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 reactions and substances named with letters (RS, RT, RZ) are fictional. Energies are in kJ mol⁻¹, per mole of the equation as written. Where a question asks for a sketch, draw it on paper, then compare it with the features listed in the solution.
Question 1 (multiple choice · foundation)
Which feature of an energy diagram shows that a process is exothermic?
- (A) The product level is below the reactant level.
- (B) There is a tall hump between the reactant and product levels.
- (C) The product level is above the reactant level.
- (D) The arrow between the levels is labelled with a positive value.
Answer and explanation
Answer: (A). Products lower than reactants means the system has lost energy to the surroundings, so ΔH is negative: exothermic.
- (B) A hump shows the activation energy. It tells you nothing about whether the process is exothermic or endothermic.
- (C) describes an endothermic process.
- (D) A positive ΔH means the system gained energy: endothermic.
Question 2 (multiple choice · core)
On an energy diagram, the reactants are at −30 kJ mol⁻¹ and the products are at +45 kJ mol⁻¹. What is ΔH for the reaction?
- (A) +45 kJ mol⁻¹
- (B) +75 kJ mol⁻¹
- (C) −75 kJ mol⁻¹
- (D) +15 kJ mol⁻¹
Answer and explanation
Answer: (B). ΔH = E(products) − E(reactants) = 45 − (−30) = +75 kJ mol⁻¹. The products are higher, so the reaction is endothermic, which matches the positive sign.
- (A) reads the product level as if the reactants were at zero. They are not.
- (C) subtracts the wrong way round (reactants − products).
- (D) adds the two levels instead of finding their difference.
Question 3 (multiple choice · core)
Which description matches an energy diagram for steam condensing to liquid water, H₂O(g) → H₂O(l)?
- (A) The H₂O(l) level is above the H₂O(g) level; the arrow points up; ΔH is positive.
- (B) The H₂O(l) level is below the H₂O(g) level; the arrow points down; ΔH is negative.
- (C) The H₂O(l) level is below the H₂O(g) level; the arrow points up from H₂O(l) to H₂O(g); ΔH is negative.
- (D) The two levels are at the same height, because condensing happens at constant temperature.
Answer and explanation
Answer: (B). Condensing lets water molecules come together and attract each other, which releases energy. The liquid is lower in energy than the gas, and the arrow runs from the start (gas) down to the end (liquid).
- (A) describes vaporisation, the reverse change.
- (C) has the right levels, but the arrow points from the end to the start. An upward arrow from liquid to gas would describe vaporisation, with a positive ΔH.
- (D) confuses temperature with energy. The temperature stays constant during a phase change, but the energy of the system still falls.
Question 4 (multiple choice · core)
Two students draw energy diagrams for the same reaction, which has ΔH = −60 kJ mol⁻¹. Student A puts the reactants at 0 and the products at −60 kJ mol⁻¹. Student B puts the reactants at 100 kJ mol⁻¹ and the products at 40 kJ mol⁻¹. Which statement is correct?
- (A) Only Student A is correct, because reactants must always be drawn at zero.
- (B) Only Student B is correct, because energy levels cannot be negative.
- (C) Both are correct, because only the difference between the levels has meaning.
- (D) Neither is correct, because an energy diagram must show the activation energy.
Answer and explanation
Answer: (C). In both diagrams the products are 60 kJ mol⁻¹ below the reactants (−60 − 0 = −60 and 40 − 100 = −60). The zero of the energy axis is a free choice.
- (A) Putting the reactants at zero is convenient, not required.
- (B) Negative levels are fine; they only mean “below the chosen zero”.
- (D) The hump is optional. The endothermic or exothermic nature depends only on the end levels.
Question 5 (constructed response · core)
The fictional reaction 2 RS(g) → R₂(g) + S₂(g) has ΔH = +58 kJ mol⁻¹.
(a) Is the reaction endothermic or exothermic? State the direction of energy transfer between the system and the surroundings. (b) Sketch an energy diagram for the reaction, to scale. State the range and tick spacing you chose for the energy axis. (c) State ΔH for R₂(g) + S₂(g) → 2 RS(g) and describe how its diagram differs from yours.
Worked solution and suggested Marlbridge rubric
(a) ΔH is positive, so the reaction is endothermic: energy moves from the surroundings into the system.
(b) One good choice: put 2 RS(g) at 0; use an axis from −20 to +80 kJ mol⁻¹ with ticks every 20. The R₂(g) + S₂(g) level is at +58, just below the +60 tick. Draw an upward arrow from the reactant level to the product level, labelled “ΔH = +58 kJ mol⁻¹”. Label the vertical axis “Energy (kJ mol⁻¹)” and the horizontal axis “Reaction progress”.
(c) ΔH = −58 kJ mol⁻¹. The two levels stay where they are; the reaction now starts at the R₂ + S₂ level and ends at the 2 RS level, so the arrow points down.
| Point | What earns it |
|---|---|
| 1 | Endothermic, with energy moving from the surroundings into the system |
| 1 | Vertical axis labelled with energy and kJ mol⁻¹, with even ticks |
| 1 | Product level above reactant level, placed at +58 on the chosen scale (within half a tick) |
| 1 | Upward arrow from reactants to products labelled with ΔH = +58 kJ mol⁻¹ |
| 1 | Reverse: ΔH = −58 kJ mol⁻¹ with the same levels and a downward arrow |
Any sensible range and tick spacing earn the second point if the +58 level fits on the axis and the ticks are evenly spaced. A diagram with a hump drawn in is fine, but the hump earns no extra credit.
Question 6 (constructed response · core)
A student is asked to draw an energy diagram for a reaction with ΔH = −80 kJ mol⁻¹. The student’s diagram has:
- a vertical axis labelled “Temperature (°C)”, with no numbers;
- the reactant level on the left and the product level on the right, with the product level drawn higher than the reactant level;
- an arrow pointing down from the product level to the reactant level, labelled “ΔH = −80”.
Identify three errors in the diagram and explain how to correct each one.
Model answer and suggested Marlbridge rubric
- Axis. The vertical axis shows the energy of the system, not temperature. It should read “Energy (kJ mol⁻¹)” and have an even scale with numbers on the ticks.
- Levels. ΔH is negative, so the reaction is exothermic and the products must be below the reactants. The student has drawn them the wrong way up.
- Arrow. The arrow must run from the start of the reaction (reactants) to the end (products). Once the levels are corrected, the arrow goes from the reactant level down to the product level.
(Also accept: the ΔH label has no unit; it should read ΔH = −80 kJ mol⁻¹.)
| Point | What earns it |
|---|---|
| 1 | Axis error: should be energy with units of kJ mol⁻¹ and a numbered, even scale |
| 1 | Level error: products should be below reactants because ΔH is negative (exothermic) |
| 1 | Arrow error: should start at the reactants and end at the products |
Award each point only if both the error and its correction are given. The missing unit can replace any one of the three points.
Question 7 (constructed response · stretch)
A fictional salt RT dissolves in water. Illustrative energy changes for three steps, per mole of RT, are:
| Step | Energy change (kJ mol⁻¹) |
|---|---|
| 1. Separate the R⁺ and T⁻ ions | +780 |
| 2. Separate water molecules to make room | +45 |
| 3. Form ion–water attractions | −805 |
(a) Sketch a step diagram for dissolving RT. Give the energy of each level, with solid RT and water at 0. (b) State whether dissolving RT is endothermic or exothermic, and predict what happens to the temperature of the water. Justify both using your diagram. (c) A second fictional salt, RZ, has the same values for steps 1 and 2. What value of step 3 would make dissolving RZ neither endothermic nor exothermic? (d) For a third salt, step 3 is −840 kJ mol⁻¹ and steps 1 and 2 are unchanged. Find the overall energy change and say how the final level moves compared with RT.
Worked solution and suggested Marlbridge rubric
(a) Levels: RT(s) + water at 0; ions separated at +780; ions and water molecules separated at 780 + 45 = +825; RT(aq) at 825 − 805 = +20 kJ mol⁻¹. Two upward arrows (steps 1 and 2), one long downward arrow (step 3).
(b) The final level (+20) is above the starting level (0), so ΔH = +20 kJ mol⁻¹ and dissolving RT is endothermic. The energy released by forming ion–water attractions (805 kJ) is less than the energy needed for steps 1 and 2 (825 kJ). The system gains energy from the water, so the temperature of the water falls.
(c) For no overall change, step 3 must exactly cancel steps 1 and 2: −825 kJ mol⁻¹.
(d) 780 + 45 − 840 = −15 kJ mol⁻¹. The final level moves from 20 above the start to 15 below it, so this salt dissolves exothermically.
| Point | What earns it |
|---|---|
| 1 | Correct levels 0, +780, +825, +20 (or a correctly scaled sketch showing them) |
| 1 | Arrows in the right directions: steps 1 and 2 up, step 3 down |
| 1 | Endothermic because the final level is above the start (energy released in step 3 is less than energy needed in steps 1 and 2) |
| 1 | Temperature of the water falls, because energy moves from the water into the system |
| 1 | (c) −825 kJ mol⁻¹ |
| 1 | (d) −15 kJ mol⁻¹, final level now below the starting level |
Carry forward an error in (a) into (b) to (d) if the method is right.
How did you do?
- Q1 or Q3 wrong: re-read “Reading endothermic and exothermic from the levels” in the study guide.
- Q2 or Q4 wrong: remember ΔH = final − start, and the zero is a free choice.
- Q5 or Q6 incomplete: go through “Scale and units” and Worked example 1 again.
- Q7 incomplete: compare your sketch with Figure 3 and Worked example 2.
Then tick off the topic checklist.