃OH(g) equilibrium is decreased by enlarging its container. Which way does the equilibrium shift?
AToward the reactants.correct
BToward the products.
This option is wrong — you shifted toward FEWER moles of gas — a pressure drop is counteracted by making MORE gas particles, and the reactant side holds 3 moles against 1.
CIt does not shift.
This option is wrong — you treated the two sides as equal — count the coefficients: 1 + 2 = 3 moles of gas on the left against 1 on the right.
DIt shifts toward the reactants and then back to the products.
This option is wrong — you let the shift bounce back — the mixture settles at a new equilibrium and stays there while the pressure holds.
Count the moles of gas: 3 on the left, 1 on the right. Decreasing the pressure shifts the equilibrium toward the side with more moles of gas. The reactant side holds more, so the shift runs toward the reactants.
Check your understanding
The pressure on the CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) equilibrium is decreased by enlarging its container. Which way does the equilibrium shift?
AIt does not shift.correct
BToward the products.
This option is wrong — you expected every pressure drop to shift something — with 2 moles of gas on each side, neither direction can raise the particle count.
CToward the reactants.
This option is wrong — you expected every pressure drop to shift something — with 2 moles of gas on each side, neither direction can raise the particle count.
DIt cannot be predicted without knowing the temperature.
This option is wrong — you reached for temperature — pressure shifts are decided by the gas-mole counts alone, and equal counts mean no shift.
Count the moles of gas: 1 + 1 = 2 on the left, 1 + 1 = 2 on the right. When both sides hold equal moles of gas, no shift occurs. Neither direction can change the particle count, so the expansion changes nothing about the amounts.
Lesson 39 of 40 · KEQ-039
A catalyst does not shift the equilibrium
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Did You Know?
Wonder this:
The rate lessons made catalysts look like the perfect upgrade: speed, with nothing used up. Ammonia factories pack their reactors with an iron catalyst. So surely the catalyst also squeezes extra NH₃ out of the equilibrium mixture? Measure it, and the amounts refuse to move.
You have seen that a catalyst speeds a reaction by providing a different path with a lower activation energy, and that an equilibrium shifts only when one direction temporarily outruns the other.
The idea
Adding a catalyst to an equilibrium mixture does not shift the equilibrium.
That is because a catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts.
The lower-energy path the catalyst provides is the same path for both directions, so both directions gain the same boost.
With both rates raised equally, neither direction outruns the other, and no shift begins.
Adding the iron catalyst to the N₂(g) + 3H₂(g) ⇌ 2NH₃(g) equilibrium leaves the amount of NH₃ unchanged.
What the catalyst does change is time: a mixture still on its way to equilibrium gets there sooner.
That is why factories still pay for catalysts — the same amounts, reached much faster.
Worked examples
Worked example 1. A vanadium(V) oxide catalyst is added to the 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) mixture at equilibrium. What happens to the amount of SO₃?
Step 1
Answer: it stays the same — a catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts.
Worked example 2. The same catalyst is added to a fresh SO₂ and O₂ mixture that has NOT yet reached equilibrium. What does the catalyst change?
Step 1
Answer: only the time — the mixture reaches equilibrium sooner, with the same final amounts.
You can now state that adding a catalyst does not shift an equilibrium, because a catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts.
Check your understanding
A nickel catalyst is added to the C₂H₄(g) + H₂(g) ⇌ C₂H₆(g) equilibrium. What happens to the amount of C₂H₆?
AIt stays the same.correct
BIt increases.
This option is wrong — you gave the catalyst a shifting job — it speeds BOTH directions equally, so the settled amounts do not move.
CIt decreases.
This option is wrong — you turned the catalyst into a brake on the forward reaction — it speeds both directions equally, and the settled amounts do not move.
DIt rises quickly at first and then falls back below its old value.
This option is wrong — you invented an overshoot — with both rates raised equally, no shift ever begins, and the amounts simply stay where they were.
A catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts. Neither direction outruns the other, so no shift begins. The amount of C₂H₆ stays exactly where it was.
Check your understanding
A catalyst is added to a reversible reaction mixture that is still on its way to equilibrium. What does the catalyst change?
AHow soon the mixture reaches equilibrium.correct
BWhich direction the equilibrium finally favors.
This option is wrong — you gave the catalyst a steering job — it speeds both directions equally, so the final balance is untouched.
CThe final amounts of reactants and products in the mixture.
This option is wrong — you let speed become amount — the mixture arrives at the SAME final amounts, only sooner.
DThe side of the equation on which heat is written.
This option is wrong — you tangled the catalyst with the heat label — heat's side follows the exothermic or endothermic classification, which a catalyst does not touch.
A catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts. Its only gift is time.
Check your understanding
Why does adding a catalyst leave an equilibrium unshifted?
ABecause a catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts.correct
BBecause a catalyst speeds up only the forward reaction, and the reverse reaction then speeds up on its own until the two rates match again, cancelling the catalyst's effect on the amounts.
This option is wrong — you gave the boost to one direction first — the catalyst's lower-energy path serves both directions from the very first moment.
CBecause a catalyst does not take part in the reaction at all, so it cannot change anything about the mixture.
This option is wrong — you overstretched 'not used up' — the catalyst DOES take part, providing a lower-energy path; it changes the speed, just equally for both directions.
DBecause the catalyst is used up before it can shift the equilibrium.
This option is wrong — you contradicted the definition — a catalyst is not used up; it is still present and unchanged when the reaction ends.
A catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts. Shifts need one direction to outrun the other, and an equal boost gives neither direction the lead.
Lesson 40 of 40 · KEQ-040
Choosing conditions for more product
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Wonder this:
A chemical plant's design team gets one job: make the reactor deliver as much product as possible. Amounts, temperature, pressure — every dial they can turn shifts the equilibrium one way or the other. Choosing the settings well is the whole game.
You have seen every single-change prediction: added and removed substances, raised and lowered temperature, increased and decreased pressure, and the catalyst that shifts nothing. This lesson runs the whole set on one equilibrium and picks the conditions that give the most product.
The idea
To harvest the most product, set every condition so that its shift points toward the products.
Substances: add more reactant, and remove the product as it forms — both shifts point toward the products.
Temperature: cool a reaction whose heat is written on the product side, and heat one whose heat is written on the reactant side.
Pressure: if the product side holds fewer moles of gas, increase the pressure; if it holds more, decrease the pressure; if the sides are equal, pressure changes do not help.
Catalyst: add one for speed — it shifts nothing, but the mixture reaches its equilibrium sooner.
Now run the set on N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + heat: add N₂ and H₂, and remove NH₃ as it forms.
Dials that point an equilibrium toward the products
Change
It shifts toward the products when…
Add a substance
the added substance is a reactant
Remove a substance
the removed substance is a product
Raise the temperature
heat is written on the reactant side
Lower the temperature
heat is written on the product side
Increase the pressure
the product side holds fewer moles of gas
Decrease the pressure
the product side holds more moles of gas
Add a catalyst
never — it changes the time, not the amounts
The whole toolkit: pick each condition so that its row points toward the products.
Heat sits on the product side, so cool the mixture.
The product side holds 2 moles of gas against 4, so increase the pressure.
Add the iron catalyst, so the mixture reaches its equilibrium sooner.
Every dial now points toward the products, and the reactor delivers its maximum NH₃.
Worked examples
Worked example 1. A reactor runs CO(g) + 2H₂(g) ⇌ CH₃OH(g) + heat. Choose the temperature change and the pressure change that give the most CH₃OH.
Step 1
Heat is written on the product side, so lowering the temperature shifts the equilibrium toward the products.
Step 2
The product side holds 1 mole of gas against 3, so increasing the pressure shifts the equilibrium toward the products.
Step 3
Lower the temperature and increase the pressure.
Worked example 2. A hydrogen plant runs heat + CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g). Choose the temperature change and the pressure change that give the most H₂.
Step 1
Heat is written on the reactant side, so raising the temperature shifts the equilibrium toward the products.
Step 2
The product side holds 4 moles of gas against 2, so decreasing the pressure shifts the equilibrium toward the products.
Step 3
Raise the temperature and decrease the pressure.
You can now predict the shift caused by each of several single changes to one gas-phase equilibrium — adding or removing a substance, raising or lowering the temperature, or changing the pressure — and select the set of conditions that gives the most product.
Check your understanding
A plant runs 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) + heat. Which ONE of these changes increases the total SO₃ produced?
ARemoving SO₃ as it forms.correct
BRaising the temperature.
This option is wrong — you heated a reaction whose written heat sits with the products — that shift points toward the reactants, costing SO₃.
CDecreasing the pressure.
This option is wrong — you shifted toward MORE moles of gas — the product side holds fewer (2 against 3), so decreasing the pressure moves away from SO₃.
DAdding a vanadium(V) oxide catalyst.
This option is wrong — you asked the catalyst for amount — it gives only speed; the settled amounts stay the same.
Check each dial against the toolkit. Removing a product shifts the equilibrium toward the products, replacing part of what is removed — more SO₃ is made. Heating shifts away from the written heat, a pressure drop shifts toward more gas moles, and a catalyst changes only the time.
Check your understanding
A reactor runs PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) + heat. Which set of conditions gives the most PCl₅?
AAdd Cl₂, remove PCl₅ as it forms, lower the temperature, increase the pressure.correct
BAdd Cl₂, remove PCl₅ as it forms, raise the temperature, decrease the pressure.
This option is wrong — you flipped both the temperature and the pressure moves — heat sits with the product, and the product side holds fewer moles of gas.
CRemove Cl₂, add extra PCl₅, lower the temperature, increase the pressure.
This option is wrong — you ran the substance moves backwards — add the REACTANT and remove the PRODUCT to point both shifts toward the products.
DAdd Cl₂, remove PCl₅ as it forms, lower the temperature, decrease the pressure.
This option is wrong — you flipped only the pressure move — the product side holds 1 mole of gas against 2, so INCREASING the pressure shifts toward it.
Substances: add the reactant Cl₂ and remove the product PCl₅. Heat sits on the product side, so lower the temperature. The product side holds 1 mole of gas against 2, so increase the pressure.
Check your understanding
A reactor runs heat + N₂O₄(g) ⇌ 2NO₂(g). Which temperature change and pressure change give the most NO₂?
ARaise the temperature and decrease the pressure.correct
BRaise the temperature and increase the pressure.
This option is wrong — you flipped the pressure move — the product side holds 2 moles of gas against 1, so a pressure INCREASE shifts away from NO₂.
CLower the temperature and decrease the pressure.
This option is wrong — you flipped the temperature move — heat sits with the reactants, so RAISING the temperature is the change that shifts toward NO₂.
DLower the temperature and increase the pressure.
This option is wrong — you flipped both moves — heating shifts away from the reactant-side heat toward NO₂, and a pressure drop shifts toward the 2-mole product side.
Heat is written on the reactant side, so raising the temperature shifts the equilibrium toward the products. The product side holds 2 moles of gas against 1, so decreasing the pressure also shifts toward the products. Raise the temperature and decrease the pressure.
Summary video — Shifting an equilibrium — substances, temperature, and pressure