AS Chemistry: Chemical Equilibria — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Chemical Equilibria revision notes

---

Section A

1. State the two features of a system at dynamic equilibrium. [2]

2. Explain why a catalyst does not change the position of equilibrium. [2]

3. State the only factor that changes the value of Kc, and explain why. [2]

---

Section B

4. For the Contact process:

2SO2(g) + O2(g)  <=>  2SO3(g)     delta-H = -196 kJ mol-1

(a) Write the expression for Kc and state its units. [2]

(b) State and explain the effect on the yield of SO₃ of (i) increasing temperature, (ii) increasing pressure. [4]

(c) In industry the process is run at about 450 °C rather than a lower temperature. Explain this choice. [3]

(d) The process operates at only 1–2 atm despite the pressure argument in (b)(ii). Suggest why. [2]

5. 2.00 mol of ethanoic acid and 3.00 mol of ethanol were mixed and allowed to reach equilibrium in a 1.00 dm³ vessel. At equilibrium, 1.60 mol of ester had formed.

CH3COOH + C2H5OH  <=>  CH3COOC2H5 + H2O

(a) Construct an ICE table and determine the equilibrium amount of each species. [3]

(b) Calculate Kc, and explain why it has no units. [3]

6. A sealed vessel at total pressure 200 kPa contains an equilibrium mixture of N₂O₄(g) ⇌ 2NO₂(g) with mole fractions 0.60 for N₂O₄ and 0.40 for NO₂.

(a) Calculate the partial pressure of each gas. [2] (b) Write the expression for Kp and calculate its value, including units. [3]

7. State which of the following change the value of K, and which only shift the position of equilibrium: (i) increasing pressure, (ii) adding a catalyst, (iii) increasing temperature. [3]

8. Explain why “no effect on K” and “no effect on yield” are not the same statement, using a pressure change as your example. [2]

---

Answers

1. The rates of the forward and reverse reactions are equal [1]; the concentrations of all species remain constant [1]. “Constant” is not “equal” — a distinction examined directly.

2. It increases the rate of the forward and reverse reactions equally [1], so equilibrium is reached sooner but at the same position [1].

3. Temperature [1], because it alters the rate constants of the forward and reverse reactions unequally [1].

4. (a) Kc = [SO₃]² ÷ ([SO₂]²[O₂]) [1]; units dm³ mol⁻¹ [1].

(b) (i) Yield decreases [1] — the forward reaction is exothermic, so raising the temperature shifts equilibrium in the endothermic (reverse) direction [1]. (ii) Yield increases [1] — there are 3 moles of gas on the left and 2 on the right, so higher pressure shifts equilibrium towards the side with fewer gas moles [1].

(c) A lower temperature would give a higher yield [1] but an unacceptably slow rate [1], so 450 °C is a compromise between yield and rate [1].

(d) The equilibrium yield at 1–2 atm already strongly favours SO₃ under typical operating conditions [1], so the modest further conversion available from raising the pressure does not justify the large cost of pressurised plant and energy [1]. (Figures like “96%” appear in some sources, but the exact percentage depends on temperature, pressure and feed composition — the economic argument only needs “already strongly favours SO₃”, not a specific number.)

5. (a)

	CH₃COOH	C₂H₅OH	ester	H₂O
Initial	2.00	3.00	0	0
Change	−1.60	−1.60	+1.60	+1.60
Equilibrium	0.40	1.40	1.60	1.60

[1 for changes in 1:1:1:1 ratio; 1 for acid and alcohol; 1 for products]

(b) Kc = (1.60 × 1.60) ÷ (0.40 × 1.40) [1] = 2.56 ÷ 0.56 = 4.57 [1]. No units because there are two moles of gas/species on each side, so the concentration units cancel [1].

6. (a) P(N₂O₄) = 0.60 × 200 = 120 kPa [1]; P(NO₂) = 0.40 × 200 = 80 kPa [1]. (b) Kp = (P_NO₂)² ÷ P_N₂O₄ [1] = 80² ÷ 120 = 6400 ÷ 120 = 53.3 kPa [1]. The mole count changes (1 → 2), so the units don’t cancel — Kp carries units of kPa¹, matching the net order of (2 − 1) = 1 [1].

7. (i) Increasing pressure — shifts position only, K unchanged [1]. (ii) Adding a catalyst — shifts nothing (equilibrium reached sooner, same position), K unchanged [1]. (iii) Increasing temperature — changes the value of K itself [1].

8. A pressure change does shift the position of equilibrium, and therefore changes the yield, by favouring the side with fewer gas moles [1]; but the numerical value of K stays the same, since only a temperature change alters K — the two statements describe different quantities [1].

---

Where marks are usually lost

• Saying concentrations become equal rather than constant.
• Claiming a catalyst increases yield.
• Giving only “compromise” without saying compromise between what and what.
• Forgetting to divide moles by volume before substituting into Kc (here volume is 1.00 dm³, so it happens to make no difference — but state it).
• Assuming Kc always has units.
• Believing pressure or concentration changes alter the value of K — they only shift the position of equilibrium; only temperature changes K.
• Forgetting Kp uses partial pressures, not concentrations, and that units depend on the net change in gas moles.