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. Data: ΔG°f at 298 K: CO₂(g) −394.39, H₂O(l) −237.14, H₂O(g) −228.58, C₂H₅OH(l) −174.8, NO(g) +86.60 kJ mol⁻¹; elements in their standard states 0. Reaction J → L and compound W use invented data, made up for practice.
Question 1 (multiple choice · foundation)
Which observation gives the best evidence that a process is under kinetic control?
- (A) A reaction with ΔG° > 0 produces no detectable product, even with a catalyst present.
- (B) A mixture with ΔG° < 0 shows no change for months, then reacts quickly once a small amount of catalyst is added.
- (C) A reaction with ΔG° < 0 is complete within a few seconds at room temperature.
- (D) In a sealed flask, the concentrations of reactants and products stay constant while both forward and reverse reactions continue.
Answer and explanation
Answer: (B). The process is favoured (ΔG° < 0) but was not happening at a measurable rate. The catalyst lowered the activation energy and the reaction then went, which shows the obstacle was kinetic.
- (A) describes a process that is not favoured. A catalyst cannot make it give much product.
- (C) is favoured and fast, so the rate is not limiting anything.
- (D) describes dynamic equilibrium (Q = K, equal rates), not kinetic control.
Question 2 (multiple choice · foundation)
A catalyst is added to a reaction that is under kinetic control. Which row correctly describes the effect on the activation energy, Eₐ, and on ΔG°?
- (A) Eₐ decreases; ΔG° does not change.
- (B) Eₐ decreases; ΔG° becomes more negative.
- (C) Eₐ does not change; ΔG° becomes more negative.
- (D) Eₐ increases; ΔG° does not change.
Answer and explanation
Answer: (A). A catalyst provides a new pathway with a lower barrier. ΔG° depends only on the free energies of the reactants and products, which the catalyst does not change.
- (B) and (C) wrongly claim the catalyst changes the thermodynamics.
- (D) has Eₐ moving the wrong way; a higher barrier would slow the reaction further.
Question 3 (multiple choice · core)
An open dish of ethanol stands in air at 25 °C. Using the data at the top of the page, which statement about C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l) is correct?
- (A) ΔG° = −1325.4 kJ mol⁻¹; the reaction is favoured, and the ethanol does not burn because the activation energy is high.
- (B) ΔG° = +1325.4 kJ mol⁻¹; the ethanol does not burn because the reaction is not favoured.
- (C) ΔG° = −456.7 kJ mol⁻¹; the reaction is favoured and fast, so the ethanol is burning slowly without a flame.
- (D) ΔG° = −1325.4 kJ mol⁻¹; the ethanol and oxygen are already at equilibrium, so no reaction is seen.
Answer and explanation
Answer: (A). ΔG° = 2(−394.39) + 3(−237.14) − (−174.8) − 3(0) = −788.78 − 711.42 + 174.8 = −1325.4 kJ mol⁻¹. The reaction is strongly favoured, but it does not happen at room temperature without a flame, so it is under kinetic control.
- (B) subtracts the wrong way round (reactants minus products).
- (C) ignores the coefficients: −394.39 − 237.14 + 174.8 = −456.7 kJ mol⁻¹. It also invents a slow, flameless burning that is not observed.
- (D) has the right ΔG° but the wrong conclusion. With ΔG° so negative, K is enormous. At equilibrium almost all the ethanol would have become CO₂ and H₂O, but the ethanol is still there, so Q ≪ K and the system is far from equilibrium.
Question 4 (multiple choice · core)
A sealed steel cylinder holds a mixture of H₂(g) and O₂(g) at 298 K. After a year, no water can be detected. For 2H₂(g) + O₂(g) → 2H₂O(g), ΔG° = −457.2 kJ mol⁻¹. Which statement is correct?
- (A) The mixture is at equilibrium, because the amounts are not changing.
- (B) K for the reaction must be very small, because so little water forms.
- (C) Q is much smaller than K, and the rate of reaction is too low to measure.
- (D) ΔG° becomes positive inside a sealed cylinder, so no reaction occurs.
Answer and explanation
Answer: (C). A large negative ΔG° means K is very large. With essentially no water, Q is close to zero, so Q ≪ K and the system is far from equilibrium. A high activation energy keeps the rate negligible: the mixture is under kinetic control.
- (A) equates “no change” with equilibrium; at equilibrium Q = K, which is not the case here.
- (B) contradicts the data: a very negative ΔG° means a very large K.
- (D) is false; a sealed container does not change the sign of ΔG°.
Question 5 (calculation · core)
For an invented reaction J → L, ΔH° = −110 kJ mol⁻¹ and ΔG° = −95 kJ mol⁻¹ at 298 K. Without a catalyst, the forward activation energy is 160 kJ mol⁻¹. With a catalyst it is 62 kJ mol⁻¹. At 298 K, no L forms without the catalyst.
(a) State whether J → L is thermodynamically favoured, with a reason. (b) Calculate the activation energy of the reverse reaction, L → J, with and without the catalyst. (Use enthalpy values, as on an enthalpy profile.) (c) Explain why the uncatalysed reaction is under kinetic control.
Worked solution
(a) ΔG° = −95 kJ mol⁻¹ < 0, so J → L is favoured.
(b) On an enthalpy profile, the products sit 110 kJ mol⁻¹ below the reactants, so the reverse barrier is the forward barrier plus 110 kJ mol⁻¹. Uncatalysed: Eₐ(reverse) = 160 + 110 = 270 kJ mol⁻¹. Catalysed: Eₐ(reverse) = 62 + 110 = 172 kJ mol⁻¹. The catalyst lowers both barriers by the same amount, 98 kJ mol⁻¹.
(c) The reaction is favoured but no L forms, so the obstacle is kinetic. With Eₐ = 160 kJ mol⁻¹, almost no J particles have enough energy at 298 K to reach the transition state, so the rate is negligible.
Suggested mark points (4): 1 for “favoured because ΔG° < 0”; 1 for 270 kJ mol⁻¹; 1 for 172 kJ mol⁻¹; 1 for linking the high Eₐ to too few successful collisions while ΔG° < 0.
Question 6 (constructed response · core)
Compound W is a white solid that can decompose: W(s) → R(s) + Y(g). For this reaction ΔH° = −35.0 kJ mol⁻¹ and ΔS° = +120 J K⁻¹ mol⁻¹. A sample of W kept in a closed jar at 25 °C for a year shows no change.
(a) Calculate ΔG° at 298 K. (b) Explain why the decomposition is favoured at every temperature, without further calculation. (c) A student claims: “Because nothing has changed in a year, the jar must be at equilibrium.” Evaluate the claim. (d) Describe an experiment that would support your answer to (c), and state the expected result.
Worked solution and suggested Marlbridge rubric
(a) ΔG° = −35.0 − 298(0.120) = −35.0 − 35.76 = −70.8 kJ mol⁻¹.
(b) ΔH° < 0 and ΔS° > 0. Both ΔH° and −TΔS° are negative, so ΔG° < 0 at every temperature.
(c) The claim is incorrect. ΔG° is negative, so K is large and an equilibrium mixture would contain mainly R and Y. The jar still contains W, so Q ≪ K and the system is far from equilibrium. The decomposition must have a high activation energy, so it is under kinetic control.
(d) Add a suitable catalyst to a sample of W (or heat a small sample strongly). Expected result: gas Y is given off and R forms. If the system were already at equilibrium, a catalyst would not change the amounts.
| Point | What earns it |
|---|---|
| 1 | −70.8 kJ mol⁻¹ with ΔS° converted to kJ |
| 1 | Both terms negative (ΔH° < 0 and −TΔS° < 0), so ΔG° < 0 at all T |
| 1 | Claim rejected: favoured process with no change means Q ≪ K, not equilibrium |
| 1 | High activation energy named as the reason (kinetic control) |
| 1 | Valid test (catalyst or strong heating) with the expected observation |
Accept any valid test that would start a kinetically hindered reaction.
Question 7 (explanation · stretch)
Air is mostly N₂ and O₂. At room temperature, no measurable NO forms from them. A mixture of H₂ and O₂ also shows no change at room temperature. A student says: “Both of these are examples of kinetic control.”
Use ΔG° values calculated from the data at the top of the page to evaluate the student’s statement for each mixture.
Model answer and suggested Marlbridge rubric
Calculations. N₂(g) + O₂(g) → 2NO(g): ΔG° = 2(+86.60) − 0 = +173.2 kJ mol⁻¹. 2H₂(g) + O₂(g) → 2H₂O(g): ΔG° = 2(−228.58) − 0 = −457.2 kJ mol⁻¹.
Evaluation. The statement is only half right.
- For H₂ and O₂, ΔG° < 0: water formation is favoured, yet it does not happen at a measurable rate. That is kinetic control, caused by a high activation energy. A spark or catalyst would start it.
- For N₂ and O₂, ΔG° > 0: forming NO is not favoured at room temperature. The absence of NO is a thermodynamic result, not a kinetic one. Kinetic control describes only favoured processes, so this mixture is not an example, and a catalyst would not produce a large amount of NO.
| Point | What earns it |
|---|---|
| 1 | Both ΔG° values correct, with coefficients and elements at zero |
| 1 | H₂/O₂ identified as kinetic control: favoured but slow because of a high Eₐ |
| 1 | N₂/O₂ identified as not favoured (ΔG° > 0), so not kinetic control |
| 1 | States that kinetic control applies only when ΔG° < 0 |
Do not award the third point for “no NO forms because it is slow”.
How did you do?
- Q1, Q2 or Q5 wrong: revisit Figure 1 in the study guide: a catalyst changes the barrier, never the start or end levels.
- Q4 or Q6(c) wrong: compare Q with K before you say “equilibrium”.
- Q3 or Q7 wrong: separate the two questions. First the sign of ΔG° (favoured or not), then the rate (fast or under kinetic control).
Then tick off the topic checklist.