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 energy values are invented for practice, in kJ mol⁻¹. Every reaction in this set is a single elementary step.
Question 1 (multiple choice · foundation)
On the energy profile of an elementary step, the reactants are at 30 kJ mol⁻¹, the transition state is at 110 kJ mol⁻¹ and the products are at 75 kJ mol⁻¹. What is the forward activation energy?
- (A) 35 kJ mol⁻¹
- (B) 45 kJ mol⁻¹
- (C) 80 kJ mol⁻¹
- (D) 110 kJ mol⁻¹
Answer and explanation
Answer: (C). Eₐ(forward) is the climb from the reactants to the transition state: 110 − 30 = 80 kJ mol⁻¹.
- (A) is the reverse activation energy, 110 − 75.
- (B) is the overall energy change, 75 − 30 = +45 kJ mol⁻¹.
- (D) is the level of the transition state on the axis, not a difference between two levels.
Question 2 (multiple choice · core)
An elementary step has a forward activation energy of 64 kJ mol⁻¹ and a reverse activation energy of 109 kJ mol⁻¹. Which describes the forward step?
- (A) ΔH = −45 kJ mol⁻¹; exothermic
- (B) ΔH = +45 kJ mol⁻¹; endothermic
- (C) ΔH = −173 kJ mol⁻¹; exothermic
- (D) ΔH = +173 kJ mol⁻¹; endothermic
Answer and explanation
Answer: (A). ΔH = Eₐ(forward) − Eₐ(reverse) = 64 − 109 = −45 kJ mol⁻¹. The reverse climb is bigger, so the products lie below the reactants: exothermic.
- (B) subtracts the wrong way round.
- (C) and (D) add the two activation energies. Both climbs end at the same peak, so they must be subtracted.
Question 3 (multiple choice · core)
Which statement about the transition state of an elementary step is correct?
- (A) It is the highest-energy arrangement along the reaction coordinate, with some bonds partly broken and others partly formed.
- (B) It is a short-lived substance that can be collected by cooling the reaction mixture quickly.
- (C) It is the arrangement of atoms with the lowest energy, so the reaction passes through it most easily.
- (D) It lies at the same energy as the products in an exothermic step.
Answer and explanation
Answer: (A). The transition state sits at the top of the single peak. Old bonds are partly broken and new ones partly formed, so it is less stable than any other arrangement along the path.
- (B) describes an intermediate, which does not exist in a single elementary step. A transition state cannot be isolated.
- (C) reverses the idea: it is the highest point, which is why reaching it needs the activation energy.
- (D) is false: the products are always below the peak (otherwise there would be no reverse activation energy).
Question 4 (multiple choice · core)
An elementary reaction is carried out at 310 K instead of 290 K. Which statement is correct?
- (A) The transition state is lower in energy, so Eₐ(forward) is smaller.
- (B) The energy profile is unchanged, but a larger fraction of collisions have enough energy to reach the transition state.
- (C) The products move to a lower energy, so the reaction becomes more exothermic.
- (D) The energy profile is unchanged, and the fraction of collisions that reach the transition state is also unchanged; only the collision frequency increases.
Answer and explanation
Answer: (B). The energy levels of reactants, transition state and products belong to the reaction, so the profile does not change. At the higher temperature, more of the Maxwell–Boltzmann curve lies beyond Eₐ, so more collisions reach the transition state and k increases (k = A e^(−Eₐ/RT)).
- (A) is a very common error: temperature does not change Eₐ. (Only a different pathway, such as one offered by a catalyst, has a different peak.)
- (C) is wrong: ΔH is the gap between two fixed levels.
- (D) names only the small effect and misses the main one.
Question 5 (constructed response · core)
An elementary step D + E → F + G has Eₐ(forward) = 58 kJ mol⁻¹ and ΔH = +26 kJ mol⁻¹.
(a) Calculate Eₐ(reverse). (b) Sketch a fully labelled energy profile for the step. (c) Assume the forward and reverse steps have similar values of A. At the same temperature, which has the larger rate constant? Justify your answer.
Worked solution and suggested Marlbridge rubric
(a) Eₐ(reverse) = Eₐ(forward) − ΔH = 58 − 26 = 32 kJ mol⁻¹.
(b) Axes: energy (y) and reaction coordinate (x). A flat level for D + E on the left; one smooth peak, labelled transition state, 58 above the reactants; a flat level for F + G on the right, 26 above the reactants. Arrows: reactants up to the peak labelled Eₐ(forward) = 58; reactants to products labelled ΔH = +26 (or products up to the peak labelled Eₐ(reverse) = 32).
(c) The reverse step (F + G → D + E). Its activation energy (32 kJ mol⁻¹) is smaller, so a larger fraction of F–G collisions have enough energy to reach the transition state. With similar A, the exponential term e^(−Eₐ/RT) is larger for the smaller Eₐ, so k(reverse) > k(forward).
| Point | What earns it |
|---|---|
| 1 | Eₐ(reverse) = 32 kJ mol⁻¹ |
| 1 | Sketch: labelled axes, one peak marked as the transition state, products higher than reactants |
| 1 | Eₐ(forward) and ΔH (or Eₐ(reverse)) shown as arrows between the correct levels |
| 1 | Reverse step chosen, justified by its smaller Eₐ giving a larger fraction of collisions able to reach the transition state |
Do not award the sketch point if two peaks are drawn or if the products are drawn below the reactants.
Question 6 (constructed response · core)
Two elementary steps, J and K, have similar values of A. Step J has Eₐ = 35 kJ mol⁻¹; step K has Eₐ = 85 kJ mol⁻¹.
(a) Which step has the larger rate constant at 300 K? Justify your answer. (b) Both reactions are warmed by 15 K. Which rate constant increases by the larger factor? Explain using the Arrhenius equation or a Maxwell–Boltzmann sketch. (c) A student says: “Heating step K lowers its activation energy until it matches step J.” Evaluate the claim.
Worked solution and suggested Marlbridge rubric
(a) Step J. Its climb from reactants to transition state is smaller, so a much larger fraction of collisions have enough energy to reach it. In k = A e^(−Eₐ/RT), with similar A and the same T, the smaller Eₐ gives the larger exponential term.
(b) Step K. Its Eₐ lies far out in the tail of the Maxwell–Boltzmann curve, where heating multiplies the area by a large factor. In Arrhenius terms, the larger Eₐ makes e^(−Eₐ/RT) change more steeply with T. (No calculation is needed.)
(c) The claim is wrong. The activation energy is the energy gap between the reactants and the transition state, which is fixed for the reaction. Heating raises the rate because more collisions have at least Eₐ, not because Eₐ falls. Step K’s activation energy stays 50 kJ mol⁻¹ larger than step J’s at every temperature.
| Point | What earns it |
|---|---|
| 1 | (a) Step J, linked to the smaller Eₐ giving a larger fraction of successful collisions (or a larger exponential term) |
| 1 | (b) Step K chosen |
| 1 | (b) Reason: a larger Eₐ means the fraction able to reach the transition state is more sensitive to temperature |
| 1 | (c) States that Eₐ does not change with temperature, and that heating increases the fraction of collisions reaching Eₐ |
Question 7 (explanation · stretch)
The gas-phase step X + Y₂ → XY + Y is elementary. It has Eₐ(forward) = 20 kJ mol⁻¹ and Eₐ(reverse) = 150 kJ mol⁻¹.
(a) Identify the bond broken and the bond formed in the forward step. (b) Describe the transition state, and state what an orientation requirement means for this step. (c) Calculate ΔH for the forward step and use it to compare the strengths of the bond broken and the bond formed. (d) Explain why, at the same temperature, the reverse step is far slower than the forward step.
Model answer and suggested Marlbridge rubric
(a) The Y–Y bond breaks; a new X–Y bond forms.
(b) In the simplest picture, the transition state has X, Y and Y in a line (X···Y···Y), with the X–Y bond partly formed and the Y–Y bond partly broken. It is the highest-energy point on the profile and cannot be isolated. Orientation: X must hit one end of Y₂, ideally along the line of the Y–Y bond; a hit that misses the end of the molecule cannot form the X–Y bond as the Y–Y bond breaks.
(c) ΔH = Eₐ(forward) − Eₐ(reverse) = 20 − 150 = −130 kJ mol⁻¹. The step releases energy overall, so more energy is released forming the X–Y bond than is used breaking the Y–Y bond: the X–Y bond is the stronger bond.
(d) The reverse step must climb 150 kJ mol⁻¹ to the same transition state, compared with only 20 kJ mol⁻¹ for the forward step. A far smaller fraction of collisions between XY and Y have enough energy to reach the peak, so (with similar A) k(reverse) is much smaller.
| Point | What earns it |
|---|---|
| 1 | Y–Y broken and X–Y formed |
| 1 | Transition state with partial X–Y and Y–Y bonds, at the peak; X must approach the end of Y₂ |
| 1 | ΔH = −130 kJ mol⁻¹ and the X–Y bond is stronger than the Y–Y bond |
| 1 | Reverse step slower because its much larger Eₐ means a far smaller fraction of collisions reach the transition state |
Accept a bent arrangement in (b) if it still shows two partial bonds to the middle Y atom.
How did you do?
- Q1, Q2 or Q5(a) wrong: practise reading levels as differences with Worked example 1 in the study guide.
- Q3 or Q7(b) wrong: re-read “The transition state”.
- Q4 or Q6 wrong: re-read “Temperature and the Arrhenius equation” and Worked example 3.
- Q5(b) incomplete: compare your sketch with Figure 1; check every label.
Then tick off the topic checklist.