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Questions on Hess's law and enthalpy cycles

Hess's law and why it holds, enthalpy cycles built from enthalpies of formation and from enthalpies of combustion, the two formulae and the direction of the arrows in each, finding the enthalpy of formation of methane and of carbon monoxide, and finding the enthalpy change for the hydration of anhydrous copper(II) sulfate from two measured enthalpies of solution, and the thermal decomposition of potassium hydrogencarbonate from its reaction and that of potassium carbonate with hydrochloric acid.

18 original questions · 60 marks · the Hess's law and enthalpy cycles notes · Energetics · Flashcards for this unit

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• 6 marks · Calculate · HSL-Q7

  Calculate the enthalpy change for CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s) from the results in the table. Each salt was dissolved in 40.0 cm³ of water in a polystyrene cup. c = 4.18 J g⁻¹ K⁻¹.

  Salt	Mass dissolved / g	Mr	Temperature change / °C
  Anhydrous CuSO₄	3.19	159.6	rise of 7.8
  CuSO₄·5H₂O	4.99	249.6	fall of 1.3

  Mark scheme

  1. M1 for the anhydrous salt, q = 40.0 × 4.18 × 7.8 = 1304 J, and the amount dissolved is 3.19 ÷ 159.6 = 0.01999 mol
  2. A1 ΔH₁ = −1.304 ÷ 0.01999 = −65.2 kJ mol⁻¹, negative because the temperature of the water rose
  3. M1 for the hydrated salt, q = 40.0 × 4.18 × 1.3 = 217 J, and the amount dissolved is 4.99 ÷ 249.6 = 0.01999 mol
  4. A1 ΔH₂ = +0.217 ÷ 0.01999 = +10.9 kJ mol⁻¹, positive because the temperature of the water fell
  5. M1 both salts end as the same solution, so from the cycle ΔH(hydration) = ΔH₁ − ΔH₂ = −65.2 − (+10.9)
  6. A1 ΔH(hydration) = −76 kJ mol⁻¹ to two significant figures, matching the 1.3 °C; allow error carried forward from either ΔH

• 5 marks · Draw · HSL-Q11

  Draw the enthalpy cycle that uses enthalpies of combustion to find the standard enthalpy of formation of propane, and use it to calculate ΔfH°(C₃H₈, g). ΔcH° / kJ mol⁻¹: C(s) −394, H₂(g) −286, C₃H₈(g) −2220.

  Mark scheme

  1. B1 3C(s) + 4H₂(g) → C₃H₈(g) across the top, with the arrow labelled ΔfH°
  2. B1 3CO₂(g) + 4H₂O(l) in a box underneath, the combustion products; allow + 5O₂(g) shown on the down arrows
  3. B1 arrows pointing down from both the elements and propane to the box, labelled 3ΔcH°(C) + 4ΔcH°(H₂) and ΔcH°(C₃H₈); arrows pointing up lose this mark
  4. M1 ΔfH° = 3(−394) + 4(−286) − (−2220) = −1182 − 1144 + 2220
  5. A1 ΔfH° = −106 kJ mol⁻¹ to three significant figures

• 5 marks · Calculate · HSL-Q17

  Calculate ΔH for 2KHCO₃(s) → K₂CO₃(s) + CO₂(g) + H₂O(l) from the enthalpy changes for reaction with excess hydrochloric acid, KHCO₃(s) +27.0 kJ mol⁻¹ and K₂CO₃(s) −34.0 kJ mol⁻¹, explaining why the first of these is multiplied by two in the cycle.

  Mark scheme

  1. B1 the measured enthalpy change for the hydrogencarbonate is per mole of KHCO₃
  2. B1 the target equation starts with two moles of KHCO₃, so two moles react with acid on the indirect route and that step contributes twice the measured value
  3. M1 both routes end at 2KCl(aq) + 2CO₂(g) + 2H₂O(l), so ΔH = 2ΔH(KHCO₃ with acid) − ΔH(K₂CO₃ with acid)
  4. M1 ΔH = 2 × (+27.0) − (−34.0) = 54.0 + 34.0
  5. A1 ΔH = +88.0 kJ mol⁻¹ to three significant figures; reject +61.0 kJ mol⁻¹, which leaves out the factor of two

• 4 marks · Calculate · HSL-Q3

  Calculate the standard enthalpy change for 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g), the first step in making nitric acid. ΔfH° / kJ mol⁻¹: NH₃(g) −46, NO(g) +90, H₂O(g) −242.

  Mark scheme

  1. M1 ΔrH° = ΣΔfH°(products) − ΣΔfH°(reactants), with ΔfH° of O₂(g) = 0 as an element in its standard state
  2. M1 products: 4(+90) + 6(−242) = 360 − 1452 = −1092
  3. M1 reactants: 4(−46) + 5(0) = −184; a coefficient left out of either sum loses that mark only
  4. A1 ΔrH° = −1092 − (−184) = −908 kJ mol⁻¹, to three significant figures

• 4 marks · Calculate · HSL-Q5

  Calculate the standard enthalpy of formation of ethanol, 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l), from these enthalpies of combustion / kJ mol⁻¹: C(s) −394, H₂(g) −286, C₂H₅OH(l) −1367.

  Mark scheme

  1. M1 a combustion cycle: ΔfH° = ΣΔcH°(reactants) − ΣΔcH°(products)
  2. M1 reactants: 2(−394) + 3(−286) = −788 − 858 = −1646
  3. M1 ΔfH° = −1646 − (−1367)
  4. A1 ΔfH° = −279 kJ mol⁻¹, to three significant figures

• 3 marks · Calculate · HSL-Q2

  Calculate the standard enthalpy change for C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l). ΔfH° / kJ mol⁻¹: C₂H₅OH(l) −277, CO₂(g) −394, H₂O(l) −286.

  Mark scheme

  1. M1 ΔrH° = ΣΔfH°(products) − ΣΔfH°(reactants), with ΔfH° of O₂ = 0
  2. M1 products: 2(−394) + 3(−286) = −1646; reactants: −277
  3. A1 ΔrH° = −1646 − (−277) = −1369 kJ mol⁻¹, to four significant figures as the data are given

• 3 marks · Calculate · HSL-Q4

  Calculate the standard enthalpy change for the thermal decomposition CaCO₃(s) → CaO(s) + CO₂(g), using the enthalpy cycle.

  Mark scheme

  1. M1 ΔrH° = ΣΔfH°(products) − ΣΔfH°(reactants)
  2. M1 = (−635 + −394) − (−1207) = −1029 + 1207
  3. A1 ΔrH° = +178 kJ mol⁻¹ to three significant figures, positive because the reaction is endothermic

• 3 marks · Calculate · HSL-Q6

  Calculate ΔrH° for the hydrogenation of ethene, C₂H₄(g) + H₂(g) → C₂H₆(g). ΔcH° / kJ mol⁻¹: C₂H₄(g) −1411, H₂(g) −286, C₂H₆(g) −1560.

  Mark scheme

  1. M1 ΔrH° = ΣΔcH°(reactants) − ΣΔcH°(products), because the data are enthalpies of combustion
  2. M1 = (−1411 + −286) − (−1560) = −1697 + 1560
  3. A1 ΔrH° = −137 kJ mol⁻¹, to three significant figures; +137 kJ mol⁻¹, from products minus reactants, scores the second mark only

• 3 marks · Explain · HSL-Q8

  Explain why the standard enthalpy of formation of carbon monoxide cannot be measured directly, and how Hess's law is used to find it.

  Mark scheme

  1. B1 burning carbon in a limited supply of oxygen gives a mixture of CO, CO₂ and unburned carbon, never pure CO
  2. B1 the enthalpies of combustion of carbon and of carbon monoxide can both be measured
  3. B1 by Hess's law, ΔfH°(CO) = ΔcH°(C) − ΔcH°(CO) = −394 − (−283) = −111 kJ mol⁻¹

• 3 marks · Explain · HSL-Q9

  Explain why the enthalpy change for the hydration of anhydrous copper(II) sulfate cannot be measured directly in a calorimeter.

  Mark scheme

  1. B1 water added to the solid cannot be made to react in exactly a 5:1 ratio
  2. B1 some of the salt dissolves in the water added, while some stays anhydrous inside lumps
  3. B1 the reaction does not go cleanly to completion, so the heat measured is not that of the hydration alone

• 3 marks · Explain · HSL-Q10

  Explain why Hess's law must be true, using the law of conservation of energy.

  Mark scheme

  1. B1 enthalpy depends only on the state of the substances, not on how they reached it
  2. B1 if two routes between the same start and finish had different enthalpy changes, going out by one and back by the other would release energy on every cycle
  3. B1 that would create energy from nothing, which the law of conservation of energy forbids, so the two routes must have the same enthalpy change

• 3 marks · Suggest · HSL-Q14

  Suggest how heat lost to the surroundings while anhydrous copper(II) sulfate dissolves would affect the value calculated for its enthalpy of hydration.

  Mark scheme

  1. B1 the temperature rise measured would be too small
  2. B1 so ΔH₁ for dissolving the anhydrous salt would be less negative than its true value
  3. B1 ΔH(hydration) = ΔH₁ − ΔH₂, so the hydration would come out less exothermic than its true value

• 3 marks · Compare · HSL-Q15

  Compare an enthalpy cycle drawn with enthalpies of formation with one drawn with enthalpies of combustion.

  Mark scheme

  1. B1 the formation cycle has the elements in the box at the bottom, while the combustion cycle has CO₂ and H₂O
  2. B1 in the formation cycle the arrows point up, out of the box, while in the combustion cycle they point down, into it
  3. B1 so ΔrH° is Σ products − Σ reactants for formation but Σ reactants − Σ products for combustion

• 3 marks · Evaluate · HSL-Q16

  Evaluate this working for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), using ΔfH° / kJ mol⁻¹ CH₄ −75, CO₂ −394, H₂O(l) −286: ΔrH° = (−394 + −286) − (−75) = −605 kJ mol⁻¹.

  Mark scheme

  1. B1 the formula used, products minus reactants, is correct for formation data, and leaving out O₂ is correct
  2. B1 the enthalpy of formation of water has not been multiplied by 2, its coefficient in the equation
  3. B1 the correct working is (−394 + 2 × −286) − (−75) = −891 kJ mol⁻¹

• 3 marks · Suggest · HSL-Q19

  Suggest why the enthalpy change of decomposition of potassium hydrogencarbonate found from its reaction and that of potassium carbonate with acid, in polystyrene cups, comes out less endothermic than the value from enthalpies of formation.

  Mark scheme

  1. B1 the hydrogencarbonate reaction cools the acid below room temperature, so heat flows in, the fall is too small and its ΔH is too small
  2. B1 the carbonate reaction warms the acid, heat is lost, the rise is too small and its ΔH is not negative enough
  3. B1 ΔH = 2ΔH₁ − ΔH₂, so both errors make the calculated decomposition less endothermic

• 2 marks · State · HSL-Q1

  State Hess's law, including the condition under which it applies.

  Mark scheme

  1. B1 the enthalpy change for a reaction is independent of the route taken
  2. B1 provided the initial and final conditions are the same

• 2 marks · Identify · HSL-Q12

  Identify the three species in this list whose standard enthalpy of formation is zero: O₂(g), O₃(g), Br₂(g), Br₂(l), C(diamond), C(graphite).

  Mark scheme

  1. B1 O₂(g) and C(graphite), the standard states of oxygen and carbon
  2. B1 Br₂(l), because bromine is a liquid at 298 K and 100 kPa; Br₂(g), O₃ and diamond are not standard states

• 2 marks · Suggest · HSL-Q13

  Suggest why enthalpies of combustion, rather than enthalpies of formation, are used to find ΔrH° for reactions of organic compounds.

  Mark scheme

  1. B1 almost all organic compounds burn completely in oxygen, so their enthalpies of combustion can be measured directly
  2. B1 organic compounds cannot be made directly from their elements, so their enthalpies of formation are not measured directly

Practise Hess's law and enthalpy cycles one question at a time