These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Every K value and every substance labelled with a letter is invented for practice. K values are written without units.
Question 1 (multiple choice · foundation)
Four reactions, each of the type X ⇌ Y, reach equilibrium at the same temperature. Which equilibrium mixture contains the smallest proportion of reactant?
- (A) P ⇌ Q, K = 3 × 10⁻⁸
- (B) R ⇌ S, K = 0.9
- (C) T ⇌ U, K = 2 × 10²
- (D) V ⇌ W, K = 5 × 10¹⁴
Answer and explanation
Answer: (D). For X ⇌ Y, K = [Y]/[X]. The larger K is, the smaller [X] is compared with [Y]. At K = 5 × 10¹⁴ there are 5 × 10¹⁴ product particles for every reactant particle, so the reaction is essentially complete.
- (A) has the smallest number, but a small K means mostly reactant: only about 3 × 10⁻⁶ % is product.
- (B) is close to 1, so the mixture holds roughly equal amounts of each (about 47 % product).
- (C) is product-favoured (about 99.5 % product), but less so than (D).
Question 2 (multiple choice · core)
For the reaction A(aq) ⇌ B(aq), K = 2.5 × 10⁻⁴ at 25 °C. Which statement best describes the equilibrium mixture at 25 °C?
- (A) It contains no B, because K is less than 1.
- (B) It contains mostly A; [A] is 4000 times [B].
- (C) It contains mostly A, because the forward reaction is very slow.
- (D) It contains equal concentrations of A and B, because the system is at equilibrium.
Answer and explanation
Answer: (B). K = [B]/[A] = 2.5 × 10⁻⁴, so [A]/[B] = 1 ÷ (2.5 × 10⁻⁴) = 4000. The mixture is mostly A.
- (A) If there were no B, [B]/[A] would be 0, not 2.5 × 10⁻⁴. A small K means a little B, not none.
- (C) K says nothing about rate. At equilibrium the forward and reverse rates are equal, whatever the size of K.
- (D) confuses equal rates with equal concentrations. Equilibrium means the concentrations stop changing, not that they are the same.
Question 3 (multiple choice · core)
For the gas-phase reaction X(g) ⇌ Y(g), K = 4.0 at a certain temperature. A closed container holds 50 particles in total (X and Y) once equilibrium is reached. How many of them are Y particles?
- (A) 4
- (B) 10
- (C) 25
- (D) 40
Answer and explanation
Answer: (D). Both gases are in the same volume, so K = [Y]/[X] = (number of Y)/(number of X) = 4.0. Split 50 in the ratio 4 : 1: Y = 50 × 4/5 = 40 and X = 10. Check: 40 ÷ 10 = 4.0 ✓.
- (A) uses the value of K as a particle count.
- (B) is the number of X particles: the ratio is the wrong way round.
- (C) assumes equilibrium means equal amounts.
Question 4 (calculation · core)
A compound A slowly converts into its isomer B in solution: A(aq) ⇌ B(aq), K = 8.0 × 10⁻³ at 30 °C. A solution starts with [A] = 0.500 M and no B.
Calculate [B] at equilibrium and the percentage of A that has been converted. State whether the reaction is reactant-favoured or product-favoured.
Worked solution
- K = [B]/[A] = 8.0 × 10⁻³, and because the reaction is 1 : 1, [A] + [B] = 0.500 M.
- [B] = 8.0 × 10⁻³ × [A], so [A](1 + 8.0 × 10⁻³) = 0.500 M, giving [A] = 0.496 M.
- [B] = 0.500 M − 0.496 M = 3.97 × 10⁻³ M (from the unrounded value 3.968 × 10⁻³ M).
- Percentage converted = 3.968 × 10⁻³ ÷ 0.500 × 100 = 0.794 %.
- Less than 1 % of A has reacted, so the reaction is reactant-favoured.
Suggested mark points (3): 1 for using [B]/[A] = K together with [A] + [B] = 0.500 M; 1 for [B] = 3.97 × 10⁻³ M; 1 for the percentage and “reactant-favoured”.
Alternative method: because K is small, [A] stays close to 0.500 M, so [B] ≈ 8.0 × 10⁻³ × 0.500 = 4.0 × 10⁻³ M (0.80 %). This approximation is within 1 % of the exact answer and is accepted if the approximation is stated.
Question 5 (constructed response · core)
Gases X and Y combine: X(g) + Y(g) ⇌ XY(g), Kc = 5.0 × 10⁸ at 300 K. A flask at equilibrium at 300 K contains [XY] = 0.30 M, and [X] = [Y].
(a) Calculate [X] at equilibrium. (b) A student claims: “Because K is so large, the flask contains only XY and no X or Y.” Evaluate the claim, using your answer to (a).
Worked solution and suggested Marlbridge rubric
(a) Kc = [XY] / ([X][Y]). With [X] = [Y] = x: 5.0 × 10⁸ = 0.30 / x², so x² = 0.30 ÷ 5.0 × 10⁸ = 6.0 × 10⁻¹⁰ and x = 2.4 × 10⁻⁵ M.
(b) The claim is partly right but wrongly worded. The large K does mean the mixture is mostly product: [XY] is about 12 000 times larger than [X] (0.30 ÷ 2.45 × 10⁻⁵). The reaction is essentially complete. But [X] and [Y] are not zero; they are 2.4 × 10⁻⁵ M. If they were zero, the expression would have zero on the bottom and could not equal 5.0 × 10⁸. The system is dynamic, so XY keeps breaking down into X and Y while X and Y keep combining.
| Point | What earns it |
|---|---|
| 1 | Correct Kc expression with [X] = [Y] substituted |
| 1 | [X] = 2.4 × 10⁻⁵ M |
| 1 | Agrees the mixture is mostly XY because K is very large (product-favoured) |
| 1 | Rejects “no X or Y”, using the non-zero value from (a) or the dynamic nature of equilibrium |
Accept an answer to (a) carried forward into (b).
Question 6 (explanation · core)
For an invented gas-phase reaction 2 J(g) + L(g) ⇌ 2 JL(g), Kp = 3 × 10³⁸ at 25 °C. A sealed flask containing only J and L is kept at 25 °C. After six months, no JL can be detected.
(a) A student says, “The value of K must be wrong.” Explain why the observation is consistent with the value of K. (b) A catalyst is then added and JL starts to form. State and explain the effect of the catalyst on Kp and on the composition of the mixture once equilibrium is reached. (c) Describe the equilibrium mixture that finally forms.
Model answer and suggested Marlbridge rubric
(a) K describes the mixture at equilibrium, not how quickly equilibrium is reached. The flask has not reached equilibrium: the reaction is extremely slow at 25 °C, probably because its activation energy is high, so very few collisions lead to reaction. A huge K and a very slow rate can both be true.
(b) A catalyst lowers the activation energy (for the forward and reverse reactions equally), so equilibrium is reached much faster. It does not change Kp, and so it does not change the composition of the mixture at equilibrium at 25 °C.
(c) Kp is very large, so at equilibrium the flask contains mostly JL, with only traces of J and L: the reaction is essentially complete.
| Point | What earns it |
|---|---|
| 1 | K describes the equilibrium position, not the rate; the system has not yet reached equilibrium |
| 1 | Slow reaction explained (for example high activation energy) |
| 1 | Catalyst: faster approach to equilibrium, Kp and equilibrium composition unchanged |
| 1 | Mixture is mostly JL with traces of J and L, linked to the very large Kp |
Question 7 (evaluation · stretch)
Two invented reactions are studied at the same temperature:
- Reaction 1: J(g) ⇌ L(g), Kc = 0.50
- Reaction 2: M(g) ⇌ 2 N(g), Kc = 0.50
A student claims: “The two reactions have the same K, so every equilibrium mixture of either reaction has the same ratio of product to reactant.”
(a) Show that both of these mixtures for Reaction 2 are at equilibrium: Mixture P, [M] = 0.080 M and [N] = 0.20 M; Mixture Q, [M] = 2.0 M and [N] = 1.0 M. (b) Use your answer to (a) to evaluate the student’s claim. (c) Explain why the ratio of product to reactant is fixed for Reaction 1 but not for Reaction 2.
Worked solution and suggested Marlbridge rubric
(a) For Reaction 2, Kc = [N]² / [M]. Mixture P: (0.20)² ÷ 0.080 = 0.040 ÷ 0.080 = 0.50 ✓. Mixture Q: (1.0)² ÷ 2.0 = 0.50 ✓. Both equal Kc, so both are at equilibrium.
(b) The claim is incorrect. For Reaction 2 the ratio [N]/[M] is 0.20 ÷ 0.080 = 2.5 in Mixture P but 1.0 ÷ 2.0 = 0.50 in Mixture Q, even though both are at equilibrium with the same Kc. For Reaction 1 the ratio [L]/[J] is always 0.50. So equal K values do not guarantee equal ratios.
(c) For Reaction 1 the expression is [L]/[J], with the same power top and bottom, so K is the ratio. For Reaction 2 the product concentration is squared, so the ratio [N]/[M] equals Kc ÷ [N], which changes when the mixture is more or less concentrated. K values can only be compared directly as ratios when the expressions have the same shape.
| Point | What earns it |
|---|---|
| 1 | Both mixtures shown to give Kc = 0.50, with the squared term |
| 1 | Claim rejected using the two different ratios (2.5 and 0.50) |
| 1 | Explains that the squared term (different powers) makes the ratio depend on concentration |
How did you do?
- Q1 or Q2 wrong: re-read “What the size of K tells you” in the study guide.
- Q3 or Q4 wrong: work through Figure 1 and Worked example 1 again.
- Q5 or Q6 incomplete: make sure you separate “mostly products” from “only products”, and position from rate.
- Q7 incomplete: see “Comparing K values fairly” in the study guide.
Then tick off the topic checklist.