Question

a) Define all the symbols in, and briefly describe the significance of, the equation
DG=Vdp — SdT

b) Given that at 100 °C and atmospheric pressure, water is in equilibrium with steam, use the equation in part (a) to deduce that at 101 °C and atmospheric pressure, water will convert spontaneously to steam.

C) Making use of van't Hoff's equation,


deduce the shape of a graph of In K versus T"' for a reaction in which the heat capacities of reactants and products are approximately equal.

d) Using the following data for the temperature variation of the equilibrium constant for the dissociation of gaseous di-iodine to iodine atoms, obtain the standard molar enthalpy change expressed in kJ mol-1.

T I K 1274 1073 973 872
K 0.16780 0.010898 0.0018012 0.00018120

Answer 1

a) The differential form of the Gibbs energy is:

$$\mathrm{d}G = V \mathrm{d}P - S \mathrm{d}T$$

This differential can be used to determine both the pressure and temperature dependence of the free energy.

G – Gibbs free energy, dG – differential form of the Gibbs free energy, S – entropy, V – volume, T – temperature, P – pressure.

b) $\Delta G = \Delta H - T\Delta S = -286000\mathrm{J/mol} - 373.15\mathrm{K} \cdot 70\mathrm{J/mol} \cdot \mathrm{K} = -312120.5\mathrm{J/mol}$

$\Delta G < 0$, the process will proceed spontaneously.

c) $\mathrm{d}\ln K_{\mathrm{eq}} = -\frac{\Delta H^0}{R} \, \mathrm{d}(1 / T)$

$$\ln \left(\frac {K _ {2}}{K _ {1}}\right) = - \frac {\Delta H ^ {0}}{R} \left(\frac {1}{T _ {2}} - \frac {1}{T _ {1}}\right)$$

From this form of the van't Hoffs equation, we see that at constant pressure, a plot with $\ln K_{\mathrm{eq}}$ on the y-axis and $1 / T$ on the x-axis has a slope given by $-\Delta H / R$. For an endothermic reaction, the slope is negative, for an exothermic reaction, the slope is positive.

d) $\ln (K_2) - \ln (K_1) = \frac{\Delta H^0}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

$$\begin{array}{l} \ln (0.010898) - \ln (0.1678) = \frac {\Delta H ^ {0}}{R} \left(\frac {1}{1274} - \frac {1}{1073}\right) \\ -4.52 + 1.78 = \frac {\Delta H ^ {0}}{R} (0.000785 - 0.000932) \\ -2.74 = \frac {\Delta H ^ {0}}{R} (-0.000147) \\ \end{array}$$

$\frac{\Delta H^0}{R} = 18639.5 \, \mathrm{kJ/mol}$ – the standard molar enthalpy change