IB Chemistry HL Reactivity 1: What drives chemical reactions? questions

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Question 1

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?

Process	ΔH° / 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)	?

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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.

Question 2

Multiple choiceMedium1 mark

A 23.0 g sample of bioethanol, C₂H₅OH, is burned completely. Assume all its carbon was fixed from atmospheric CO₂ by recent photosynthesis, and ignore emissions from production and transport. The combustion equation is C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O. Use M(C₂H₅OH) = 46.0 g mol⁻¹ and M(CO₂) = 44.0 g mol⁻¹. What are the mass of CO₂ released and the net change in atmospheric CO₂ from fixation and combustion together? Treat an increase as positive.

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Answer: C. The equation gives 2 mol CO₂ per mole of ethanol. The ethanol amount is 23.0/46.0 = 0.500 mol, so combustion releases 1.00 mol CO₂ × 44.0 g mol⁻¹ = 44.0 g. The same 1.00 mol of atmospheric CO₂ had been fixed to provide the carbon in the ethanol, so the net change is +44.0 − 44.0 = 0.0 g.

A: uses a 1:1 rather than 1:2 ethanol-to-CO₂ mole ratio, giving 22.0 g. B: calculates the released CO₂ but omits the earlier photosynthetic uptake. D: counts the uptake as a negative change but omits the CO₂ released on combustion.

Remember: for the stated carbon balance, recently fixed carbon returns to the atmosphere on combustion.

Question 3

Multiple choiceMedium1 mark

For the reaction shown, the standard molar entropies are S°(H₂O₂(l)) = 109.6, S°(H₂O(l)) = 69.9 and S°(O₂(g)) = 205.0 J K⁻¹ mol⁻¹. Which option gives ΔᵣS° per mole of reaction as written and the resulting change in the system’s entropy?

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Answer: C. For a reaction, ΔᵣS° = ΣνS°(products) − ΣνS°(reactants). Using the coefficients read from the figure: [2(69.9) + 205.0] − [2(109.6)] = +125.6 J K⁻¹ mol⁻¹. The positive value means the system’s entropy increases.

A: reverses products minus reactants, giving −125.6 J K⁻¹ mol⁻¹ and a decrease.
B: omits the coefficient 2 for water, giving +55.7 J K⁻¹ mol⁻¹.
D: omits the coefficient 2 for hydrogen peroxide, giving +235.2 J K⁻¹ mol⁻¹.

Remember: multiply each standard molar entropy by its stoichiometric coefficient before summing.

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