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. Use these standard enthalpies of formation at 298 K, in kJ mol⁻¹, for every question:
| Substance | ΔH°f | Substance | ΔH°f |
|---|---|---|---|
| CO₂(g) | −393.5 | SO₂(g) | −296.8 |
| H₂O(l) | −285.8 | SO₃(g) | −395.7 |
| H₂O(g) | −241.8 | Fe₂O₃(s) | −824.2 |
| C₂H₆(g) | −83.7 | Al₂O₃(s) | −1675.5 |
| CH₃OH(l) | to be found (Q6) | NO₂(g) | +33.2 |
| N₂O₄(g) | +9.2 |
Elements in their standard states have ΔH°f = 0. Molar masses: H 1.008, C 12.01 g mol⁻¹.
Question 1 (multiple choice · foundation)
Which substance has a standard enthalpy of formation of exactly zero?
- (A) O₂(g)
- (B) O₃(g)
- (C) H₂O(l)
- (D) C(diamond)
Answer and explanation
Answer: (A). O₂(g) is the standard state of oxygen. Forming it from itself involves no change, so ΔH°f = 0.
- (B) Ozone is an element, but not in its standard state. Forming O₃ from O₂ needs energy, so its ΔH°f is positive.
- (C) Water is a compound formed from H₂ and O₂, with a large negative ΔH°f.
- (D) The standard state of carbon is graphite, not diamond, so diamond has a small positive ΔH°f.
Question 2 (multiple choice · foundation)
Which equation has a ΔH° equal to the standard enthalpy of formation of ammonia, NH₃(g)?
- (A) N₂(g) + 3H₂(g) → 2NH₃(g)
- (B) ½N₂(g) + 3/2 H₂(g) → NH₃(g)
- (C) N(g) + 3H(g) → NH₃(g)
- (D) NH₃(g) → ½N₂(g) + 3/2 H₂(g)
Answer and explanation
Answer: (B). A formation equation makes exactly 1 mol of the substance from its elements in their standard states.
- (A) forms 2 mol of NH₃, so its ΔH° is twice ΔH°f.
- (C) starts from gaseous atoms, which are not the standard states of nitrogen and hydrogen.
- (D) is the reverse of formation, so its ΔH° has the opposite sign.
Question 3 (multiple choice · core)
Calculate ΔH° for 2SO₂(g) + O₂(g) → 2SO₃(g).
- (A) −197.8 kJ mol⁻¹
- (B) −98.9 kJ mol⁻¹
- (C) +197.8 kJ mol⁻¹
- (D) −692.5 kJ mol⁻¹
Answer and explanation
Answer: (A). Products: 2(−395.7) = −791.4 kJ. Reactants: 2(−296.8) + 0 = −593.6 kJ. ΔH° = (−791.4) − (−593.6) = −197.8 kJ mol⁻¹.
- (B) ignores the coefficients (one SO₂ → one SO₃). It is the value per mole of SO₃, not for the equation as written.
- (C) calculates reactants − products.
- (D) adds the two ΔH°f values instead of subtracting.
Question 4 (multiple choice · core)
Calculate ΔH° for the thermite reaction, Fe₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Fe(s).
- (A) +851.3 kJ mol⁻¹
- (B) −851.3 kJ mol⁻¹
- (C) −1675.5 kJ mol⁻¹
- (D) −2499.7 kJ mol⁻¹
Answer and explanation
Answer: (B). Al and Fe are elements in their standard states (0). ΔH° = (−1675.5) − (−824.2) = −851.3 kJ mol⁻¹. The reaction is strongly exothermic.
- (A) calculates reactants − products.
- (C) uses only the product and forgets to subtract the reactant Fe₂O₃.
- (D) adds the two values: −1675.5 + (−824.2).
Question 5 (constructed response · core)
Ethane burns completely: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l).
(a) Calculate ΔH° for the equation as written. (b) Calculate ΔH° per mole of ethane. (c) Calculate the heat released when 10.0 g of ethane burns completely. (d) Calculate ΔH° for the equation if the water is formed as H₂O(g) instead, and explain the difference.
Worked solution and suggested Marlbridge rubric
(a)
| Substance | n | n × ΔH°f (kJ) |
|---|---|---|
| CO₂(g) | 4 | −1574.0 |
| H₂O(l) | 6 | −1714.8 |
| Products | −3288.8 | |
| C₂H₆(g) | 2 | −167.4 |
| O₂(g) | 7 | 0 |
| Reactants | −167.4 |
ΔH° = (−3288.8) − (−167.4) = −3121.4 kJ mol⁻¹ (per mole of the equation as written).
(b) −3121.4 ÷ 2 = −1560.7 kJ per mol C₂H₆.
(c) M(C₂H₆) = 2(12.01) + 6(1.008) = 30.07 g mol⁻¹. n = 10.0 ÷ 30.07 = 0.3326 mol. Heat released = 0.3326 × 1560.7 = 519 kJ.
(d) Products: 4(−393.5) + 6(−241.8) = −3024.8 kJ. ΔH° = (−3024.8) − (−167.4) = −2857.4 kJ mol⁻¹. This is 264.0 kJ less negative, which is 6 × 44.0 kJ: the energy that would be released if the 6 mol of water vapour condensed to liquid.
| Point | What earns it |
|---|---|
| 1 | Coefficients applied to every ΔH°f, with O₂ = 0 |
| 1 | −3121.4 kJ mol⁻¹, as products − reactants |
| 1 | −1560.7 kJ per mol ethane |
| 1 | About 519 kJ released, with moles of ethane shown |
| 1 | −2857.4 kJ mol⁻¹ with H₂O(g) |
| 1 | Difference explained as condensation of water (6 × 44.0 kJ), or as H₂O(g) being higher in energy than H₂O(l) |
Carry forward errors from (a) into (b) and (c).
Question 6 (constructed response · core)
The standard enthalpy of combustion of liquid methanol is −726.7 kJ mol⁻¹: CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l).
(a) Calculate ΔH°f of CH₃OH(l). (b) A student leaves O₂ out of the calculation completely. Explain why this does not change the answer. (c) Would you expect ΔH°f of CH₃OH(g) to be more negative or less negative than ΔH°f of CH₃OH(l)? Justify your answer. (d) ΔH°f of CH₃OH(g) is −201.0 kJ mol⁻¹. Use this value and your answer to (a) to calculate ΔH° for the physical change CH₃OH(l) → CH₃OH(g).
Worked solution and suggested Marlbridge rubric
(a) Products: (−393.5) + 2(−285.8) = −965.1 kJ. Let ΔH°f(CH₃OH(l)) = x. Then −726.7 = −965.1 − x, so x = −965.1 + 726.7 = −238.4 kJ mol⁻¹.
(b) O₂(g) is an element in its standard state, so its ΔH°f is 0. Multiplying by 3/2 still gives 0, so it adds nothing to the reactants total.
(c) Less negative. Turning liquid methanol into gas is endothermic, because energy is needed to overcome the attractions (including hydrogen bonds) between the molecules. So the gas lies higher in energy than the liquid, and its ΔH°f is closer to zero than −238.4 kJ mol⁻¹.
(d) ΔH° = ΔH°f(CH₃OH(g)) − ΔH°f(CH₃OH(l)) = (−201.0) − (−238.4) = +37.4 kJ mol⁻¹. The positive sign agrees with (c): vaporisation is endothermic.
| Point | What earns it |
|---|---|
| 1 | Products total −965.1 kJ, with 2 × ΔH°f of H₂O(l) |
| 1 | Correct equation with the unknown on the reactant side, and x = −238.4 kJ mol⁻¹ |
| 1 | O₂ has ΔH°f = 0 because it is an element in its standard state |
| 1 | Less negative, because vaporisation is endothermic (the gas is higher in energy than the liquid) |
| 1 | +37.4 kJ mol⁻¹ for vaporisation, as products − reactants (carry forward from (a)) |
A common error is to forget the 2 in front of H₂O, which gives x = +47.4 kJ mol⁻¹; a positive value for a stable liquid like methanol should make you check your working.
Question 7 (constructed response · stretch)
Nitrogen dioxide forms dinitrogen tetroxide: 2NO₂(g) → N₂O₄(g).
(a) Calculate ΔH° for the reaction. (b) Describe an energy diagram that shows this reaction with the elements, N₂(g) + 2O₂(g), at 0. Give the energy of each level. (c) Both NO₂ and N₂O₄ have positive enthalpies of formation, yet the reaction is exothermic. Explain how this is possible. (d) State ΔH° for N₂O₄(g) → 2NO₂(g).
Worked solution and suggested Marlbridge rubric
(a) ΔH° = (+9.2) − 2(+33.2) = 9.2 − 66.4 = −57.2 kJ mol⁻¹.
(b) Vertical axis “Energy (kJ mol⁻¹)”. Elements N₂(g) + 2O₂(g) at 0. Reactants 2NO₂(g) at +66.4. Product N₂O₄(g) at +9.2. Arrow from 2NO₂ down to N₂O₄, labelled ΔH° = −57.2 kJ mol⁻¹. Both levels lie above the elements.
(c) ΔH° depends on where the products lie relative to the reactants, not relative to zero. Both lie above the elements, but N₂O₄ (+9.2) lies below 2NO₂ (+66.4), so energy is released when NO₂ turns into N₂O₄.
(d) +57.2 kJ mol⁻¹ (the reverse reaction has the opposite sign).
| Point | What earns it |
|---|---|
| 1 | Coefficient 2 applied to NO₂ (66.4 kJ) |
| 1 | −57.2 kJ mol⁻¹ |
| 1 | Diagram levels: elements 0, 2NO₂ +66.4, N₂O₄ +9.2, with a downward arrow from 2NO₂ to N₂O₄ |
| 1 | Explanation: the sign depends on products relative to reactants, and N₂O₄ is lower than 2NO₂ |
| 1 | +57.2 kJ mol⁻¹ for the reverse |
How did you do?
- Q1 or Q2 wrong: re-read “What a standard enthalpy of formation is” in the study guide.
- Q3, Q4 or Q5 wrong: use the table layout in Worked examples 1 and 2, and check every coefficient and state.
- Q6 or Q7 incomplete: go through Worked example 3 and Figure 1.
Then tick off the topic checklist.