Skip to content
MarkSchemeMarkScheme
Sign up — no card

9701 · 5.2

Hess's law — practice questions

Practice and worked examples for 9701 Hess's law. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Calculate the standard enthalpy change of reaction, ΔHr⊖\Delta H_r^\ominus, for the thermal decomposition of calcium carbonate. CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})

Use the following standard enthalpy of formation data: ΔHf⊖[CaCO3(s)]=−1207 kJ mol−1\Delta H_f^\ominus[\text{CaCO}_3(\text{s})] = -1207 \text{ kJ mol}^{-1} ΔHf⊖[CaO(s)]=−635 kJ mol−1\Delta H_f^\ominus[\text{CaO}(\text{s})] = -635 \text{ kJ mol}^{-1} ΔHf⊖[CO2(g)]=−394 kJ mol−1\Delta H_f^\ominus[\text{CO}_2(\text{g})] = -394 \text{ kJ mol}^{-1}

Show solution outline
  1. Identify the formula: We are using formation data, so we use ΔHr⊖=∑ΔHf⊖(products)−∑ΔHf⊖(reactants)\Delta H_r^\ominus = \sum \Delta H_f^\ominus(\text{products}) - \sum \Delta H_f^\ominus(\text{reactants}).
  2. Sum the products: ∑ΔHf⊖(products)=(1×ΔHf⊖[CaO(s)])+(1×ΔHf⊖[CO2(g)])\sum \Delta H_f^\ominus(\text{products}) = (1 \times \Delta H_f^\ominus[\text{CaO}(\text{s})]) + (1 \times \Delta H_f^\ominus[\text{CO}_2(\text{g})]) =(1×−635)+(1×−394)=−1029 kJ mol−1= (1 \times -635) + (1 \times -394) = -1029 \text{ kJ mol}^{-1}
  3. Sum the reactants: ∑ΔHf⊖(reactants)=1×ΔHf⊖[CaCO3(s)]=−1207 kJ mol−1\sum \Delta H_f^\ominus(\text{reactants}) = 1 \times \Delta H_f^\ominus[\text{CaCO}_3(\text{s})] = -1207 \text{ kJ mol}^{-1}
  4. Calculate ΔHr⊖\Delta H_r^\ominus: ΔHr⊖=(−1029)−(−1207)\Delta H_r^\ominus = (-1029) - (-1207) ΔHr⊖=−1029+1207=+178 kJ mol−1\Delta H_r^\ominus = -1029 + 1207 = +178 \text{ kJ mol}^{-1}

Answer: The standard enthalpy change of reaction is +178 kJ mol⁻¹.

Worked example 2

Calculate the standard enthalpy change for the hydrogenation of ethene to ethane. C2H4(g)+H2(g)→C2H6(g)\text{C}_2\text{H}_4(\text{g}) + \text{H}_2(\text{g}) \rightarrow \text{C}_2\text{H}_6(\text{g})

Use the following standard enthalpy of combustion data: ΔHc⊖[C2H4(g)]=−1411 kJ mol−1\Delta H_c^\ominus[\text{C}_2\text{H}_4(\text{g})] = -1411 \text{ kJ mol}^{-1} ΔHc⊖[H2(g)]=−286 kJ mol−1\Delta H_c^\ominus[\text{H}_2(\text{g})] = -286 \text{ kJ mol}^{-1} ΔHc⊖[C2H6(g)]=−1560 kJ mol−1\Delta H_c^\ominus[\text{C}_2\text{H}_6(\text{g})] = -1560 \text{ kJ mol}^{-1}

Show solution outline
  1. Identify the formula: We are using combustion data, so we use ΔHr⊖=∑ΔHc⊖(reactants)−∑ΔHc⊖(products)\Delta H_r^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products}).
  2. Sum the reactants: ∑ΔHc⊖(reactants)=(1×ΔHc⊖[C2H4(g)])+(1×ΔHc⊖[H2(g)])\sum \Delta H_c^\ominus(\text{reactants}) = (1 \times \Delta H_c^\ominus[\text{C}_2\text{H}_4(\text{g})]) + (1 \times \Delta H_c^\ominus[\text{H}_2(\text{g})]) =(1×−1411)+(1×−286)=−1697 kJ mol−1= (1 \times -1411) + (1 \times -286) = -1697 \text{ kJ mol}^{-1}
  3. Sum the products: ∑ΔHc⊖(products)=1×ΔHc⊖[C2H6(g)]=−1560 kJ mol−1\sum \Delta H_c^\ominus(\text{products}) = 1 \times \Delta H_c^\ominus[\text{C}_2\text{H}_6(\text{g})] = -1560 \text{ kJ mol}^{-1}
  4. Calculate ΔHr⊖\Delta H_r^\ominus: ΔHr⊖=(−1697)−(−1560)\Delta H_r^\ominus = (-1697) - (-1560) ΔHr⊖=−1697+1560=−137 kJ mol−1\Delta H_r^\ominus = -1697 + 1560 = -137 \text{ kJ mol}^{-1}

Answer: The standard enthalpy change of reaction is -137 kJ mol⁻¹.

Mark a 9701 question on this topicFull lessonOpen in course studio