Answer key · Unit 8

Equilibrium.

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 20 minutes

Reversible reactions

A reaction that runs both ways and never reaches an end: a brown gas that pairs up and splits back, until the mixture stops changing.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Both ways
Practice · Choose
Question
You seal brown $\mathrm{NO_2}$ ⁠ gas in a flask. $\mathrm{NO_2}$ ⁠ molecules pair up into $\mathrm{N_2O_4}$ ⁠, a colourless gas, and each pair can also split back: $\mathrm{2NO_2 \rightleftharpoons N_2O_4}$ ⁠. What will the flask hold after an hour?

Correct answer
- Correct A mixture of $\mathrm{NO_2}$ ⁠ and $\mathrm{N_2O_4}$ ⁠ that no longer changes
Setup, choices and feedback
The Scene at this step
Temperature
298 K
NO₂ sealed in
40 mmol/L
Push in more NO₂
offdefault
Time
Start
Choices
Shown in a random order.
- Correct A mixture of $\mathrm{NO_2}$ ⁠ and $\mathrm{N_2O_4}$ ⁠ that no longer changes
- Wrong Only $\mathrm{N_2O_4}$ ⁠: all the $\mathrm{NO_2}$ ⁠ will pair up
  Misconception: Thinks every reaction runs until one reactant is used up, even a reversible one.
  Feedback from the catalogue
  In a reversible reaction the product can break back down to the reactants. The more $\mathrm{N_2O_4}$ ⁠ forms, the more of it splits each second, until splitting catches up with pairing. From then on the amounts stop changing, and the flask holds both $\mathrm{NO_2}$ ⁠ and $\mathrm{N_2O_4}$ ⁠.
- Wrong Only $\mathrm{NO_2}$ ⁠: every pair that forms will split
  Feedback
  Pairs do split, but new ones form at the same time. As long as there is $\mathrm{NO_2}$ ⁠, some of it pairs up.
Feedback for a right answer
A reversible reaction does not reach an end: it reaches equilibrium, a mixture of reactants and products whose amounts are fixed. In the Scene the flask grows paler, but never clear.

Hint
The double arrow, $\rightleftharpoons$ ⁠, says the reaction runs both ways.

Worked solution
1. At first there is only $\mathrm{NO_2}$ ⁠, so only pairing happens.
2. As $\mathrm{N_2O_4}$ ⁠ builds up it starts to split back, at a rate that grows the more of it there is. Pairing slows, because the $\mathrm{NO_2}$ ⁠ thins out.
3. When the rates are equal, as much forms as splits, and the mixture stops changing.
If you chose “only $\mathrm{N_2O_4}$ ⁠”, you thought of a reaction that runs only forwards, like burning. A reversible reaction settles in between.
Step 2 of 5
Open this step
Watching the colour
Practice · Predict, then look
Question
Time is at Start, just sealed: the flask is all brown. What will happen to the colour as time runs on?

Correct answer
- Correct It will pale to a certain shade, then stay that way
Setup, choices and feedback
The Scene at this step
Temperature
298 Kfixed
NO₂ sealed in
40 mmol/Lfixed
Push in more NO₂
offdefault
Time
Start
Choices
- Correct It will pale to a certain shade, then stay that way
- Wrong It will pale until the flask is clear
  Misconception: Thinks every reaction runs until one reactant is used up, even a reversible one.
  Feedback from the catalogue
  In a reversible reaction the product can break back down to the reactants. The more $\mathrm{N_2O_4}$ ⁠ forms, the more of it splits each second, until splitting catches up with pairing. From then on the amounts stop changing, and the flask holds both $\mathrm{NO_2}$ ⁠ and $\mathrm{N_2O_4}$ ⁠.
- Wrong It will not change
  Feedback
  Pairing is very fast: at first there is only $\mathrm{NO_2}$ ⁠, and it pairs up at once.
Once the prediction is locked in, the Scene changes
Time
0.9 (Start → Settled)
The Scene then shows: 23.88 · Share of the nitrogen in brown NO₂, per cent/nodes/flask/outputs/no2Percent
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The colour pales and stops: about $24\%$ ⁠ of the nitrogen stays in brown $\mathrm{NO_2}$ ⁠, the rest in colourless $\mathrm{N_2O_4}$ ⁠. In the concentrations chart both lines flatten out.

Worked solution
1. At first only $\mathrm{NO_2}$ ⁠, so it pairs up fast and the colour fades.
2. As $\mathrm{N_2O_4}$ ⁠ forms, its splitting gives back $\mathrm{NO_2}$ ⁠.
3. At equilibrium pairing and splitting are equal, and the colour stays fixed, a pale brown.
If you predicted a clear flask, you treated the reaction as if it ran only forwards.
Step 3 of 5
Open this step
Counting the nitrogen
Practice · Calculate
Question
You sealed in 40 mmol/L of $\mathrm{NO_2}$ ⁠. At equilibrium the chart shows about 15.2 mmol/L of $\mathrm{N_2O_4}$ ⁠. How much $\mathrm{NO_2}$ ⁠ is left?

Correct answer
9.551 mmol/L ± 0.3 mmol/L
Setup, choices and feedback
The Scene at this step
Temperature
298 Kfixed
NO₂ sealed in
40 mmol/Lfixed
Push in more NO₂
offdefault
Time
0.9 (Start → Settled)fixed
Accepted within
± 0.3 mmol/L
Computed from the Scene at this step’s setting
NO₂ concentration (mol/m³ is mmol/L)/nodes/flask/outputs/no2
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 24.8 mmol/L ± 0.3 mmol/L
  Misconception: Forgets that each N₂O₄ took two NO₂, subtracting one NO₂ for each N₂O₄ formed.
  Feedback from the catalogue
  Each $\mathrm{N_2O_4}$ ⁠ molecule forms from two $\mathrm{NO_2}$ ⁠ molecules: $\mathrm{2NO_2 \rightleftharpoons N_2O_4}$ ⁠. So when 15 mmol/L of $\mathrm{N_2O_4}$ ⁠ forms, the $\mathrm{NO_2}$ ⁠ drops by 30 mmol/L. Check with the nitrogen: $[\mathrm{NO_2}] + 2[\mathrm{N_2O_4}]$ ⁠ always equals what was sealed in at the start.
Feedback for a right answer
$40 - 2 \times 15.2 = 9.6$ ⁠, about 9.6 mmol/L: the brown line in the chart. The nitrogen is conserved: $9.6 + 2 \times 15.2 = 40$ ⁠.

Hint
Each $\mathrm{N_2O_4}$ ⁠ molecule took two $\mathrm{NO_2}$ ⁠ molecules.

Worked solution
1. $\mathrm{2NO_2 \rightleftharpoons N_2O_4}$ ⁠: each $\mathrm{N_2O_4}$ ⁠ formed uses $2$ ⁠ $\mathrm{NO_2}$ ⁠.
2. About $15.2$ ⁠ formed, so $2 \times 15.2 = 30.4$ ⁠ were used.
3. Left: $40 - 30.4 = 9.6\ \mathrm{mmol/L}$ ⁠.
If you answered $24.8$ ⁠, you subtracted one $\mathrm{NO_2}$ ⁠ for each $\mathrm{N_2O_4}$ ⁠.
Step 4 of 5
Open this step
More in, more out
Practice · Reach the target
Question
Change NO₂ sealed in so that at equilibrium the flask holds 24 mmol/L of $\mathrm{N_2O_4}$ ⁠, then press Check.

Goal
- 24 mmol/L ± 0.4 mmol/L
  At the start of the step: 15.22 mmol/L
Setup, choices and feedback
The Scene at this step
Temperature
298 Kfixed
NO₂ sealed in
40 mmol/L
Push in more NO₂
offdefault
Time
0.9 (Start → Settled)fixed
Read from the Scene at
N₂O₄ concentration (mol/m³ is mmol/L)/nodes/flask/outputs/n2o4
One setting that reaches it
NO₂ sealed in
60 mmol/L
Push in more NO₂
off
Feedback for a right answer
At 60 mmol/L: 24 mmol/L of $\mathrm{N_2O_4}$ ⁠ and 12 mmol/L of $\mathrm{NO_2}$ ⁠. One and a half times the nitrogen did not give one and a half times of each: $\mathrm{N_2O_4}$ ⁠ grew more, because pairing goes as the square of $[\mathrm{NO_2}]$ ⁠.

Feedback for a wrong answer
Not yet. Watch the blue line in the concentrations chart.

Hint
With 40 mmol/L there is 15.2. To get more $\mathrm{N_2O_4}$ ⁠, seal in more $\mathrm{NO_2}$ ⁠, and watch the chart.

Worked solution
1. More $\mathrm{NO_2}$ ⁠ at the start gives more $\mathrm{N_2O_4}$ ⁠ at equilibrium.
2. Try: at 50 mmol/L there is about 19.6; at 60 mmol/L there is 24.0.
3. Check: $12 + 2 \times 24 = 60$ ⁠, and $\dfrac{0.024}{0.012^2} = 167$ ⁠, close to $K$ ⁠.
Step 5 of 5
Open this step
From the other side
Practice · Choose
Question
Instead of 40 mmol/L of $\mathrm{NO_2}$ ⁠ you seal in, at the same temperature, 20 mmol/L of pure $\mathrm{N_2O_4}$ ⁠: the same nitrogen. What will the flask hold at equilibrium?

Correct answer
- Correct The same mixture: about 9.6 mmol/L of $\mathrm{NO_2}$ ⁠ and 15.2 mmol/L of $\mathrm{N_2O_4}$ ⁠
Setup, choices and feedback
The Scene at this step
Temperature
298 K
NO₂ sealed in
40 mmol/L
Push in more NO₂
offdefault
Time
0.9 (Start → Settled)
Choices
Shown in a random order.
- Correct The same mixture: about 9.6 mmol/L of $\mathrm{NO_2}$ ⁠ and 15.2 mmol/L of $\mathrm{N_2O_4}$ ⁠
- Wrong Only $\mathrm{N_2O_4}$ ⁠: the reaction makes $\mathrm{N_2O_4}$ ⁠, it does not split it
  Misconception: Thinks equilibrium can only be reached from the reactants, or that starting from the products gives a different mixture.
  Feedback from the catalogue
  An equilibrium does not remember where it started. Seal pure $\mathrm{N_2O_4}$ ⁠ holding the same nitrogen and it splits until the rates are equal, reaching the same mixture exactly: because $K$ ⁠ depends only on the temperature, and $\mathrm{NO_2 + 2 \times N_2O_4}$ ⁠ stays fixed.
- Wrong A different mixture, with more $\mathrm{N_2O_4}$ ⁠, because we started from it
  Misconception: Thinks equilibrium can only be reached from the reactants, or that starting from the products gives a different mixture.
  Feedback from the catalogue
  An equilibrium does not remember where it started. Seal pure $\mathrm{N_2O_4}$ ⁠ holding the same nitrogen and it splits until the rates are equal, reaching the same mixture exactly: because $K$ ⁠ depends only on the temperature, and $\mathrm{NO_2 + 2 \times N_2O_4}$ ⁠ stays fixed.
Feedback for a right answer
The same mixture. The $\mathrm{N_2O_4}$ ⁠ splits until the rates are equal, reaching the same point from the other side. An equilibrium does not remember where it started.

Hint
The reaction runs both ways, and $K$ ⁠ depends only on the temperature.

Worked solution
1. At first only $\mathrm{N_2O_4}$ ⁠, so only splitting: $\mathrm{NO_2}$ ⁠ forms.
2. As $\mathrm{NO_2}$ ⁠ forms, pairing grows stronger, until it equals splitting.
3. The same nitrogen and the same $K$ ⁠ give the same mixture: $\dfrac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2} = K$ ⁠ and $[\mathrm{NO_2}] + 2[\mathrm{N_2O_4}] = 40$ ⁠ have one solution.

Lesson 2 of 4 · About 25 minutes

Dynamic equilibrium

Two rates that meet: why nothing changes though the molecules never stop, what is equal at equilibrium and what is not, and the equilibrium constant K.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Two rates
Practice · Predict, then look
Question
Just after sealing, pairing is 28 times faster than splitting. What will happen to the two rates as time runs on?

Correct answer
- Correct Pairing will slow and splitting speed up, until they are equal
Setup, choices and feedback
The Scene at this step
Temperature
298 Kfixed
NO₂ sealed in
40 mmol/Lfixed
Push in more NO₂
offdefault
Time
0.04 (Start → Settled)
Choices
- Correct Pairing will slow and splitting speed up, until they are equal
- Wrong Pairing will slow until it stops, and the reaction will be over
  Misconception: Thinks the reactions stop once equilibrium is reached.
  Feedback from the catalogue
  At equilibrium nothing visible changes, but the molecules do not stop: pairs keep forming and splitting, at exactly the same rate. That is dynamic equilibrium. In the rates chart neither line falls to zero; they meet and run on together.
- Wrong Both will stay as they are
  Feedback
  The rates depend on the concentrations, and the concentrations are changing: $\mathrm{NO_2}$ ⁠ is used up and $\mathrm{N_2O_4}$ ⁠ builds up.
Once the prediction is locked in, the Scene changes
Time
0.9 (Start → Settled)
The Scene then shows: 1.035 · Forward (pairing) rate over back (splitting) rate; 1 at equilibrium/nodes/flask/outputs/rateRatio
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
In the rates chart the orange line falls and the blue line rises until they meet: the ratio of the rates reaches $1$ ⁠. From then on both run on together, and neither falls to zero.

Worked solution
1. The pairing rate goes as $[\mathrm{NO_2}]^2$ ⁠, and the $\mathrm{NO_2}$ ⁠ thins out: the rate falls.
2. The splitting rate goes as $[\mathrm{N_2O_4}]$ ⁠, which builds up: the rate rises.
3. They meet when they are equal, and then the concentrations stop changing, and so do the rates.
The reaction does not stop: every second billions of pairs still form and split, in equal numbers.
Step 2 of 5
Open this step
The molecules never stop
Practice · Choose
Question
The flask is at equilibrium, and its colour no longer changes. What are the molecules doing?

Correct answer
- Correct They keep pairing and splitting, at the same rate
Setup, choices and feedback
The Scene at this step
Temperature
298 K
NO₂ sealed in
40 mmol/L
Push in more NO₂
offdefault
Time
0.9 (Start → Settled)
Choices
Shown in a random order.
- Correct They keep pairing and splitting, at the same rate
- Wrong They have stopped reacting: everything that could pair has paired
  Misconception: Thinks the reactions stop once equilibrium is reached.
  Feedback from the catalogue
  At equilibrium nothing visible changes, but the molecules do not stop: pairs keep forming and splitting, at exactly the same rate. That is dynamic equilibrium. In the rates chart neither line falls to zero; they meet and run on together.
- Wrong Only splitting, very slowly
  Feedback
  If only splitting went on, $\mathrm{NO_2}$ ⁠ would build up and the flask would darken.
Feedback for a right answer
This is dynamic equilibrium: two opposite processes at the same rate. In the Scene the molecules keep swapping partners, and in the rates chart both lines stay above zero, side by side.

Hint
Watch the molecules themselves in the Scene, not just the colour.

Worked solution
1. At equilibrium the pairing rate equals the splitting rate.
2. Each second as many pairs form as split, so the amounts do not change.
3. The molecules themselves never rest: each one goes back and forth between $\mathrm{NO_2}$ ⁠ and $\mathrm{N_2O_4}$ ⁠.
Step 3 of 5
Open this step
What is equal
Practice · Choose
Question
What is equal at equilibrium?

Correct answer
- Correct The pairing rate and the splitting rate
Setup, choices and feedback
The Scene at this step
Temperature
298 K
NO₂ sealed in
40 mmol/L
Push in more NO₂
offdefault
Time
0.9 (Start → Settled)
Choices
Shown in a random order.
- Correct The pairing rate and the splitting rate
- Wrong The concentration of $\mathrm{NO_2}$ ⁠ and the concentration of $\mathrm{N_2O_4}$ ⁠
  Misconception: Thinks that at equilibrium the reactants and products are present in equal amounts.
  Feedback from the catalogue
  At equilibrium the rates are equal, not the amounts. At 25 °C the flask holds about 9.5 mmol/L of $\mathrm{NO_2}$ ⁠ against 15 mmol/L of $\mathrm{N_2O_4}$ ⁠, and in boiling water almost nothing but $\mathrm{NO_2}$ ⁠. Where the mix settles is set by $K$ ⁠, which can be very large or very small.
- Wrong The nitrogen in $\mathrm{NO_2}$ ⁠ and the nitrogen in $\mathrm{N_2O_4}$ ⁠
  Misconception: Thinks that at equilibrium the reactants and products are present in equal amounts.
  Feedback from the catalogue
  At equilibrium the rates are equal, not the amounts. At 25 °C the flask holds about 9.5 mmol/L of $\mathrm{NO_2}$ ⁠ against 15 mmol/L of $\mathrm{N_2O_4}$ ⁠, and in boiling water almost nothing but $\mathrm{NO_2}$ ⁠. Where the mix settles is set by $K$ ⁠, which can be very large or very small.
Feedback for a right answer
The rates. The concentrations are steady, but not equal: here 9.5 mmol/L of $\mathrm{NO_2}$ ⁠ against 15.2 mmol/L of $\mathrm{N_2O_4}$ ⁠. In the rates chart the lines meet, and in the concentrations chart they only flatten.

Hint
Compare the two charts: in which do the lines meet?

Worked solution
1. At equilibrium the forward rate equals the back rate.
2. The concentrations stop changing, but each stays at its own value.
3. The ratio between them is fixed: $\dfrac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2} = K$ ⁠, and $K$ ⁠ can be any number.
Step 4 of 5
Open this step
The equilibrium constant
Practice · Calculate
Question
At 348 K the flask has settled at $[\mathrm{NO_2}] = 0.0284$ ⁠ and $[\mathrm{N_2O_4}] = 0.0058$ ⁠, in mol/L. Work out the equilibrium constant, $K$ ⁠, in L/mol, then compare with the line under the chart.

Correct answer
7.18 ± 3%
Setup, choices and feedback
The Scene at this step
Temperature
348 Kfixed
NO₂ sealed in
40 mmol/Lfixed
Push in more NO₂
offdefault
Time
0.9 (Start → Settled)fixed
Accepted within
± 3%
Computed from the Scene at this step’s setting
Reaction quotient [N₂O₄]/[NO₂]², L/mol (equals K once settled)/nodes/flask/outputs/Q
Wrong answers it recognises
- 0.204 ± 3%
  Misconception: Leaves the coefficients out of the equilibrium expression, writing [N₂O₄]/[NO₂] instead of [N₂O₄]/[NO₂]².
  Feedback from the catalogue
  In the expression for $K$ ⁠ each concentration is raised to the power of its coefficient in the equation. In $\mathrm{2NO_2 \rightleftharpoons N_2O_4}$ ⁠ the coefficient of $\mathrm{NO_2}$ ⁠ is $2$ ⁠, so $K = \dfrac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2}$ ⁠. The power comes from each pairing needing two $\mathrm{NO_2}$ ⁠ molecules to meet.
- 0.139 ± 3%
  Misconception: Writes the equilibrium constant with the reactants on top and the products underneath.
  Feedback from the catalogue
  In the equilibrium constant the products go on top and the reactants underneath: $K = \dfrac{[\text{products}]}{[\text{reactants}]}$ ⁠, each to the power of its coefficient. So a large $K$ ⁠ means plenty of product at equilibrium.
Feedback for a right answer
$\dfrac{0.0058}{0.0284^2} = 7.2$ ⁠. At 298 K the constant is about 170: heating shrank it 24 times, which is why the mix is so much browner.

Hint
$K = \dfrac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2}$ ⁠.

Worked solution
1. The expression: $K = \dfrac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2}$ ⁠, because the coefficient of $\mathrm{NO_2}$ ⁠ is $2$ ⁠.
2. $[\mathrm{NO_2}]^2 = 0.0284^2 = 8.07 \times 10^{-4}$ ⁠.
3. $K = \dfrac{0.0058}{8.07 \times 10^{-4}} = 7.2\ \mathrm{L/mol}$ ⁠.
If you got $0.204$ ⁠, you forgot the square. If $0.139$ ⁠, you turned the fraction upside down: products go on top.
Step 5 of 5
Open this step
Which way the mix moves
Practice · Choose
Question
Shortly after sealing, the line under the chart shows $Q = 52$ ⁠ and $K = 170$ ⁠. Which way will the mix move?

Correct answer
- Correct Towards $\mathrm{N_2O_4}$ ⁠: pairing is still faster than splitting
Setup, choices and feedback
The Scene at this step
Temperature
298 Kfixed
NO₂ sealed in
40 mmol/Lfixed
Push in more NO₂
offdefault
Time
0.2 (Start → Settled)fixed
Choices
Shown in a random order.
- Correct Towards $\mathrm{N_2O_4}$ ⁠: pairing is still faster than splitting
- Wrong Towards $\mathrm{NO_2}$ ⁠: $Q$ ⁠ is small, so the reaction goes backwards
  Misconception: Reads Q below K as meaning the mixture will move back towards the reactants.
  Feedback from the catalogue
  $Q$ ⁠ is built like $K$ ⁠, from the concentrations right now. When $Q < K$ ⁠ there is still too little product for equilibrium, so the forward reaction is faster and the mix moves towards the products, until $Q$ ⁠ reaches $K$ ⁠. When $Q > K$ ⁠ it moves back towards the reactants.
- Wrong Nowhere: the flask is already at equilibrium
  Feedback
  At equilibrium $Q = K$ ⁠. Here $Q$ ⁠ is a third of $K$ ⁠.
Feedback for a right answer
When $Q < K$ ⁠ there is too little $\mathrm{N_2O_4}$ ⁠, and the mix carries on towards it until $Q$ ⁠ reaches $K$ ⁠. In the rates chart the orange line is still above the blue.

Hint
$Q$ ⁠ is built like $K$ ⁠, from the concentrations right now. A small $Q$ ⁠ says: little product for the reactant.

Worked solution
1. $Q = \dfrac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2}$ ⁠ right now, and $K$ ⁠ is its value at equilibrium.
2. $Q = 52 < K = 170$ ⁠: the top is too small, or the bottom too large.
3. For $Q$ ⁠ to grow, you need more $\mathrm{N_2O_4}$ ⁠ and less $\mathrm{NO_2}$ ⁠: the reaction carries on forwards.
Rule of thumb: $Q < K$ ⁠ forwards, $Q > K$ ⁠ backwards, $Q = K$ ⁠ equilibrium.

Lesson 3 of 4 · About 25 minutes

Le Chatelier's principle

What happens when you disturb an equilibrium: add a reactant, heat it or cool it. The system shifts to reduce the disturbance, and only temperature changes K.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Adding a reactant
Practice · Predict, then look
Question
The flask is at equilibrium, with 9.5 mmol/L of $\mathrm{NO_2}$ ⁠. A syringe pushes in more $\mathrm{NO_2}$ ⁠, and $[\mathrm{NO_2}]$ ⁠ jumps to 29.5 mmol/L. What will happen next?

Correct answer
- Correct Some of the added $\mathrm{NO_2}$ ⁠ will pair up, and $[\mathrm{NO_2}]$ ⁠ will settle between 9.5 and 29.5
Setup, choices and feedback
The Scene at this step
Temperature
298 Kfixed
NO₂ sealed in
40 mmol/Lfixed
Push in more NO₂
offdefault
Time
0.9 (Start → Settled)
Choices
- Correct Some of the added $\mathrm{NO_2}$ ⁠ will pair up, and $[\mathrm{NO_2}]$ ⁠ will settle between 9.5 and 29.5
- Wrong All of the extra will pair up, and $[\mathrm{NO_2}]$ ⁠ will go back to 9.5
  Misconception: Thinks that after a disturbance the system returns exactly to its old concentrations.
  Feedback from the catalogue
  Le Chatelier’s principle says the system reduces the change, not that it undoes it. After $\mathrm{NO_2}$ ⁠ is added some of it pairs up, but at the new equilibrium there is more $\mathrm{NO_2}$ ⁠ than before the push, and more $\mathrm{N_2O_4}$ ⁠ too. Only the ratio $\dfrac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2}$ ⁠ goes back to $K$ ⁠.
- Wrong Nothing: the flask was already at equilibrium, and $[\mathrm{NO_2}]$ ⁠ will stay at 29.5
  Misconception: Thinks the reactions stop once equilibrium is reached.
  Feedback from the catalogue
  At equilibrium nothing visible changes, but the molecules do not stop: pairs keep forming and splitting, at exactly the same rate. That is dynamic equilibrium. In the rates chart neither line falls to zero; they meet and run on together.
Once the prediction is locked in, the Scene changes
Push in more NO₂
on
Time
Settled
The Scene then shows: 11.88 mmol/L · NO₂ concentration (mol/m³ is mmol/L)/nodes/flask/outputs/no2
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$[\mathrm{NO_2}]$ ⁠ settles at about 11.9 mmol/L: more than before, less than at the push. The mix “fought” the addition and moved towards $\mathrm{N_2O_4}$ ⁠, which rose to 24 mmol/L. That is Le Chatelier’s principle.

Worked solution
1. Just after the push $Q$ ⁠ is very small, because the bottom, $[\mathrm{NO_2}]^2$ ⁠, grew nine times.
2. $Q < K$ ⁠, so pairing outruns splitting, and the mix moves towards $\mathrm{N_2O_4}$ ⁠.
3. It stops when $Q = K$ ⁠ again. That happens before all the extra has paired, because $[\mathrm{N_2O_4}]$ ⁠ on top rises too.
So the system reduces the change rather than undoing it.
Step 2 of 5
Open this step
The same constant
Practice · Choose
Question
After the push the flask has settled at a new equilibrium, with more $\mathrm{NO_2}$ ⁠ and more $\mathrm{N_2O_4}$ ⁠. What happened to the equilibrium constant, $K$ ⁠?

Correct answer
- Correct It did not change: $Q$ ⁠ went back to the same $K$ ⁠
Setup, choices and feedback
The Scene at this step
Temperature
298 K
NO₂ sealed in
40 mmol/L
Push in more NO₂
on
Time
Settled
Choices
Shown in a random order.
- Correct It did not change: $Q$ ⁠ went back to the same $K$ ⁠
- Wrong It grew, because there is more $\mathrm{N_2O_4}$ ⁠
  Misconception: Thinks adding a reactant or product changes the equilibrium constant.
  Feedback from the catalogue
  The equilibrium constant depends only on the reaction and the temperature. Adding $\mathrm{NO_2}$ ⁠ changes $Q$ ⁠, not $K$ ⁠, and the mix moves until $Q$ ⁠ is back at the same $K$ ⁠. Only a change of temperature changes $K$ ⁠ itself.
- Wrong It shrank, because there is more $\mathrm{NO_2}$ ⁠
  Misconception: Thinks adding a reactant or product changes the equilibrium constant.
  Feedback from the catalogue
  The equilibrium constant depends only on the reaction and the temperature. Adding $\mathrm{NO_2}$ ⁠ changes $Q$ ⁠, not $K$ ⁠, and the mix moves until $Q$ ⁠ is back at the same $K$ ⁠. Only a change of temperature changes $K$ ⁠ itself.
Feedback for a right answer
$K$ ⁠ depends only on the temperature. The push changed $Q$ ⁠, and the mix moved until $Q$ ⁠ was back at the same $K$ ⁠: $\dfrac{0.0241}{0.0119^2} \approx 170$ ⁠, as before.

Hint
Compare the line under the chart at Settled with Push in more NO₂ off and on.

Worked solution
1. $K$ ⁠ is set by the reaction and the temperature, which did not change.
2. The push dropped $Q$ ⁠ below $K$ ⁠, and the reaction ran forwards until $Q = K$ ⁠.
3. Check at the new equilibrium: $\dfrac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2} = \dfrac{0.0241}{0.0119^2} = 170$ ⁠.
Step 3 of 5
Open this step
The moment of the push
Practice · Choose
Question
Push in more NO₂ is on and Time is at Start: the moment of the push. Which rate jumped, by the rates chart?

Correct answer
- Correct The pairing rate, about ten times; the splitting rate did not change
Setup, choices and feedback
The Scene at this step
Temperature
298 K
NO₂ sealed in
40 mmol/L
Push in more NO₂
onfixed
Time
Startfixed
Choices
Shown in a random order.
- Correct The pairing rate, about ten times; the splitting rate did not change
- Wrong Both rates jumped by the same amount
  Feedback
  Splitting depends only on $[\mathrm{N_2O_4}]$ ⁠, and the syringe added no $\mathrm{N_2O_4}$ ⁠. So it does not jump.
- Wrong The splitting rate, to get rid of the extra $\mathrm{NO_2}$ ⁠
  Misconception: Reads Q below K as meaning the mixture will move back towards the reactants.
  Feedback from the catalogue
  $Q$ ⁠ is built like $K$ ⁠, from the concentrations right now. When $Q < K$ ⁠ there is still too little product for equilibrium, so the forward reaction is faster and the mix moves towards the products, until $Q$ ⁠ reaches $K$ ⁠. When $Q > K$ ⁠ it moves back towards the reactants.
Feedback for a right answer
$[\mathrm{NO_2}]$ ⁠ tripled, so pairing, which goes as its square, rose nine to ten times. Splitting did not change at first. That is exactly Le Chatelier’s principle: the change speeds up the direction that reduces it.

Hint
Pairing goes as $[\mathrm{NO_2}]^2$ ⁠, and splitting as $[\mathrm{N_2O_4}]$ ⁠. Which did the syringe change?

Worked solution
1. The pairing rate: $k_f[\mathrm{NO_2}]^2$ ⁠. $[\mathrm{NO_2}]$ ⁠ rose from $9.5$ ⁠ to $29.5$ ⁠, $3.1$ ⁠ times, so the rate rose $3.1^2 \approx 9.6$ ⁠ times.
2. The splitting rate: $k_b[\mathrm{N_2O_4}]$ ⁠, and $[\mathrm{N_2O_4}]$ ⁠ did not change at the push.
3. Pairing is now faster, and the mix moves towards $\mathrm{N_2O_4}$ ⁠.
A reaction does not “decide” to get rid of an excess: the rates are simply set by the concentrations.
Step 4 of 5
Open this step
Heating it
Practice · Predict, then look
Question
Pairing, $\mathrm{2NO_2 \rightarrow N_2O_4}$ ⁠, gives out 57 kJ/mol, and splitting takes it in. What will happen to the colour at equilibrium when you heat the flask from 298 K to 348 K?

Correct answer
- Correct It will darken: heating favours splitting, which takes heat in
Setup, choices and feedback
The Scene at this step
Temperature
298 K
NO₂ sealed in
40 mmol/Lfixed
Push in more NO₂
offdefault
Time
0.9 (Start → Settled)fixed
Choices
- Correct It will darken: heating favours splitting, which takes heat in
- Wrong It will pale: heat speeds the reaction forwards
  Misconception: Thinks heating always shifts an equilibrium towards the products, because heat speeds reactions up.
  Feedback from the catalogue
  Heating speeds up both directions, but not equally: it favours the direction that takes in heat. In $\mathrm{2NO_2 \rightleftharpoons N_2O_4}$ ⁠ pairing gives out 57 kJ/mol, so splitting takes it in, and heating moves the mix back, to brown $\mathrm{NO_2}$ ⁠. In the Haber reaction too, which gives out heat, heating lowers the amount of ammonia at equilibrium.
- Wrong It will not change: heating only makes equilibrium come sooner
  Misconception: Thinks temperature changes only how fast equilibrium is reached, not where it lies.
  Feedback from the catalogue
  Temperature changes $K$ ⁠ itself, and so the mix the system settles at. In the flask: in ice water about $9\%$ ⁠ of the nitrogen is left in $\mathrm{NO_2}$ ⁠ at equilibrium, and in boiling water about $87\%$ ⁠. A catalyst, by contrast, changes only the speed.
Once the prediction is locked in, the Scene changes
Temperature
348 K
The Scene then shows: 71.02 · Share of the nitrogen in brown NO₂, per cent/nodes/flask/outputs/no2Percent
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The share of the nitrogen in $\mathrm{NO_2}$ ⁠ jumps from $24\%$ ⁠ to $71\%$ ⁠, and the flask turns dark brown. Heating is a “change” the system reduces: it soaks up heat by splitting $\mathrm{N_2O_4}$ ⁠. $K$ ⁠ falls from 170 to 7.

Worked solution
1. Le Chatelier’s principle: a system at equilibrium that is disturbed shifts so as to reduce the disturbance.
2. The disturbance is heat coming in. The direction that takes heat in is splitting, $\mathrm{N_2O_4 \rightarrow 2NO_2}$ ⁠.
3. So the mix moves to brown $\mathrm{NO_2}$ ⁠, and $K$ ⁠ falls.
Heating speeds up both directions, but the one that takes in heat more. And temperature, unlike a catalyst, changes $K$ ⁠ itself.
Step 5 of 5
Open this step
Half and half
Practice · Reach the target
Question
Change Temperature so that at equilibrium exactly half the nitrogen is in $\mathrm{NO_2}$ ⁠ and half in $\mathrm{N_2O_4}$ ⁠, then press Check.

Goal
- 50 ± 1
  At the start of the step: 23.88
Setup, choices and feedback
The Scene at this step
Temperature
298 K
NO₂ sealed in
40 mmol/Lfixed
Push in more NO₂
offdefault
Time
0.9 (Start → Settled)fixed
Read from the Scene at
Share of the nitrogen in brown NO₂, per cent/nodes/flask/outputs/no2Percent
One setting that reaches it
Temperature
326 K
Push in more NO₂
off
Feedback for a right answer
At about 326 K, that is 53 °C. At that temperature $K = \dfrac{0.010}{0.020^2} = 25$ ⁠. A few degrees change the mix a lot, which is why this flask works as a colour thermometer.

Feedback for a wrong answer
Not yet. Heat to move nitrogen into $\mathrm{NO_2}$ ⁠, cool to move it into $\mathrm{N_2O_4}$ ⁠.

Hint
Half the nitrogen in $\mathrm{NO_2}$ ⁠ means $[\mathrm{NO_2}] = 20$ ⁠ and $[\mathrm{N_2O_4}] = 10$ ⁠, in mmol/L. Watch the concentrations chart.

Worked solution
1. At 298 K only $24\%$ ⁠ of the nitrogen is in $\mathrm{NO_2}$ ⁠: you need more, so heat.
2. At 348 K it is already $71\%$ ⁠: come back down a little.
3. Around 326 K you reach $[\mathrm{NO_2}] = 20$ ⁠ and $[\mathrm{N_2O_4}] = 10$ ⁠, half and half.

Lesson 4 of 4 · About 20 minutes

Equilibrium in industry

Pressure, temperature and a catalyst in the Haber process, which makes the ammonia for fertilisers: why squeeze, why not cool all the way, and what the catalyst does not do.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Squeezing
Practice · Predict, then look
Question
The flask holds 20 mmol/L, and about $32\%$ ⁠ of the nitrogen is in $\mathrm{NO_2}$ ⁠. What will happen to that share when you seal in gas four times as dense, 80 mmol/L, at the same temperature?

Correct answer
- Correct It will shrink: in a denser gas a bigger share pairs up
Setup, choices and feedback
The Scene at this step
Temperature
298 Kfixed
NO₂ sealed in
20 mmol/L
Push in more NO₂
offdefault
Time
0.9 (Start → Settled)fixed
Choices
- Correct It will shrink: in a denser gas a bigger share pairs up
- Wrong It will stay $32\%$ ⁠: the temperature and $K$ ⁠ have not changed
  Misconception: Thinks raising the pressure (or concentration) of a gas mixture cannot move its equilibrium.
  Feedback from the catalogue
  When the gas is squeezed, the mix moves to the side with fewer gas molecules, because that lowers the pressure. In $\mathrm{2NO_2 \rightleftharpoons N_2O_4}$ ⁠ two molecules become one, so a denser gas holds more of its nitrogen in $\mathrm{N_2O_4}$ ⁠. Pressure matters only when the two sides have different numbers of molecules.
- Wrong It will grow: there is more $\mathrm{NO_2}$ ⁠
  Feedback
  There is more $\mathrm{NO_2}$ ⁠ in number, but the question is about the share. In a denser gas pairs form more often, because pairing goes as the square of the concentration.
Once the prediction is locked in, the Scene changes
NO₂ sealed in
80 mmol/L
The Scene then shows: 17.59 · Share of the nitrogen in brown NO₂, per cent/nodes/flask/outputs/no2Percent
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The share drops to $18\%$ ⁠. In a denser gas, as in a squeezed one, the mix moves to the side with fewer molecules: two $\mathrm{NO_2}$ ⁠ molecules become one. $K$ ⁠ has not changed; only the way to satisfy it.

Worked solution
1. $K = \dfrac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2}$ ⁠ does not change when every concentration grows four times.
2. But if both grew four times, $Q$ ⁠ would be four times smaller: $\dfrac{4}{4^2} = \tfrac14$ ⁠.
3. For $Q$ ⁠ to get back to $K$ ⁠, a bigger share of the $\mathrm{NO_2}$ ⁠ pairs up.
Pressure affects only gas reactions whose two sides have different numbers of molecules.
Step 2 of 5
Open this step
A quarter
Practice · Reach the target
Question
Change NO₂ sealed in so that at equilibrium a quarter of the nitrogen, $25\%$ ⁠, is in $\mathrm{NO_2}$ ⁠, then press Check.

Goal
- 25 ± 0.5
  At the start of the step: 17.59
Setup, choices and feedback
The Scene at this step
Temperature
298 Kfixed
NO₂ sealed in
80 mmol/L
Push in more NO₂
offdefault
Time
0.9 (Start → Settled)fixed
Read from the Scene at
Share of the nitrogen in brown NO₂, per cent/nodes/flask/outputs/no2Percent
One setting that reaches it
NO₂ sealed in
35 mmol/L
Push in more NO₂
off
Feedback for a right answer
At 35 mmol/L. For less to pair up, the gas has to be thinner: like lowering the pressure.

Feedback for a wrong answer
Not yet. The denser the gas, the smaller the share in $\mathrm{NO_2}$ ⁠.

Hint
At 80 mmol/L the share is $18\%$ ⁠, and at 20 mmol/L $32\%$ ⁠. A lower concentration leaves more $\mathrm{NO_2}$ ⁠.

Worked solution
1. The share in $\mathrm{NO_2}$ ⁠ grows as the gas gets thinner.
2. At 40 mmol/L it is $23.9\%$ ⁠, a little too little: thin it.
3. At 35 mmol/L it is $25.3\%$ ⁠.
Step 3 of 5
Open this step
The Haber process
Practice · Choose
Question
Ammonia, the raw material of most fertilisers, is made by the Haber process: $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$ ⁠, $\Delta H = -92\ \mathrm{kJ}$ ⁠. Under which conditions is there the most ammonia at equilibrium?

Correct answer
- Correct High pressure and low temperature
Setup, choices and feedback
The Scene at this step
Temperature
298 K
NO₂ sealed in
40 mmol/L
Push in more NO₂
offdefault
Time
0.9 (Start → Settled)
Choices
Shown in a random order.
- Correct High pressure and low temperature
- Wrong High pressure and high temperature
  Misconception: Thinks heating always shifts an equilibrium towards the products, because heat speeds reactions up.
  Feedback from the catalogue
  Heating speeds up both directions, but not equally: it favours the direction that takes in heat. In $\mathrm{2NO_2 \rightleftharpoons N_2O_4}$ ⁠ pairing gives out 57 kJ/mol, so splitting takes it in, and heating moves the mix back, to brown $\mathrm{NO_2}$ ⁠. In the Haber reaction too, which gives out heat, heating lowers the amount of ammonia at equilibrium.
- Wrong Low pressure and low temperature
  Misconception: Thinks raising the pressure (or concentration) of a gas mixture cannot move its equilibrium.
  Feedback from the catalogue
  When the gas is squeezed, the mix moves to the side with fewer gas molecules, because that lowers the pressure. In $\mathrm{2NO_2 \rightleftharpoons N_2O_4}$ ⁠ two molecules become one, so a denser gas holds more of its nitrogen in $\mathrm{N_2O_4}$ ⁠. Pressure matters only when the two sides have different numbers of molecules.
Feedback for a right answer
Four gas molecules on the left and two on the right, so high pressure shifts right. Joining gives out heat, so cooling shifts right. Just as in $\mathrm{2NO_2 \rightleftharpoons N_2O_4}$ ⁠: there cold and dense gives more $\mathrm{N_2O_4}$ ⁠.

Hint
Count the gas molecules on each side, and see which direction gives out heat.

Worked solution
1. Pressure: $1 + 3 = 4$ ⁠ molecules become $2$ ⁠. High pressure favours the side with fewer molecules: the ammonia.
2. Temperature: $\Delta H < 0$ ⁠, so making ammonia gives out heat. Cooling favours the direction that gives out heat: the ammonia.
3. So high pressure and low temperature.
Step 4 of 5
Open this step
A compromise
Practice · Choose
Question
If cooling gives more ammonia, why do plants run the Haber process at about 450 °C and not at room temperature?

Correct answer
- Correct At a low temperature the reaction is far too slow, and the plant would make very little each hour
Setup, choices and feedback
The Scene at this step
Temperature
298 K
NO₂ sealed in
40 mmol/L
Push in more NO₂
offdefault
Time
0.9 (Start → Settled)
Choices
Shown in a random order.
- Correct At a low temperature the reaction is far too slow, and the plant would make very little each hour
- Wrong At a high temperature there is more ammonia at equilibrium
  Misconception: Thinks heating always shifts an equilibrium towards the products, because heat speeds reactions up.
  Feedback from the catalogue
  Heating speeds up both directions, but not equally: it favours the direction that takes in heat. In $\mathrm{2NO_2 \rightleftharpoons N_2O_4}$ ⁠ pairing gives out 57 kJ/mol, so splitting takes it in, and heating moves the mix back, to brown $\mathrm{NO_2}$ ⁠. In the Haber reaction too, which gives out heat, heating lowers the amount of ammonia at equilibrium.
- Wrong The temperature does not change the amount of ammonia at all, only the speed
  Misconception: Thinks temperature changes only how fast equilibrium is reached, not where it lies.
  Feedback from the catalogue
  Temperature changes $K$ ⁠ itself, and so the mix the system settles at. In the flask: in ice water about $9\%$ ⁠ of the nitrogen is left in $\mathrm{NO_2}$ ⁠ at equilibrium, and in boiling water about $87\%$ ⁠. A catalyst, by contrast, changes only the speed.
Feedback for a right answer
It is a compromise. In the cold there is plenty of ammonia at equilibrium, but it takes a very long time to get there, because breaking $\mathrm{N \equiv N}$ ⁠ has a high barrier. At 450 °C, with an iron catalyst and about 200 atmospheres, each pass gives about $15\%$ ⁠ ammonia, fast. Whatever did not react goes back into the reactor.

Hint
Equilibrium says how much ammonia there will be in the end. What does collision theory say about when?

Worked solution
1. Equilibrium: a low temperature gives more ammonia, because the reaction gives out heat.
2. Rate: a low temperature gives a very slow reaction, because few collisions clear the barrier.
3. A plant needs both a lot and fast, so it picks a middling temperature, high pressure and a catalyst, and recycles the gases that did not react.
Step 5 of 5
Open this step
The catalyst
Practice · Choose
Question
What does the iron catalyst in the Haber process do to the amount of ammonia at equilibrium?

Correct answer
- Correct Nothing: equilibrium arrives sooner, but with the same amount of ammonia
Setup, choices and feedback
The Scene at this step
Temperature
298 K
NO₂ sealed in
40 mmol/L
Push in more NO₂
offdefault
Time
0.9 (Start → Settled)
Choices
Shown in a random order.
- Correct Nothing: equilibrium arrives sooner, but with the same amount of ammonia
- Wrong It increases it, because it speeds up making ammonia
  Misconception: Thinks a catalyst moves an equilibrium towards the products and so increases the yield.
  Feedback from the catalogue
  A catalyst lowers the barrier for both directions by the same amount, so it speeds the forward and the back reaction by the same factor. Equilibrium arrives sooner, but it is the same equilibrium, with the same $K$ ⁠ and the same amount of product.
- Wrong It decreases it, because it speeds up the breakdown too
  Misconception: Thinks a catalyst moves an equilibrium towards the products and so increases the yield.
  Feedback from the catalogue
  A catalyst lowers the barrier for both directions by the same amount, so it speeds the forward and the back reaction by the same factor. Equilibrium arrives sooner, but it is the same equilibrium, with the same $K$ ⁠ and the same amount of product.
Feedback for a right answer
A catalyst lowers the barrier for both directions by the same amount, so it speeds up making and breaking ammonia by the same factor. $K$ ⁠ does not change, and nor does the final mix. What the catalyst gives is time: the same ammonia, in minutes instead of years.

Hint
A catalyst lowers the hump on the energy profile. From which side is it climbed?

Worked solution
1. On the energy profile, the hump of the back reaction is the same hump, seen from the products’ side.
2. A catalyst lowers the hump, and so both barriers, speeding up both directions.
3. The balance between the rates at equilibrium, $K$ ⁠, does not change.

Across the unit · 14 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

Reversible reactions

Both ways

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Watching the colour

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Counting the nitrogen

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

More in, more out

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

From the other side

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Dynamic equilibrium

Two rates

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The molecules never stop

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

What is equal

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

The equilibrium constant

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Which way the mix moves

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Le Chatelier's principle

Adding a reactant

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution