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 (\(M_r = 159.6\)), 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 \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\) 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 \(q = m \times c \times \Delta T = 50.0 \times 4.18 \times 11.5 = 2403.5\text{ J} = 2.4035\text{ kJ}\). (Allow \(m = 53.99\text{ g}\), giving \(q = 2595.3\text{ J} = 2.5953\text{ kJ}\)). [1]
- Moles of \(\text{CuSO}_4 = \text{mass} / M_r = 3.99 / 159.6 = 0.0250\text{ mol}\). [1]
- \(\Delta H = -q / n = -2.4035 / 0.0250 = -96.1\text{ kJ mol}^{-1}\). (If \(m = 53.99\text{ g}\) is used, ΔH = −104 kJ mol⁻¹). [1]
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 \(\Delta H = \sum(\Delta_c H\text{ of reactants}) - \sum(\Delta_c H\text{ of products})\):
ΔH = [4 × (−393.5) + 5 × (−285.8)] − [−2876.5] [1] - ΔH = [−1574.0 − 1429.0] + 2876.5 = −126.5 kJ mol⁻¹ [1]
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.
• \(\text{N}\equiv\text{N}\): 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 \(= 1 \times (\text{N}\equiv\text{N}) + 3 \times (\text{H}-\text{H}) = 945 + 3 \times 436 = 2253\text{ kJ}\). [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]
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