IB Chemistry HL R1.2 Energy cycles in reactions questions

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Type
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Question 1

Paper 2Easy4 marks

Ethanol, C₂H₅OH(l), is used as a fuel. Its complete combustion produces carbon dioxide and liquid water. The standard enthalpies of formation at 298 K are:

SubstanceΔH°f / kJ mol⁻¹
C₂H₅OH(l)−277.7
O₂(g)0
CO₂(g)−393.5
H₂O(l)−285.8
(a)

Write the balanced equation for the complete combustion of ethanol, including state symbols.

[2]
(b)

Using the equation in (a) and the equation for enthalpies of formation in the data booklet, determine the standard enthalpy change for the combustion of 1 mol of ethanol. Give your answer in kJ mol⁻¹.

[2]

Question 2

Multiple choiceMedium1 mark

The following standard enthalpy changes are given for a Born–Haber cycle for MgCl₂(s). The chloride electron-affinity step occurs twice. Which value and interpretation of the final step are consistent with the cycle?

ProcessΔH° / kJ mol⁻¹
Mg(s) → Mg(g)+148
Mg(g) → Mg⁺(g) + e⁻+738
Mg⁺(g) → Mg²⁺(g) + e⁻+1451
Cl₂(g) → 2Cl(g)+243
Cl(g) + e⁻ → Cl⁻(g)−349
Mg(s) + Cl₂(g) → MgCl₂(s)−642
Mg²⁺(g) + 2Cl⁻(g) → MgCl₂(s)?
Show answer and explanation

Answer: B. The enthalpy changes around the cycle sum to the standard enthalpy of formation: 148 + 738 + 1451 + 243 + 2(−349) + ΔH_latt = −642, so ΔH_latt = −2524 kJ mol⁻¹. The negative value means energy is released as gaseous ions form MgCl₂(s).

A: counts only one electron-affinity step, giving −2873 kJ mol⁻¹.
C: omits the second ionization energy, giving −1073 kJ mol⁻¹.
D: reverses the sign of the lattice step, making formation from gaseous ions endothermic.

Remember: the enthalpy changes around a complete Hess cycle sum to zero.

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