5 exam-style question sets on Chemical equilibrium (Paper 2), each with a hint and a fully worked answer. The app holds 25 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.
Practise this section in the app →
Question 1
Carbon monoxide gas reacts with chlorine gas in a sealed container at constant temperature T °C and reaches equilibrium according to the balanced equation:
CO(g) + Cl₂(g) ⇌ COCl₂(g) ΔH = −108 kJ·mol⁻¹
The graph shows the concentrations of the three gases, represented by X, Y and Z, as a function of time. At t₁ more chlorine gas is pumped into the container. At t₂ the volume of the container is changed at constant temperature.
A DIFFERENT gaseous equilibrium is studied in QUESTIONS f and g: 1,2 mol of PCl₅(g) is sealed in a 2 dm³ container at temperature T and the system reaches equilibrium according to:
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
At equilibrium the container holds 0,4 mol of Cl₂(g).
- aDefine the term chemical equilibrium. (2)
- bWrite down the letter (X, Y or Z) that represents CO(g). Give a reason for the answer by referring to the graph at t₁. (2)
- cWas the volume of the container INCREASED or DECREASED at t₂? Give a reason for the answer by referring to the graph. (2)
- dAfter the new equilibrium is established, the temperature of the container is increased at constant volume. How will the number of moles of COCl₂(g) change? Write down INCREASES, DECREASES or REMAINS THE SAME. Use Le Chatelier's principle to explain the answer. (4)
- eHow will the change in QUESTION d affect the value of the equilibrium constant, Kc? Write down INCREASES, DECREASES or REMAINS THE SAME. (1)
- fWrite down the expression for the equilibrium constant, Kc, for the PCl₅(g) equilibrium. (2)
- gCalculate the value of Kc for the PCl₅(g) equilibrium at temperature T. (6)
This question has a diagram, shown in the app.
Hint
The gas that is added shows an instantaneous jump; for the Kc calculation set up a RICE table using the 1 : 1 : 1 mole ratio and divide the equilibrium moles by 2 dm³.
Worked answer
a) Chemical equilibrium is the stage in a reversible reaction, in a closed system, at which the rate of the forward reaction equals the rate of the reverse reaction (concentrations remain constant). ✓✓
b) Y. ✓ At t₁ the concentration of Z jumps up instantaneously (Cl₂ was added) while Y decreases gradually — Y is the other reactant, CO, being used up as the equilibrium shifts forward. ✓
c) DECREASED. ✓ At t₂ the concentrations of ALL three gases increase instantaneously (sudden jump), which happens when the same amounts of gas occupy a smaller volume. ✓
d) DECREASES. ✓ The forward reaction is exothermic (ΔH < 0). ✓ According to Le Chatelier's principle, an increase in temperature favours the reaction that absorbs heat, i.e. the endothermic reverse reaction. ✓ COCl₂ is used up, so its number of moles decreases. ✓
e) DECREASES ✓ (only a temperature change alters Kc; the shift is towards reactants).
f) ✓✓
g) RICE table (moles): ratio 1 : 1 : 1.
| PCl₅ | PCl₃ | Cl₂ | |
|---|---|---|---|
| Initial (mol) | 1,2 | 0 | 0 |
| Change (mol) | −0,4 | +0,4 | +0,4 ✓ |
| Equilibrium (mol) | 0,8 ✓ | 0,4 | 0,4 |
| ÷ 2 dm³ (mol·dm⁻³) | 0,4 | 0,2 | 0,2 ✓✓ |
✓ = 0,10 ✓ (no unit)
Modelled on Nov 2024 Paper 2, Question 6 — same skills, new scenario.
Question 2
Nitrogen and hydrogen are sealed in a container in a 1 : 3 mole ratio, with no ammonia present initially. The system reaches equilibrium at constant temperature:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Which ONE of the following is ALWAYS TRUE at equilibrium?
Hint
The gases start in the mole ratio of their coefficients and are consumed in that same ratio.
Worked answer
B. N₂ and H₂ start in a 1 : 3 ratio and are used up in a 1 : 3 ratio, so at every moment — including equilibrium — [H₂] remains three times [N₂]. A would require equal starting amounts. C and D depend on how far the reaction proceeds (the value of Kc), so they are not guaranteed. Modelled on Nov 2024 Paper 2, Question 1.5 — same skill, new scenario.
Question 3
Consider the following reaction at equilibrium in a closed container:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ·mol⁻¹
Which ONE of the following changes will INCREASE the yield of NH₃(g)?
Hint
Count the moles of gas on each side and remember the forward reaction is exothermic.
Worked answer
C. Decreasing the volume increases the pressure; the equilibrium shifts to the side with fewer moles of gas (4 mol → 2 mol), i.e. forward, so more NH₃ forms. A is wrong: a catalyst speeds up both reactions equally and does not shift the equilibrium. B is wrong: higher temperature favours the endothermic reverse reaction, decreasing the yield. D is wrong: a larger volume (lower pressure) favours the side with more moles of gas — the reverse reaction.
Question 4
Part 1: At a salt-recovery plant, a saturated sodium chloride solution is kept in contact with excess solid salt in a closed tank at 25 °C. Equilibrium is established according to:
Part 2: In a separate investigation, hydrogen gas and iodine gas react in a sealed container according to:
The table below shows the equilibrium constant, Kc, for this reaction at different temperatures.
| Temperature (°C) | Kc |
|---|---|
| 400 | 49 |
| 450 | 36 |
| 500 | 27 |
- aState Le Chatelier's principle. (2)
- bA few drops of concentrated hydrochloric acid are added to the tank in Part 1. What effect does this addition have on the mass of NaCl(s)? Choose from INCREASES, DECREASES or REMAINS THE SAME. (1)
- cExplain the answer to QUESTION b by using Le Chatelier's principle. (2)
- dIs the FORWARD reaction in Part 2 EXOTHERMIC or ENDOTHERMIC? (1)
- eExplain the answer to QUESTION d by referring to the data in the table and using Le Chatelier's principle. (3)
- fInitially 2,0 mol of H₂(g) and 2,0 mol of I₂(g) are sealed in a 2 dm³ container at 450 °C and allowed to reach equilibrium. Calculate the mass of HI(g) present in the container at equilibrium. (8)
Hint
For the last part set up a mole (RICE) table with change −x, −x, +2x; because the initial amounts are equal, Kc gives a perfect square, so take the square root of both sides.
Worked answer
a) When the equilibrium in a closed system is disturbed, the system will re-instate a new equilibrium by favouring the reaction that opposes (cancels) the disturbance. ✓✓
b) INCREASES ✓
c) The concentrated HCl adds Cl⁻ ions (a common ion), so [Cl⁻] increases ✓. By Le Chatelier's principle the system opposes this increase by favouring the REVERSE reaction, so more solid NaCl forms and its mass increases ✓.
d) EXOTHERMIC ✓
e) The table shows that Kc DECREASES as the temperature increases ✓. By Le Chatelier's principle an increase in temperature favours the endothermic reaction ✓. Since Kc (ratio of products to reactants) decreases, the REVERSE reaction is favoured, so the reverse reaction is endothermic and the forward reaction is exothermic ✓.
f) RICE table (mol):
| H₂ | I₂ | HI | |
|---|---|---|---|
| Initial | 2,0 | 2,0 | 0 |
| Change | −x | −x | +2x |
| Equilibrium | 2,0 − x | 2,0 − x | 2x |
Divide by V = 2 dm³ for concentrations ✓✓
✓✓
mol ✓
mol ✓
✓ g ✓
Modelled on Nov 2025 Paper 2, Question 6 — same skills, new scenario.
Question 5
The reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (ΔH < 0) is at equilibrium in a closed container. The graph shows the forward and reverse rates against time. At t₁ both rates jump instantly to the same higher value and remain equal afterwards. What change was made at t₁?
This question has a diagram, shown in the app.
Hint
Which single change increases BOTH rates equally without shifting the equilibrium?
Worked answer
A. A catalyst speeds up the forward and reverse reactions EQUALLY, so both rates jump to the same higher value and the equilibrium position is unchanged. B: halving the volume would raise the forward rate (4 mol gas) more than the reverse (2 mol gas). C: cooling would DECREASE both rates. D: extra N₂ would spike only the forward rate at first.
Modelled on Nov 2025 Paper 2, Question 1.5 — same skills, new scenario.