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. All substances named with letters (P, Q, R, S, T, U, V) are fictional, and all rate constants and data are invented. Every system is closed and at constant temperature. Where a question says a reaction is elementary, you may write each direction’s rate law from the equation.
Question 1 (multiple choice · foundation)
At one moment in a reversible reaction, the forward rate is 4.5 × 10⁻³ M s⁻¹ and the reverse rate is 1.5 × 10⁻³ M s⁻¹. What happens next?
- (A) The concentrations of the products increase, and the system moves towards equilibrium.
- (B) The concentrations of the reactants increase, because the reverse reaction is slower.
- (C) Nothing changes, because both reactions are happening.
- (D) The forward rate keeps increasing until all the reactants are used up.
Answer and explanation
Answer: (A). Forward > reverse, so there is a net conversion of reactants into products, at 4.5 × 10⁻³ − 1.5 × 10⁻³ = 3.0 × 10⁻³ M s⁻¹. As this happens the forward rate falls and the reverse rate rises, until they are equal.
- (B) gets the direction backwards. A slower reverse reaction means products are made faster than they are used.
- (C) Both reactions are happening, but at different rates, so there is a net change. No net change needs equal rates.
- (D) As reactants are used up, the forward rate decreases. The reaction reaches equilibrium; it does not run to completion.
Question 2 (multiple choice · core)
A flask is filled with only the products of a reversible reaction. Which statement describes the rates at the moment the flask is filled?
- (A) The forward rate is at its maximum and the reverse rate is zero.
- (B) The forward rate is zero and the reverse rate is at its maximum.
- (C) The forward and reverse rates are equal, because the reaction has not started.
- (D) Both rates are zero until some reactant has formed.
Answer and explanation
Answer: (B). With no reactants present, nothing can react in the forward direction, so the forward rate is zero. The products are at their highest concentration, so the reverse rate is at its largest. There is a net change from products to reactants.
- (A) describes a flask that starts with only reactants.
- (C) Equal rates would mean equilibrium. Here one rate is zero and the other is not.
- (D) The reverse reaction needs only products, which are present, so it starts at once.
Question 3 (multiple choice · core)
The elementary reversible reaction 2R(g) ⇌ R₂(g) has k_f = 0.50 M⁻¹ s⁻¹ (forward rate = k_f[R]²) and k_r = 0.020 s⁻¹ (reverse rate = k_r[R₂]). A mixture contains [R] = 0.10 M and [R₂] = 0.40 M. Which statement is correct?
- (A) Forward rate 0.050 M s⁻¹, reverse rate 0.0080 M s⁻¹: net formation of R₂.
- (B) Forward rate 0.0050 M s⁻¹, reverse rate 0.0080 M s⁻¹: net formation of R.
- (C) Forward rate 0.0050 M s⁻¹, reverse rate 0.0080 M s⁻¹: net formation of R₂.
- (D) The direction cannot be predicted because k_f is larger than k_r.
Answer and explanation
Answer: (B). Forward rate = 0.50 × (0.10)² = 0.50 × 0.010 = 0.0050 M s⁻¹. Reverse rate = 0.020 × 0.40 = 0.0080 M s⁻¹. Reverse > forward, so R₂ breaks up into R: [R] increases and [R₂] decreases.
- (A) forgets to square [R]. The coefficient 2 in an elementary step gives the power 2.
- (C) has the right rates but the wrong direction. The larger rate (reverse) wins.
- (D) The rate constants have different units here, so they cannot be compared directly. And even with the same units, the direction depends on the rates, which include concentrations.
Question 4 (calculation · core)
The isomerisation U(aq) ⇌ V(aq) is elementary in both directions, with k_f = 2.4 × 10⁻³ s⁻¹ and k_r = 8.0 × 10⁻⁴ s⁻¹. A solution is made with [U] = 0.050 M and [V] = 0.250 M.
(a) Calculate the forward and reverse rates in this solution. (b) State the direction of net change and calculate the net rate. (c) Calculate [U] and [V] when the solution reaches equilibrium.
Worked solution
(a) Forward rate = 2.4 × 10⁻³ s⁻¹ × 0.050 M = 1.2 × 10⁻⁴ M s⁻¹. Reverse rate = 8.0 × 10⁻⁴ s⁻¹ × 0.250 M = 2.0 × 10⁻⁴ M s⁻¹.
(b) Reverse > forward, so the net change is to the left: V → U. Net rate = 2.0 × 10⁻⁴ − 1.2 × 10⁻⁴ = 8.0 × 10⁻⁵ M s⁻¹ (V used up, U made).
(c)
- Each V that changes makes one U, so [U] + [V] stays at 0.300 M.
- At equilibrium the rates are equal: 2.4 × 10⁻³[U] = 8.0 × 10⁻⁴[V], so [V] = 3[U].
- [U] + 3[U] = 0.300 M, so [U] = 0.075 M and [V] = 0.225 M.
- Check: forward rate = 2.4 × 10⁻³ × 0.075 = 1.8 × 10⁻⁴ M s⁻¹; reverse rate = 8.0 × 10⁻⁴ × 0.225 = 1.8 × 10⁻⁴ M s⁻¹. Equal.
Suggested mark points (5): 1 for each rate in (a); 1 for the direction with reference to the larger rate; 1 for using equal rates at equilibrium; 1 for both equilibrium concentrations. [U] rising from 0.050 M to 0.075 M is consistent with the net change to the left found in (b).
Question 5 (constructed response · core)
The fictional molecules P and Q are isomers: P(g) ⇌ Q(g). A computer model shows a small sealed box at four times:
| Snapshot | Time (min) | Number of P | Number of Q |
|---|---|---|---|
| 1 | 0 | 4 | 16 |
| 2 | 5 | 7 | 13 |
| 3 | 10 | 8 | 12 |
| 4 | 15 | 8 | 12 |
(a) State the direction of net change between snapshots 1 and 2, and compare the forward and reverse rates in that interval. (b) A student says: “Snapshot 1 shows that Q is the more stable isomer, so the reaction must go forward.” Use the data to evaluate this. (c) Is the system at equilibrium at 10 min? Justify your answer. (d) Describe one limitation of this model for showing what happens at equilibrium.
Worked solution and suggested Marlbridge rubric
(a) P rises from 4 to 7 while Q falls from 16 to 13, so the net change is Q → P (to the left). In that interval the reverse rate (Q → P) is greater than the forward rate (P → Q).
(b) The claim is not supported. The data show a net change from Q to P, the reverse direction. Snapshot 1 started with more Q than the equilibrium amount, so the reverse rate was larger. The direction depends on which rate is larger at the time, not on which species is present in greater amount. The starting counts were simply chosen when the model was set up, so they say nothing about which isomer is more stable.
(c) Probably yes. The counts are the same at 10 min and 15 min (8 P, 12 Q), so there is no net change, which means the forward and reverse rates are equal. (The model cannot show whether equilibrium was reached exactly at 10 min or a little before.)
(d) Any one: the snapshots do not show that P and Q are still converting both ways at equilibrium; with only 20 molecules, random fluctuations could change the counts even at equilibrium; snapshots show amounts but not rates, so the rates can only be inferred from changes between snapshots.
| Point | What earns it |
|---|---|
| 1 | Net change Q → P, with the reverse rate greater than the forward rate |
| 1 | Refutes the claim using the observed direction (Q decreasing) |
| 1 | Constant counts at 10 and 15 min, so no net change and equal rates |
| 1 | A valid limitation of the snapshot model |
Question 6 (constructed response · core)
For the fictional reaction S(aq) ⇌ T(aq), a student measures the forward and reverse rates at several times. At the start, both S and T are present.
| Time (s) | 0 | 20 | 40 | 60 | 100 | 150 |
|---|---|---|---|---|---|---|
| Forward rate (10⁻³ M s⁻¹) | 2.00 | 3.52 | 4.08 | 4.28 | 4.38 | 4.40 |
| Reverse rate (10⁻³ M s⁻¹) | 5.00 | 4.62 | 4.48 | 4.43 | 4.40 | 4.40 |
(a) State the direction of net change at 20 s, with a reason. (b) Explain why the forward rate increases and the reverse rate decreases between 0 and 100 s. (c) At what time in the table is the system first at equilibrium? Justify your answer. (d) Calculate the net rate at 0 s, and say what happens to [S] during the first 20 s.
Worked solution and suggested Marlbridge rubric
(a) At 20 s the reverse rate (4.62 × 10⁻³) is greater than the forward rate (3.52 × 10⁻³), so the net change is to the left, T → S.
(b) The net change is T → S, so [S] increases and [T] decreases. The forward rate depends on [S], so it rises; the reverse rate depends on [T], so it falls.
(c) At 150 s. It is the first time in the table at which the two rates are equal (4.40 × 10⁻³ M s⁻¹ each). At 100 s the rates still differ slightly (4.38 vs 4.40), so there is still a small net change. Equilibrium was reached somewhere between 100 s and 150 s.
(d) Net rate = forward − reverse = 2.00 × 10⁻³ − 5.00 × 10⁻³ = −3.00 × 10⁻³ M s⁻¹. The negative sign means net change to the left: [S] increases in the first 20 s.
| Point | What earns it |
|---|---|
| 1 | Direction at 20 s, with the comparison of rates |
| 1 | Links rising forward rate to rising [S] and falling reverse rate to falling [T] |
| 1 | 150 s, with the reason that the rates have become equal (not yet equal at 100 s) |
| 1 | −3.00 × 10⁻³ M s⁻¹ (or “3.00 × 10⁻³ M s⁻¹ to the left”) and [S] increasing |
Question 7 (explanation · stretch)
The solution from Question 4 has reached equilibrium: [U] = 0.075 M and [V] = 0.225 M. A chemist quickly adds some V, without changing the volume noticeably, so that [V] becomes 0.450 M. The rate constants are unchanged.
(a) Calculate the forward and reverse rates immediately after the addition. (b) Explain, in terms of rates, what happens next and why the system reaches a new equilibrium. (c) Calculate [U] and [V] at the new equilibrium.
Model answer and suggested Marlbridge rubric
(a) Forward rate = 2.4 × 10⁻³ × 0.075 = 1.8 × 10⁻⁴ M s⁻¹ (unchanged, because [U] has not changed). Reverse rate = 8.0 × 10⁻⁴ × 0.450 = 3.6 × 10⁻⁴ M s⁻¹ (doubled, because [V] doubled).
(b) Reverse > forward, so there is a net change V → U. As V is used up the reverse rate falls, and as U is made the forward rate rises. When the two rates are equal again, there is no further net change: a new equilibrium.
(c) Total [U] + [V] = 0.075 + 0.450 = 0.525 M. At equilibrium [V] = 3[U] (as in Question 4), so 4[U] = 0.525 M: [U] = 0.131 M and [V] = 0.394 M (0.13125 M and 0.39375 M before rounding). Both rates are then 3.15 × 10⁻⁴ M s⁻¹.
| Point | What earns it |
|---|---|
| 1 | Both rates in (a), with the forward rate unchanged |
| 1 | Net change V → U because the reverse rate is now larger |
| 1 | Describes the rates converging (reverse falls, forward rises) until equal |
| 1 | Correct new equilibrium concentrations |
This question previews Topics 7.9 and 7.10, where you will meet this kind of disturbance as Le Châtelier’s principle.
How did you do?
- Q1 or Q2 wrong: re-read “Two rates, one net direction” and Figure 1 in the study guide.
- Q3 or Q4 wrong: check the rate laws for each direction in “Calculating the two rates for an elementary step”, and Worked example 1.
- Q5 or Q6 incomplete: your answer needs a direction and a comparison of rates; see Worked example 2.
- Q7: if you got the direction but not the numbers, practise the “equal rates at equilibrium” step from Question 4.
Then tick off the topic checklist.