Answer key · Unit 7

Rates and energy changes.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 25 minutes

Collision theory

Why only some collisions lead to a reaction, why twice the concentration makes it four times faster, and why a hundred degrees speed it up far more than the molecules' speed suggests.

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The 6 steps and their answers

Step 1 of 6
Open this step
Molecules have to meet
Practice · Choose
Question
The flask holds $\mathrm{NO}$ ⁠ molecules and ozone molecules, $\mathrm{O_3}$ ⁠. What has to happen for a pair of them to react to $\mathrm{NO_2}$ ⁠ and $\mathrm{O_2}$ ⁠?

Correct answer
- Correct They must collide hard enough, and the right way round
Setup, choices and feedback
The Scene at this step
Temperature
298 K
Concentration
60 %
Labels
offdefault
Highlight energetic molecules
Off
Choices
Shown in a random order.
- Correct They must collide hard enough, and the right way round
- Wrong It is enough for them to touch
  Misconception: Thinks molecules react whenever they touch.
  Feedback from the catalogue
  Most collisions do not lead to a reaction. For bonds to break and re-form, the two molecules must meet hard enough, with at least the activation energy, and often the right way round too. At room temperature only about one collision in hundreds clears the barrier, and in the Scene most molecules just bounce apart.
- Wrong They must stay side by side long enough
  Feedback
  Gas molecules do not stay side by side: they bounce apart after a hit that lasts less than a trillionth of a second. What counts is how hard the hit is.
Feedback for a right answer
That is collision theory: molecules react only when they meet with at least the activation energy, and in a direction that lets the ozone’s oxygen atom reach the nitrogen. Most collisions in the flask fail one test or the other, and bounce off without reacting.

Hint
Watch the Scene: how many of the collisions end in a flash?

Worked solution
1. To react, an $\mathrm{O-O}$ ⁠ bond in the ozone must break and a new $\mathrm{N-O}$ ⁠ bond must form.
2. Breaking a bond takes energy, and it comes from the kinetic energy of the collision. Below the activation energy the molecules just bounce apart.
3. The direction matters too: if the nitrogen meets the wrong side of the molecule, no bond can form.
If every touch led to a reaction, any gases that met would react almost at once. At room temperature only about one collision in hundreds clears the barrier.
Step 2 of 6
Open this step
A more crowded flask
Practice · Predict, then look
Question
Concentration stands at 30%, and Temperature is fixed. What will happen to the rate when Concentration rises to 60%, that is, with twice as many $\mathrm{NO}$ ⁠ molecules and twice as many $\mathrm{O_3}$ ⁠ molecules?

Correct answer
- Correct It will rise four times
Setup, choices and feedback
The Scene at this step
Temperature
298 Kfixed
Concentration
30 %
Labels
offdefault
Highlight energetic molecules
Off
Choices
- Correct It will rise four times
- Wrong It will rise twice
  Misconception: Thinks doubling the concentration of both reactants doubles the rate.
  Feedback from the catalogue
  To react, a molecule of A has to meet a molecule of B. Double both concentrations and each A molecule meets twice as many B molecules, and there are twice as many A molecules: $2 \times 2 = 4$ ⁠. So the rate, $k[\mathrm{A}][\mathrm{B}]$ ⁠, rises four times.
- Wrong It will not change: each molecule still needs the same energy
  Feedback
  True, each collision needs the same energy, so the share of fruitful collisions does not change. But the number of collisions each second grows, and with it the number of reactions.
Once the prediction is locked in, the Scene changes
Concentration
60 %
The Scene then shows: 0.36 · Gas modes: rate relative to the same reaction at 298 K, with every molecule a reactant and no catalyst/nodes/flask/outputs/rate
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Four times. Each $\mathrm{NO}$ ⁠ molecule now meets twice as many $\mathrm{O_3}$ ⁠ molecules, and there are twice as many $\mathrm{NO}$ ⁠ molecules doing the meeting: $2 \times 2 = 4$ ⁠. The flask shows many more flashes.

Worked solution
1. The rate is the number of reactions per second, in proportion to the number of meetings between $\mathrm{NO}$ ⁠ and $\mathrm{O_3}$ ⁠.
2. The number of meetings goes as both concentrations together: $\text{rate} = k[\mathrm{NO}][\mathrm{O_3}]$ ⁠.
3. Double each concentration: $k(2[\mathrm{NO}])(2[\mathrm{O_3}]) = 4k[\mathrm{NO}][\mathrm{O_3}]$ ⁠.
If you predicted twice, you thought about only one of the reactants. Both got more crowded.
Step 3 of 6
Open this step
Thinning it out
Practice · Reach the target
Question
Concentration stands at 100%, and the flask is full of reactants. Change Concentration so that the reaction runs at a quarter of the rate it has now, then press Check.

Goal
- 0.25 ± 0.005
  At the start of the step: 1
Setup, choices and feedback
The Scene at this step
Temperature
298 Kfixed
Concentration
100 %
Labels
offdefault
Highlight energetic molecules
Off
Read from the Scene at
Gas modes: rate relative to the same reaction at 298 K, with every molecule a reactant and no catalyst/nodes/flask/outputs/rate
One setting that reaches it
Concentration
50 %
Labels
off
Feedback for a right answer
At 50%: half of each reactant gives $\tfrac12 \times \tfrac12 = \tfrac14$ ⁠ of the rate. That is how a plant slows a dangerous reaction, or how diluting the air puts out a fire.

Feedback for a wrong answer
Not yet. If each reactant drops to a fraction $f$ ⁠ of its concentration, the rate drops to $f^2$ ⁠.

Hint
The rate goes as the square of the concentration, because both reactants change together.

Worked solution
1. $\text{rate} \propto [\mathrm{NO}][\mathrm{O_3}]$ ⁠, and both go as Concentration.
2. So the rate goes as the square of Concentration: $f^2 = \tfrac14$ ⁠.
3. $f = \tfrac12$ ⁠: half of 100%, that is 50%.
If you set 25%, the rate fell to $\tfrac14 \times \tfrac14 = \tfrac{1}{16}$ ⁠.
Step 4 of 6
Open this step
Turning up the heat
Practice · Predict, then look
Question
You heat the flask from 298 K to 398 K: 1.34 times in kelvin. How many times faster will the reaction be?

Correct answer
- Correct About three times
Setup, choices and feedback
The Scene at this step
Temperature
298 K
Concentration
60 %fixed
Labels
offdefault
Highlight energetic molecules
Off
Choices
- Correct About three times
- Wrong 1.34 times, like the temperature
  Misconception: Thinks the rate grows in proportion to the absolute temperature.
  Feedback from the catalogue
  The rate does not grow in step with the temperature in kelvin. From 298 K to 398 K the temperature rises only 1.34 times, but the share of molecules with more than the activation energy roughly triples. The right-hand tail of the energy distribution swells much faster than the average.
- Wrong Hardly at all: the molecules are only a little faster
  Misconception: Thinks heating speeds a reaction only because the molecules collide more often.
  Feedback from the catalogue
  Heating does speed the molecules up and make collisions more frequent, but only a little: from 298 K to 398 K by about $16\%$ ⁠. The main reason is another: more of the collisions carry enough energy to clear the barrier. In the energy chart the shaded share above $E_a$ ⁠ triples.
Once the prediction is locked in, the Scene changes
Temperature
398 K
Highlight energetic molecules
Off
The Scene then shows: 3.245 · Gas modes: rate relative to this flask at 298 K, all else the same/nodes/flask/outputs/rateVsRoom
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About 3.2 times. The molecules are only $16\%$ ⁠ faster, but the share of them that clears the barrier rises from 1.8% to 5.6%: compare the shaded area in the chart.

Worked solution
1. A molecule’s mean speed goes as $\sqrt{T}$ ⁠: $\sqrt{\tfrac{398}{298}} = 1.16$ ⁠. That raises the number of collisions by only $16\%$ ⁠.
2. The share of collisions that clear the barrier grows much more, because it goes as $e^{-E_a/RT}$ ⁠: the tail of the distribution swells fast.
3. Together: the reaction is about 3.2 times faster.
With a higher barrier the effect is stronger still: with a 50 kJ/mol barrier, every ten degrees near room temperature doubles the rate.
Step 5 of 6
Open this step
The energetic few
Practice · Reach the target
Question
The chart shows how many of the molecules carry more than the activation energy, 12.5 kJ/mol. Change Temperature until 5% of the molecules clear the barrier, then press Check.

Goal
- 0.05 ± 0.002
  At the start of the step: 0.01781
Setup, choices and feedback
The Scene at this step
Temperature
298 K
Concentration
60 %fixed
Labels
offdefault
Highlight energetic molecules
Onfixed
Read from the Scene at
Share of molecules with more energy than the activation energy, 0..1 (the distribution chart)/nodes/flask/outputs/aboveBarrier
One setting that reaches it
Temperature
381 K
Labels
off
Feedback for a right answer
At about 385 K, that is 112 °C. At room temperature only 1.8% clear it; another 87 degrees nearly triples that share. The whole curve shifts right and flattens, and the tail beyond the line swells.

Feedback for a wrong answer
Not yet. Heat or cool, and watch the line under the chart.

Hint
The line under the chart gives the share above $E_a$ ⁠.

Worked solution
1. At room temperature, 298 K, the line shows about 1.8%.
2. To reach 5% you need more fast molecules: raise Temperature.
3. Around 385 K the line shows 5.0%.
The barrier does not move when you heat: the $E_a$ ⁠ line stays put, and it is the molecules that change.
Step 6 of 6
Open this step
Why heating works
Practice · Choose
Question
From 298 K to 398 K the molecules are only $16\%$ ⁠ faster, but the reaction is three times faster. What is the main reason?

Correct answer
- Correct A much larger share of the collisions carries enough energy to clear the barrier
Setup, choices and feedback
The Scene at this step
Temperature
398 K
Concentration
60 %
Labels
offdefault
Highlight energetic molecules
On
Choices
Shown in a random order.
- Correct A much larger share of the collisions carries enough energy to clear the barrier
- Wrong The molecules collide more often
  Misconception: Thinks heating speeds a reaction only because the molecules collide more often.
  Feedback from the catalogue
  Heating does speed the molecules up and make collisions more frequent, but only a little: from 298 K to 398 K by about $16\%$ ⁠. The main reason is another: more of the collisions carry enough energy to clear the barrier. In the energy chart the shaded share above $E_a$ ⁠ triples.
- Wrong The heat lowers the activation energy
  Misconception: Thinks heating lowers the activation energy.
  Feedback from the catalogue
  Heating does not change the barrier: the activation energy belongs to the reaction and its route, and the $E_a$ ⁠ line in the chart does not move when you change the temperature. What changes is the molecules: at a higher temperature more of them have energy above the barrier. Only a catalyst, which opens another route, lowers the barrier.
Feedback for a right answer
The tail of the distribution. More frequent collisions add only $16\%$ ⁠; the rest comes from the share clearing the barrier tripling. That is why food keeps in the fridge: spoiling is reactions, and the cold trims the tail.

Hint
Compare in the chart: what moves when you change the temperature, the curve or the $E_a$ ⁠ line?

Worked solution
1. The number of collisions grows as $\sqrt{T}$ ⁠: by only $16\%$ ⁠.
2. The activation energy belongs to the reaction, and its line in the chart does not move.
3. What grows a lot is the share of molecules beyond the line: from 1.8% to 5.6%, about three times.
So temperature affects reaction rates far more than the molecules’ speed suggests.

Lesson 2 of 4 · About 20 minutes

Activation energy and catalysts

The hump every reaction has to get over, and how a catalyst opens a lower route without giving any energy and without running out: the platinum in a catalytic converter.

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The 5 steps and their answers

Step 1 of 5
Open this step
The barrier
Practice · Choose
Question
In the second chart, the hump between reactants and products is marked $E_a$ ⁠. What is the activation energy?

Correct answer
- Correct The least energy a collision needs for the molecules to react
Setup, choices and feedback
The Scene at this step
Temperature
550 K
Platinum catalyst
off
Labels
offdefault
Highlight energetic molecules
Off
Choices
Shown in a random order.
- Correct The least energy a collision needs for the molecules to react
- Wrong The energy the reaction gives out
  Misconception: Confuses the activation energy with the energy the reaction gives out (ΔH).
  Feedback from the catalogue
  These are two different quantities on the energy profile. The activation energy is the height of the hump above the reactants: how much energy it takes to get going. The enthalpy change, $\Delta H$ ⁠, is the difference between the products’ level and the reactants’ level: how much the reaction gives out or takes in overall. A reaction can give out a lot and still be very slow, if the hump is high.
- Wrong The energy needed to heat the flask to its temperature
  Feedback
  Heating the flask gives the molecules kinetic energy, but the activation energy is measured per collision, per mole of molecules: how much energy a pair needs in order to react.
Feedback for a right answer
The activation energy is the height of the hump above the reactants, here 50 kJ/mol. The drop from reactants to products is something else: the energy the reaction gives out, 283 kJ/mol per $\mathrm{CO_2}$ ⁠. A reaction can give out a lot and still be slow, if the hump is high.

Hint
Where on the chart is $E_a$ ⁠: between the reactants and the top of the hump, or between the reactants and the products?

Worked solution
1. On the energy profile the reactants are on the left, the products on the right, and a hump between them.
2. The hump is the state in which old bonds are already weakened and new ones not yet formed. To reach it, the collision has to supply energy.
3. Its height above the reactants is the activation energy, $E_a$ ⁠.
The difference between reactants and products is $\Delta H$ ⁠, not $E_a$ ⁠.
Step 2 of 5
Open this step
Putting in the platinum
Practice · Predict, then look
Question
The flask holds carbon monoxide and oxygen at 550 K, without a catalyst: almost nothing reacts. What will happen when you put the Platinum catalyst in?

Correct answer
- Correct The reaction will speed up enormously, because on the platinum there is a route with a lower barrier
Setup, choices and feedback
The Scene at this step
Temperature
550 Kfixed
Platinum catalyst
off
Labels
offdefault
Highlight energetic molecules
Off
Choices
- Correct The reaction will speed up enormously, because on the platinum there is a route with a lower barrier
- Wrong The reaction will speed up, because the platinum hands energy to the molecules
  Misconception: Thinks a catalyst speeds a reaction by giving the molecules extra energy, as heating does.
  Feedback from the catalogue
  A catalyst adds no energy to any molecule and does not heat the flask. It opens another route for the reaction, with a lower barrier: on the platinum $\mathrm{O_2}$ ⁠ splits into atoms, and $\mathrm{CO}$ ⁠ reacts with them easily. The same molecules, at the same temperature, and hundreds of times more of them clear the lower barrier.
- Wrong The rate will not change, but more $\mathrm{CO_2}$ ⁠ will form in the end
  Misconception: Thinks a catalyst increases how much product a reaction makes.
  Feedback from the catalogue
  A catalyst changes only the rate, not the amount: it brings the reaction to its end sooner, but the end itself is set by the reactants, the limiting reagent and the equilibrium. In a catalytic converter each $\mathrm{CO}$ ⁠ molecule gives one $\mathrm{CO_2}$ ⁠ molecule, with or without the catalyst.
Once the prediction is locked in, the Scene changes
Platinum catalyst
on
Highlight energetic molecules
Off
The Scene then shows: 15 kJ/mol · Activation energy of the route in use (the catalyst route when the catalyst is in)/nodes/flask/outputs/barrier
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The barrier drops from 50 to 15 kJ/mol, and at the same temperature hundreds of times more molecules clear it. On the platinum $\mathrm{O_2}$ ⁠ splits into oxygen atoms, and $\mathrm{CO}$ ⁠ reacts with them easily. In the top chart the area beyond the dashed line is far bigger.

Worked solution
1. Without a catalyst, each $\mathrm{CO} + \mathrm{O_2}$ ⁠ collision needs at least 50 kJ/mol, and only about 0.007% of the molecules have that at 550 K.
2. With platinum the reaction takes another route, through oxygen atoms on the metal, with a barrier of 15 kJ/mol, which about 9% of the molecules clear.
3. The temperature did not change, and the molecules gained no energy. The route changed.
A catalyst adds no product either: it only reaches the same end sooner.
Step 3 of 5
Open this step
A lower hump
Practice · Calculate
Question
Read it off the Activation barrier chart: what is the activation energy on the platinum route?

Correct answer
15 kJ/mol ± 0.5 kJ/mol
Setup, choices and feedback
The Scene at this step
Temperature
550 Kfixed
Platinum catalyst
onfixed
Labels
offdefault
Highlight energetic molecules
Off
Accepted within
± 0.5 kJ/mol
Computed from the Scene at this step’s setting
Activation energy of the route in use (the catalyst route when the catalyst is in)/nodes/flask/outputs/barrier
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 50 kJ/mol ± 0.5 kJ/mol
  Misconception: Thinks a catalyst speeds a reaction by giving the molecules extra energy, as heating does.
  Feedback from the catalogue
  A catalyst adds no energy to any molecule and does not heat the flask. It opens another route for the reaction, with a lower barrier: on the platinum $\mathrm{O_2}$ ⁠ splits into atoms, and $\mathrm{CO}$ ⁠ reacts with them easily. The same molecules, at the same temperature, and hundreds of times more of them clear the lower barrier.
- 283 kJ/mol ± 0.5 kJ/mol
  Misconception: Confuses the activation energy with the energy the reaction gives out (ΔH).
  Feedback from the catalogue
  These are two different quantities on the energy profile. The activation energy is the height of the hump above the reactants: how much energy it takes to get going. The enthalpy change, $\Delta H$ ⁠, is the difference between the products’ level and the reactants’ level: how much the reaction gives out or takes in overall. A reaction can give out a lot and still be very slow, if the hump is high.
Feedback for a right answer
15 kJ/mol, instead of 50. Notice the products stay at the same level: the catalyst does not change the energy the reaction gives out, only the hump on the way.

Hint
The dashed line in the chart is the route with the catalyst.

Worked solution
1. The chart has two routes: the solid line, without a catalyst, with $E_a = 50$ ⁠; and the dashed line, with the catalyst.
2. The dashed line’s label: with catalyst 15.
3. The activation energy with the catalyst: 15 kJ/mol.
If you answered $50$ ⁠, you read the route without the catalyst. The catalyst does not change the molecules; it opens a lower route. And $283$ ⁠ is $\Delta H$ ⁠, what the reaction gives out.
Step 4 of 5
Open this step
Heating instead
Practice · Predict, then look
Question
With the platinum at 550 K, about 9% of the molecules clear the barrier. If you take the catalyst out and heat instead, to the top of the scale, 800 K, will a similar share of molecules clear the barrier?

Correct answer
- Correct No, not even 1%
Setup, choices and feedback
The Scene at this step
Temperature
550 K
Platinum catalyst
on
Labels
offdefault
Highlight energetic molecules
Off
Choices
- Correct No, not even 1%
- Wrong Yes: heating does what the catalyst does, it adds energy
  Misconception: Thinks a catalyst speeds a reaction by giving the molecules extra energy, as heating does.
  Feedback from the catalogue
  A catalyst adds no energy to any molecule and does not heat the flask. It opens another route for the reaction, with a lower barrier: on the platinum $\mathrm{O_2}$ ⁠ splits into atoms, and $\mathrm{CO}$ ⁠ reacts with them easily. The same molecules, at the same temperature, and hundreds of times more of them clear the lower barrier.
- Wrong More than 9%: 800 K is much hotter
  Misconception: Thinks heating lowers the activation energy.
  Feedback from the catalogue
  Heating does not change the barrier: the activation energy belongs to the reaction and its route, and the $E_a$ ⁠ line in the chart does not move when you change the temperature. What changes is the molecules: at a higher temperature more of them have energy above the barrier. Only a catalyst, which opens another route, lowers the barrier.
Once the prediction is locked in, the Scene changes
Temperature
800 K
Platinum catalyst
off
Highlight energetic molecules
Off
The Scene then shows: 0.001787 · Share of molecules with more energy than the activation energy, 0..1 (the distribution chart)/nodes/flask/outputs/aboveBarrier
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Only about 0.18%, fifty times less. The 50 kJ/mol barrier is too high for heat to make up for it. That is why cars need a catalytic converter: without it the $\mathrm{CO}$ ⁠ in the exhaust would nearly all go out into the air.

Worked solution
1. The share that clears a barrier depends on the ratio $\dfrac{E_a}{RT}$ ⁠.
2. With the catalyst at 550 K: $\dfrac{15}{0.008314 \times 550} = 3.3$ ⁠, and about 9% clear it.
3. Without it at 800 K: $\dfrac{50}{0.008314 \times 800} = 7.5$ ⁠, far higher, and only about 0.18% clear it.
Heating does not lower the barrier and is no substitute for a catalyst. The catalyst changes $E_a$ ⁠ itself; heating only moves the molecules.
Step 5 of 5
Open this step
It never runs out
Practice · Choose
Question
A catalytic converter works in a car for ten years and more. What happens to its platinum in each of the reactions it speeds?

Correct answer
- Correct The molecules react on it and leave, and it stays as it was
Setup, choices and feedback
The Scene at this step
Temperature
550 K
Platinum catalyst
on
Labels
offdefault
Highlight energetic molecules
Off
Choices
Shown in a random order.
- Correct The molecules react on it and leave, and it stays as it was
- Wrong A little is used up in each reaction, so it slowly runs out
  Misconception: Thinks a catalyst is used up in the reaction it speeds.
  Feedback from the catalogue
  A catalyst takes part in the reaction, but comes out as it went in. Molecules bind to the platinum, react on it and leave as $\mathrm{CO_2}$ ⁠, and the platinum stays platinum, ready for the next pair. That is why a little catalyst lasts for years: a catalytic converter holds only a few grams of platinum.
- Wrong It goes into the $\mathrm{CO_2}$ ⁠ and leaves through the exhaust with it
  Misconception: Thinks a catalyst is used up in the reaction it speeds.
  Feedback from the catalogue
  A catalyst takes part in the reaction, but comes out as it went in. Molecules bind to the platinum, react on it and leave as $\mathrm{CO_2}$ ⁠, and the platinum stays platinum, ready for the next pair. That is why a little catalyst lasts for years: a catalytic converter holds only a few grams of platinum.
Feedback for a right answer
A catalyst takes part in the reaction and comes out unchanged, so a few grams of platinum handle tonnes of exhaust. Converters do wear out over the years, but for other reasons: deposits coat the metal, and great heat makes it clump.

Hint
In the Scene, the plate at the bottom of the flask: does it shrink as $\mathrm{CO_2}$ ⁠ forms?

Worked solution
1. $\mathrm{O_2}$ ⁠ and $\mathrm{CO}$ ⁠ bind to the platinum’s surface.
2. On the metal $\mathrm{O_2}$ ⁠ splits into atoms, and an oxygen atom reacts with $\mathrm{CO}$ ⁠ to make $\mathrm{CO_2}$ ⁠.
3. The $\mathrm{CO_2}$ ⁠ comes off, and the platinum site is free again for the next pair.
In the overall equation, $\mathrm{2CO + O_2 \rightarrow 2CO_2}$ ⁠, the platinum does not appear at all: none of its atoms goes into the product.

Lesson 3 of 4 · About 20 minutes

Exothermic and endothermic

Where the energy in a reaction goes: burning that gives out heat, the power of a flame, the sign of ΔH, and a cold pack that takes heat in from your hand.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Heat coming out
Practice · Choose
Question
The flame lights only after the igniter sparks. Is burning methane an exothermic reaction (giving out heat) or an endothermic one (taking it in)?

Correct answer
- Correct Exothermic: the products hold less energy than the reactants, and the difference comes out as heat and light
Setup, choices and feedback
The Scene at this step
Gas flow
4 cm³/s
Oxygen in the air
21 %
Water mist
0 mg/s
Zoom to the molecules
Flame
Choices
Shown in a random order.
- Correct Exothermic: the products hold less energy than the reactants, and the difference comes out as heat and light
- Wrong Endothermic: it needs the heat of the spark
  Misconception: Thinks a reaction that needs heat or a spark to start must be endothermic.
  Feedback from the catalogue
  Starting and the overall balance are two different things. The spark gives a few molecules the activation energy, to get them over the hump. From there the burning gives out far more energy than it took in, and its heat lights the next molecules. A reaction is exothermic when the products sit lower in energy than the reactants, $\Delta H < 0$ ⁠, even if it needs a push to start.
- Wrong Neither: the heat comes from the gas itself, not from a reaction
  Feedback
  Methane on its own is not hot; the heat appears only when it reacts with oxygen. Turn Oxygen in the air to zero, and the flame goes out.
Feedback for a right answer
Burning is exothermic: each mole of methane gives out about 802 kJ. The spark only gives a few molecules the activation energy; from there the flame’s own heat lights the next molecules.

Hint
The spark only starts the reaction. What happens next: does the flame keep going on its own, or need more sparks?

Worked solution
1. In an exothermic reaction the products sit lower in energy than the reactants: $\Delta H < 0$ ⁠, and the substances hand energy to the surroundings.
2. Burning warms the chimney and the air, so it gives out heat.
3. Needing a spark is about the activation energy, the hump on the way, not about the reaction’s overall balance.
Even a reaction that gives out heat can need a push to start: a match does not light by itself, but once lit it burns on its own.
Step 2 of 5
Open this step
The flame's power
Practice · Calculate
Question
Gas flow is fixed at 4 cm³/s of methane. A mole of gas fills 24,055 cm³ here, and each mole of methane burned gives out 802 kJ. What is the flame’s power, that is, how much heat does it give out each second?

Correct answer
133.4 W ± 2%
Setup, choices and feedback
The Scene at this step
Gas flow
4 cm³/sfixed
Oxygen in the air
21 %fixed
Water mist
0 mg/sfixed
Zoom to the molecules
Flame
Accepted within
± 2%
Computed from the Scene at this step’s setting
Heat released, 0 when out/nodes/flame/outputs/power
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 3209 W ± 2%
  Misconception: Uses a volume of gas, in cm³, as if it were a number of moles.
  Feedback from the catalogue
  A cubic centimetre of gas is not a mole. To go from a volume to an amount of substance, divide by the molar volume: at room temperature and atmospheric pressure a mole of any gas fills about 24,000 cm³. 4 cm³ of methane is only $\dfrac{4}{24\,055} = 1.66 \times 10^{-4}$ ⁠ mol.
Feedback for a right answer
About 133 W: $\dfrac{4}{24\,055} = 1.66 \times 10^{-4}$ ⁠ mol per second, times 802,000 J. About as much as a bright incandescent bulb, from a very small flame.

Hint
First moles per second: divide the flow by the molar volume.

Worked solution
1. Moles per second: $\dfrac{4\ \mathrm{cm^3/s}}{24\,055\ \mathrm{cm^3/mol}} = 1.663 \times 10^{-4}\ \mathrm{mol/s}$ ⁠.
2. Heat per mole: $802\ \mathrm{kJ} = 802\,000\ \mathrm{J}$ ⁠.
3. The power: $1.663 \times 10^{-4} \times 802\,000 = 133\ \mathrm{W}$ ⁠.
If you got about $3209$ ⁠, you multiplied the flow in cubic centimetres straight by $802$ ⁠, as if each cubic centimetre were a mole.
Step 3 of 5
Open this step
A bigger flame
Practice · Reach the target
Question
Change Gas flow so that the flame gives out 200 W, then press Check.

Goal
- 200 W ± 1.5 W
  At the start of the step: 133.4 W
Setup, choices and feedback
The Scene at this step
Gas flow
4 cm³/s
Oxygen in the air
21 %fixed
Water mist
0 mg/sfixed
Zoom to the molecules
Flame
Read from the Scene at
Heat released, 0 when out/nodes/flame/outputs/power
One setting that reaches it
Gas flow
6 cm³/s
Feedback for a right answer
At 6 cm³/s: 1.5 times the methane each second gives 1.5 times the heat. The flame’s temperature does not change, only its size: each mole burns with the same oxygen and gives out the same energy.

Feedback for a wrong answer
Not yet. Each 1 cm³/s of methane adds about 33 W.

Hint
At 4 cm³/s the flame gives out 133 W. The power goes in proportion to the flow.

Worked solution
1. The heat per second goes with the moles per second, so with the flow.
2. 133 W at 4 cm³/s: about 33.4 W for each 1 cm³/s.
3. $\dfrac{200}{33.4} = 6.0$ ⁠: set Gas flow to 6 cm³/s.
Step 4 of 5
Open this step
The sign of ΔH
Practice · Calculate
Question
Burning a mole of methane to $\mathrm{CO_2}$ ⁠ and water vapour gives out 802 kJ to the surroundings. What is $\Delta H$ ⁠ for the reaction $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$ ⁠, with its sign?

Correct answer
-802.3 kJ/mol ± 2 kJ/mol
Setup, choices and feedback
The Scene at this step
Gas flow
4 cm³/s
Oxygen in the air
21 %
Water mist
0 mg/s
Zoom to the molecules
Flame
Accepted within
± 2 kJ/mol
Computed from the Scene at this step’s setting
Heat given out per mole of methane burned to CO₂ and water vapour (the lower heating value)/nodes/flame/outputs/heatPerMole
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 802.3 kJ/mol ± 2 kJ/mol
  Misconception: Gives ΔH a positive sign for a reaction that gives out heat.
  Feedback from the catalogue
  $\Delta H$ ⁠ is measured from the substances’ point of view: the products’ energy minus the reactants’. When a reaction gives out heat the substances are left with less energy, so $\Delta H$ ⁠ is negative: $\Delta H = -802\ \mathrm{kJ}$ ⁠ for burning a mole of methane. A reaction that takes in heat has a positive $\Delta H$ ⁠.
Feedback for a right answer
$\Delta H = -802\ \mathrm{kJ}$ ⁠. The sign says the substances lost energy: it went to the surroundings, as heat warming the chimney and as light.

Hint
$\Delta H$ ⁠ is the products’ energy minus the reactants’ energy.

Worked solution
1. $\Delta H = H_{\text{products}} - H_{\text{reactants}}$ ⁠.
2. The substances gave 802 kJ to the surroundings, so the products hold 802 kJ less.
3. $\Delta H = -802\ \mathrm{kJ}$ ⁠.
If you answered $+802$ ⁠, you took the point of view of the surroundings, which gained heat. $\Delta H$ ⁠ is written from the substances’ point of view.
Step 5 of 5
Open this step
Heat going in
Practice · Choose
Question
An instant cold pack for injuries holds ammonium nitrate, $\mathrm{NH_4NO_3}$ ⁠, and water. Squeeze it and the solid dissolves and the pack quickly turns cold. Why?

Correct answer
- Correct Dissolving takes heat in from the water, the pack and your hand
Setup, choices and feedback
The Scene at this step
Gas flow
4 cm³/s
Oxygen in the air
21 %
Water mist
0 mg/s
Zoom to the molecules
Flame
Choices
Shown in a random order.
- Correct Dissolving takes heat in from the water, the pack and your hand
- Wrong Dissolving gives out cold
  Misconception: Thinks an endothermic reaction gives out "cold".
  Feedback from the catalogue
  There is no such thing as “cold” coming out of a substance. Cold is a lack of heat. An endothermic reaction takes in heat from its surroundings, from the water, from the pack and from the hand holding it, so the surroundings cool down.
- Wrong Dissolving gives out heat, so the substance itself cools down
  Feedback
  If dissolving gave out heat, the water would warm up, as in a hand warmer. Here the water cools, so it has given up heat.
Feedback for a right answer
This is an endothermic process: $\Delta H$ ⁠ is positive, and the energy it needs comes from the surroundings. The surroundings, your skin included, lose heat and cool down. No “cold” flows: only heat flowing out of you.

Hint
Who loses heat, and who gains it?

Worked solution
1. Pulling the crystal apart into ions in water takes energy, more than is released when the ions bind to the water.
2. The missing energy is taken from the water’s thermal energy, so it cools.
3. The cold water takes heat from your hand: that is what you feel.
Other endothermic processes: photosynthesis, which takes in light, and breaking down limestone by heating.

Lesson 4 of 4 · About 25 minutes

Bond energies

Where a flame's heat comes from: breaking bonds takes energy in and making them gives it out, and working out the ΔH of burning from a table of bond energies.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Breaking and making
Practice · Choose
Question
Zoom to the molecules is at Molecules: in the flame bonds break in methane and oxygen, and new bonds form in $\mathrm{CO_2}$ ⁠ and water. What is true about the energy?

Correct answer
- Correct Breaking bonds takes energy in, and making bonds gives energy out
Setup, choices and feedback
The Scene at this step
Gas flow
4 cm³/s
Oxygen in the air
21 %
Water mist
0 mg/s
Zoom to the molecules
Molecules
Choices
Shown in a random order.
- Correct Breaking bonds takes energy in, and making bonds gives energy out
- Wrong Breaking bonds gives energy out: that is where the flame’s heat comes from
  Misconception: Thinks breaking a chemical bond releases energy.
  Feedback from the catalogue
  Breaking a bond always takes energy: the atoms pull on each other, and it takes work to part them, like pulling two magnets apart. Energy is released when new bonds form. In burning, the heat comes from the bonds in the products, $\mathrm{C=O}$ ⁠ and $\mathrm{O-H}$ ⁠, being stronger than the bonds that broke.
- Wrong Both breaking and making bonds give energy out
  Misconception: Thinks breaking a chemical bond releases energy.
  Feedback from the catalogue
  Breaking a bond always takes energy: the atoms pull on each other, and it takes work to part them, like pulling two magnets apart. Energy is released when new bonds form. In burning, the heat comes from the bonds in the products, $\mathrm{C=O}$ ⁠ and $\mathrm{O-H}$ ⁠, being stronger than the bonds that broke.
Feedback for a right answer
Breaking takes energy, and making releases it. In burning, the bonds that form are stronger than the ones that break, so more comes out than goes in: that is the flame’s heat.

Hint
Think of two magnets stuck together: does it take work to part them, or do they give energy as they part?

Worked solution
1. Bonded atoms attract each other. Parting them takes work: breaking takes energy in.
2. When atoms bond, the attraction does work and releases energy: making gives energy out.
3. A reaction’s $\Delta H$ ⁠ is what breaking takes in minus what making gives out.
The mistake “breaking bonds releases energy” is very common, perhaps because “burning fuel” sounds like taking things apart. But the energy in burning comes from the new bonds.
Step 2 of 6
Open this step
The bonds that break
Practice · Calculate
Question
Average bond energies, in kJ/mol: $\mathrm{C-H}$ ⁠ 413, $\mathrm{O=O}$ ⁠ 498. How much energy does it take to break all the bonds in the reactants of $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$ ⁠?

Correct answer
2648 kJ ± 3 kJ
Setup, choices and feedback
The Scene at this step
Gas flow
4 cm³/s
Oxygen in the air
21 %
Water mist
0 mg/s
Zoom to the molecules
Molecules
Accepted within
± 3 kJ
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 2150 kJ ± 3 kJ
  Misconception: Counts the bonds of one molecule of each substance, forgetting the coefficients of the equation.
  Feedback from the catalogue
  The coefficients say how many molecules of each substance take part, and every molecule brings all its bonds. In $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$ ⁠ two $\mathrm{O=O}$ ⁠ bonds break and four $\mathrm{O-H}$ ⁠ bonds form, two in each of the two water molecules. Make a list: the number of bonds of each kind, times its energy.
- 911 kJ ± 3 kJ
  Misconception: Counts the bonds of one molecule of each substance, forgetting the coefficients of the equation.
  Feedback from the catalogue
  The coefficients say how many molecules of each substance take part, and every molecule brings all its bonds. In $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$ ⁠ two $\mathrm{O=O}$ ⁠ bonds break and four $\mathrm{O-H}$ ⁠ bonds form, two in each of the two water molecules. Make a list: the number of bonds of each kind, times its energy.
Feedback for a right answer
$4 \times 413 + 2 \times 498 = 1652 + 996 = 2648$ ⁠, that is 2648 kJ going in.

Hint
Methane has four $\mathrm{C-H}$ ⁠ bonds, and there are two $\mathrm{O_2}$ ⁠ molecules.

Worked solution
1. A $\mathrm{CH_4}$ ⁠ molecule has four $\mathrm{C-H}$ ⁠ bonds: $4 \times 413 = 1652$ ⁠.
2. An $\mathrm{O_2}$ ⁠ molecule has one double bond, $\mathrm{O=O}$ ⁠, and there are two: $2 \times 498 = 996$ ⁠.
3. Total: $1652 + 996 = 2648\ \mathrm{kJ}$ ⁠.
If you got $2150$ ⁠, you counted only one oxygen molecule. If $911$ ⁠, you counted one $\mathrm{C-H}$ ⁠ bond and one $\mathrm{O=O}$ ⁠ bond.
Step 3 of 6
Open this step
The bonds that form
Practice · Calculate
Question
$\mathrm{CO_2}$ ⁠ has two $\mathrm{C=O}$ ⁠ bonds, 805 kJ/mol each, and each water molecule has two $\mathrm{O-H}$ ⁠ bonds, 464 kJ/mol each. How much energy is released when all the bonds in the products, $\mathrm{CO_2 + 2H_2O}$ ⁠, form?

Correct answer
3466 kJ ± 3 kJ
Setup, choices and feedback
The Scene at this step
Gas flow
4 cm³/s
Oxygen in the air
21 %
Water mist
0 mg/s
Zoom to the molecules
Molecules
Accepted within
± 3 kJ
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 2538 kJ ± 3 kJ
  Misconception: Counts the bonds of one molecule of each substance, forgetting the coefficients of the equation.
  Feedback from the catalogue
  The coefficients say how many molecules of each substance take part, and every molecule brings all its bonds. In $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$ ⁠ two $\mathrm{O=O}$ ⁠ bonds break and four $\mathrm{O-H}$ ⁠ bonds form, two in each of the two water molecules. Make a list: the number of bonds of each kind, times its energy.
Feedback for a right answer
$2 \times 805 + 4 \times 464 = 1610 + 1856 = 3466$ ⁠, that is 3466 kJ coming out: more than went in.

Hint
Two water molecules, two bonds in each: four $\mathrm{O-H}$ ⁠ bonds.

Worked solution
1. $\mathrm{CO_2}$ ⁠: two $\mathrm{C=O}$ ⁠ bonds, $2 \times 805 = 1610$ ⁠.
2. $\mathrm{2H_2O}$ ⁠: two molecules times two $\mathrm{O-H}$ ⁠ bonds, $4 \times 464 = 1856$ ⁠.
3. Total: $1610 + 1856 = 3466\ \mathrm{kJ}$ ⁠.
If you got $2538$ ⁠, you counted one water molecule instead of two.
Step 4 of 6
Open this step
The difference
Practice · Calculate
Question
From the last two answers: 2648 kJ go in to break bonds, and 3466 kJ come out from making them. What is $\Delta H$ ⁠ for burning methane by bond energies, with its sign?

Correct answer
-818 kJ ± 3 kJ
Setup, choices and feedback
The Scene at this step
Gas flow
4 cm³/s
Oxygen in the air
21 %
Water mist
0 mg/s
Zoom to the molecules
Molecules
Accepted within
± 3 kJ
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 818 kJ ± 3 kJ
  Misconception: Gives ΔH a positive sign for a reaction that gives out heat.
  Feedback from the catalogue
  $\Delta H$ ⁠ is measured from the substances’ point of view: the products’ energy minus the reactants’. When a reaction gives out heat the substances are left with less energy, so $\Delta H$ ⁠ is negative: $\Delta H = -802\ \mathrm{kJ}$ ⁠ for burning a mole of methane. A reaction that takes in heat has a positive $\Delta H$ ⁠.
- 6114 kJ ± 3 kJ
  Misconception: Thinks breaking a chemical bond releases energy.
  Feedback from the catalogue
  Breaking a bond always takes energy: the atoms pull on each other, and it takes work to part them, like pulling two magnets apart. Energy is released when new bonds form. In burning, the heat comes from the bonds in the products, $\mathrm{C=O}$ ⁠ and $\mathrm{O-H}$ ⁠, being stronger than the bonds that broke.
Feedback for a right answer
$\Delta H = 2648 - 3466 = -818\ \mathrm{kJ}$ ⁠: the reaction gives out heat, because the new bonds are stronger. That is very close to the 802 kJ the flame in the Scene gives out.

Hint
$\Delta H = \text{(energy to break)} - \text{(energy released making)}$ ⁠.

Worked solution
1. $\Delta H = \sum E(\text{bonds broken}) - \sum E(\text{bonds made})$ ⁠.
2. Substitute: $\Delta H = 2648 - 3466$ ⁠.
3. $\Delta H = -818\ \mathrm{kJ}$ ⁠.
If you answered $+818$ ⁠, you subtracted the wrong way round. If you added the two numbers, $6114$ ⁠, you treated breaking as if it gave energy out too.
Step 5 of 6
Open this step
Averages
Practice · Choose
Question
Bond energies gave $\Delta H = -818\ \mathrm{kJ}$ ⁠, but a measurement of burning methane gives $-802\ \mathrm{kJ}$ ⁠. Where does the difference come from?

Correct answer
- Correct The bond energies in the table are averages over many molecules, not the exact values in methane, oxygen and water
Setup, choices and feedback
The Scene at this step
Gas flow
4 cm³/s
Oxygen in the air
21 %
Water mist
0 mg/s
Zoom to the molecules
Flame
Choices
Shown in a random order.
- Correct The bond energies in the table are averages over many molecules, not the exact values in methane, oxygen and water
- Wrong Some of the energy is destroyed in the flame
  Misconception: Thinks some energy is destroyed or used up in a reaction, so it gives out less than it should.
  Feedback from the catalogue
  Energy is not destroyed: it only moves from place to place. Everything the substances lose in burning goes out to the surroundings as heat and light. If a calculation from bond energies differs from the measurement, it is because the bond energies in the table are averages over many different molecules, not the exact values in this reaction’s molecules.
- Wrong The activation energy has to be subtracted too
  Misconception: Confuses the activation energy with the energy the reaction gives out (ΔH).
  Feedback from the catalogue
  These are two different quantities on the energy profile. The activation energy is the height of the hump above the reactants: how much energy it takes to get going. The enthalpy change, $\Delta H$ ⁠, is the difference between the products’ level and the reactants’ level: how much the reaction gives out or takes in overall. A reaction can give out a lot and still be very slow, if the hump is high.
Feedback for a right answer
A value like 413 kJ/mol is an average of $\mathrm{C-H}$ ⁠ bonds in many different molecules, each a little different. So a calculation from bond energies is a good estimate, to within a few per cent, not an exact result.

Hint
A $\mathrm{C-H}$ ⁠ bond in methane and a $\mathrm{C-H}$ ⁠ bond in sugar: are they exactly the same bond?

Worked solution
1. A bond’s energy depends on its neighbours: a $\mathrm{C-H}$ ⁠ bond in methane differs a little from one in ethane.
2. Tables give an average, so the calculation is approximate: $-818$ ⁠ against $-802$ ⁠, a $2\%$ ⁠ difference.
3. Energy is not destroyed, and the activation energy does not enter $\Delta H$ ⁠: what goes into climbing the hump comes back on the way down.
Step 6 of 6
Open this step
Another reaction
Practice · Calculate
Question
Hydrogen and chlorine react: $\mathrm{H_2 + Cl_2 \rightarrow 2HCl}$ ⁠. Bond energies, in kJ/mol: $\mathrm{H-H}$ ⁠ 436, $\mathrm{Cl-Cl}$ ⁠ 243, $\mathrm{H-Cl}$ ⁠ 432. What is the reaction’s $\Delta H$ ⁠, with its sign?

Correct answer
-185 kJ ± 2 kJ
Setup, choices and feedback
The Scene at this step
Gas flow
4 cm³/s
Oxygen in the air
21 %
Water mist
0 mg/s
Zoom to the molecules
Flame
Accepted within
± 2 kJ
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 247 kJ ± 2 kJ
  Misconception: Counts the bonds of one molecule of each substance, forgetting the coefficients of the equation.
  Feedback from the catalogue
  The coefficients say how many molecules of each substance take part, and every molecule brings all its bonds. In $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$ ⁠ two $\mathrm{O=O}$ ⁠ bonds break and four $\mathrm{O-H}$ ⁠ bonds form, two in each of the two water molecules. Make a list: the number of bonds of each kind, times its energy.
- 185 kJ ± 2 kJ
  Misconception: Gives ΔH a positive sign for a reaction that gives out heat.
  Feedback from the catalogue
  $\Delta H$ ⁠ is measured from the substances’ point of view: the products’ energy minus the reactants’. When a reaction gives out heat the substances are left with less energy, so $\Delta H$ ⁠ is negative: $\Delta H = -802\ \mathrm{kJ}$ ⁠ for burning a mole of methane. A reaction that takes in heat has a positive $\Delta H$ ⁠.
Feedback for a right answer
$\Delta H = (436 + 243) - 2 \times 432 = 679 - 864 = -185\ \mathrm{kJ}$ ⁠. A hydrogen and chlorine mixture explodes in bright light: the light supplies the activation energy, and the reaction gives out heat.

Hint
Two $\mathrm{HCl}$ ⁠ molecules form.

Worked solution
1. Broken: one $\mathrm{H-H}$ ⁠ and one $\mathrm{Cl-Cl}$ ⁠: $436 + 243 = 679$ ⁠.
2. Made: two $\mathrm{H-Cl}$ ⁠: $2 \times 432 = 864$ ⁠.
3. $\Delta H = 679 - 864 = -185\ \mathrm{kJ}$ ⁠.
If you got $+247$ ⁠, you counted one $\mathrm{H-Cl}$ ⁠ bond, forgetting the coefficient $2$ ⁠.

Across the unit · 16 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Collision theory

Molecules have to meet

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

A more crowded flask

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Thinning it out

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Turning up the heat

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The energetic few

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Why heating works

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Activation energy and catalysts

The barrier

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Putting in the platinum

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

A lower hump

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Heating instead

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

It never runs out

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Exothermic and endothermic