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Topic 3.1.4: Energetics - Exam Practice

AQA A-Level Chemistry • Physical Chemistry • Exam Practice

Exam Practice

Topic 3.1.4: Energetics Exam Practice

Test your understanding of enthalpy changes, calorimetry experiments, Hess's Law cycles, and mean bond enthalpy calculations.

Energetics › Exam Practice

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📋 Structured Questions

Complete each question on paper, then check your answers against the mark scheme.

Question 1: Calorimetry and Enthalpy calculation

6 marks

(a) State the difference between the standard enthalpy of combustion and the standard enthalpy of formation. [2]

(b) A student carried out an experiment to determine the enthalpy change of solution of anhydrous copper(II) sulfate. 50.0 cm3 of water was placed in a polystyrene cup. The initial temperature of the water was 19.5 °C. After adding 3.99 g of anhydrous copper(II) sulfate (Mr = 159.6), the temperature rose to a maximum of 31.0 °C. Calculate the molar enthalpy of solution of copper(II) sulfate. Give your answer to 3 significant figures. (Assume specific heat capacity c = 4.18 J g^-1 K^-1 and density of solution = 1.00 g cm^-3). [4]

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(a)

  • Standard enthalpy of combustion: Enthalpy change when one mole of a substance is burned completely in oxygen, under standard conditions. [1]
  • Standard enthalpy of formation: Enthalpy change when one mole of a compound is formed from its constituent elements in their standard states, under standard conditions. [1]

(b)

  • Temperature rise ΔT = 31.0 - 19.5 = 11.5 K (or °C). [1]
  • Heat energy change q = m * c * ΔT = 50.0 * 4.18 * 11.5 = 2403.5 J = 2.4035 kJ. (Allow m = 53.99 g, giving q = 2595.3 J = 2.5953 kJ). [1]
  • Moles of CuSO4 = mass / Mr = 3.99 / 159.6 = 0.0250 mol. [1]
  • ΔH = -q / n = -2.4035 / 0.0250 = -96.1 kJ mol^-1. (If m = 53.99 g is used, ΔH = -104 kJ mol^-1). [1]
Examiner tip: The negative sign on the final ΔH value is essential since the temperature increased, meaning the reaction is exothermic. Marks are frequently lost for forgetting this negative sign. Standard practice in AQA allows using either the mass of water alone (50.0 g) or the mass of water + solid (53.99 g); both are accepted as long as the steps are clear.

Question 2: Hess's Law Cycle

4 marks

(a) State Hess's Law. [1]

(b) Calculate the standard enthalpy of formation of butane, C4H10(g), using the following standard enthalpies of combustion:
• C(s) = -393.5 kJ mol^-1
• H2(g) = -285.8 kJ mol^-1
• C4H10(g) = -2876.5 kJ mol^-1 [3]

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(a)

  • The enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final states are the same. [1]

(b)

  • Reaction equation: 4C(s) + 5H2(g) -> C4H10(g) [1]
  • Using ΔH = Σ(ΔcH of reactants) - Σ(ΔcH of products):
    ΔH = [4 * (-393.5) + 5 * (-285.8)] - [-2876.5] [1]
  • ΔH = [-1574.0 - 1429.0] + 2876.5 = -126.5 kJ mol^-1 [1]
Examiner tip: Draw a Hess's Law cycle if you get confused. In a combustion cycle, the arrows point down towards the combustion products (CO2 and H2O). Therefore, the route is reactants to products, which goes with the reactant arrows and against the product arrow.

Question 3: Bond Enthalpies

5 marks

(a) Define the term mean bond enthalpy. [2]

(b) Consider the following reaction: N2(g) + 3H2(g) -> 2NH3(g) (ΔH = -92.0 kJ mol^-1).
Use the following bond enthalpies to calculate the N-H bond enthalpy in ammonia.
• N≡N: 945 kJ mol^-1
• H-H: 436 kJ mol^-1 [3]

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(a)

  • The average enthalpy change required to break one mole of a covalent bond [1]
  • in the gaseous state, averaged over a range of different compounds. [1]

(b)

  • Bonds broken = 1 * (N≡N) + 3 * (H-H) = 945 + 3 * 436 = 2253 kJ. [1]
  • Bonds formed = 6 * (N-H). So, ΔH = bonds broken - bonds formed:
    -92.0 = 2253 - 6 * (N-H) [1]
  • 6 * (N-H) = 2253 + 92.0 = 2345
    N-H bond enthalpy = 2345 / 6 = 390.8 kJ mol^-1 (allow 391 kJ mol^-1). [1]
Examiner tip: Make sure you count the bonds correctly. One ammonia molecule has 3 N-H bonds, so 2 moles of NH3 contain 6 moles of N-H bonds. Double check your algebra when shifting terms across the equals sign.

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