Question 1 – Thermodynamics

(a) Signs of $\Delta G^\circ$ and spontaneity

(i) Exothermic with $\Delta S > 0$ gives $\Delta G^\circ = \Delta H^\circ - T\Delta S < 0$ at all $T$ therefore reaction is spontaneous at every temperature.

(ii) Endothermic with $\Delta S > 0$ gives $\Delta G^\circ$ negative only when $T\Delta S > \Delta H^\circ$ so reaction is spontaneous only at sufficiently high $T$ and non-spontaneous at low $T$

(iii) The first process is always spontaneous while the second becomes spontaneous on heating.

(b) Oxidation of glucose

(i) Standard enthalpy of formation of $\mathrm O_2(g)$

By definition $\Delta_\mathrm f H^\circ$ of an element in its reference state is $0\text{ kJ mol}^{-1}$.

(ii) Standard enthalpy change of reaction

$\Delta_\mathrm r H^\circ = [6(-393.5) + 6(-285.8)] - [-1268 + 6(0)] = -2807.8\text{ kJ}$

The negative sign confirms the reaction is strongly exothermic.

(iii) Comment already covered in part (ii).

(iv) Entropy changes

$\Delta_\mathrm r S^\circ = [6(213.8) + 6(69.9)] - [209.2 + 6(205.2)] = +261.8\text{ J K}^{-1}$

$\Delta S_\text{surr} = -\frac{\Delta_\mathrm r H^\circ}{T} = -\frac{-2807.8\times10^{3}\text{ J}}{298\text{ K}} = +9.43\times10^{3}\text{ J K}^{-1}$

$\Delta S_\text{univ} = \Delta_\mathrm r S^\circ + \Delta S_\text{surr} \approx +9.69\times10^{3}\text{ J K}^{-1}$

A positive $\Delta S_\text{univ}$ confirms spontaneity.

(v) Standard Gibbs energy at $298\text{ K}$

$\Delta_\mathrm r G^\circ = \Delta_\mathrm r H^\circ - T\Delta_\mathrm r S^\circ = -2807.8\text{ kJ} - 298(0.2618\text{ kJ K}^{-1}) = -2.89\times10^{3}\text{ kJ}$

Negative value indicates a thermodynamically favourable process.