Chemistry 2720 Fall 2000 Assignment 5 Solutions

1.

At constant temperature and pressure, questions of spontaneity can be resolved by computing the Gibbs free energy change. The reaction is

$$\mathrm{CO_{(g)}} + 2\mathrm{H_{2(g)}} \rightarrow \mathrm{CH_3OH_{(l)}}.$$

The standard free energy change is therefore

$$\begin{eqnarray*}\Delta\bar{G}^\circ & = & \Delta\bar{G}^\circ_{f(\mathrm{CH_3OH... ...- (-137.17)\,\mathrm{kJ/mol}\\ & = & -26.13\,\mathrm{kJ/mol}. \end{eqnarray*}$$

From the pressures of CO and of H₂, we obtain $a_\mathrm{CO} = a_\mathrm{H_2} = (0.5\,\mathrm{bar})/(1\,\mathrm{bar}) = 0.5$. The free energy change under the stated conditions is therefore

$$\begin{eqnarray*}\Delta\bar{G} & = & -26.13\,\mathrm{kJ/mol}\\ && \mbox{}+ (8... ...frac{1}{(0.5)(0.5^2)}\right)\\ & = & -20.98\,\mathrm{kJ/mol}. \end{eqnarray*}$$

Since the free energy change is negative, the reaction would be spontaneous. There is therefore no thermodynamic reason why this would not work.

2.

(a)

Since the reaction is isothermal and electrical work is clearly not pressure-volume work, we want to calculate the change in Gibbs free energy. The reaction is

$$\mathrm{H_{2(g)}} + \frac{1}{2}\mathrm{O_{2(g)}} \rightarrow \mathrm{H_2O_{(l)}}.$$

For this reaction, $\Delta\bar{G}^\circ = \Delta\bar{G}^\circ_{f\mathrm{(H_2O)}} = -237.140\,\mathrm{kJ/mol}$. The activity of oxygen is

$$a_\mathrm{O_2} = \frac{P}{P^\circ} = \frac{0.2\,\mathrm{atm}}... ....2(101\,325\,\mathrm{Pa})}{100\,000\,\mathrm{Pa}} = 0.202\,65.$$

Similarly, $a_\mathrm{H_2} = 1.013\,25$. Therefore

$$\begin{eqnarray*}\Delta\bar{G} & = & -237.140\,\mathrm{kJ/mol}\\ && \mbox{}+ ... ...^{1/2})(1.013\,25)}\right)\\ & = & -235.194\,\mathrm{kJ/mol}. \end{eqnarray*}$$

The molar mass of molecular hydrogen is 2.02g/mol, so

$$\Delta\tilde{G} = \frac{-235.194\,\mathrm{kJ/mol}}{2.02\,\mathrm{g/mol}} = -116\,\mathrm{kJ/g} \equiv -116\,\mathrm{MJ/kg}.$$

The maximum work is therefore 116MJ/kg.  

(b)

To calculate the free energy change at different temperatures, we need $\Delta\bar{H}^\circ$ and $\Delta\bar{S}^\circ$. Using a thermodynamic table, we find $\Delta\bar{H}^\circ = -285.830\,\mathrm{kJ/mol}$. Therefore

$$\begin{eqnarray*}\Delta\bar{S}^\circ & = & \frac{\Delta\bar{H}^\circ-\Delta\bar... ...,\mathrm{K}}\\ & = & -0.163\,31\,\mathrm{kJ\,K^{-1}mol^{-1}}. \end{eqnarray*}$$

Accordingly,

$$\begin{eqnarray*}\Delta\bar{G}^\circ_{0^\circ\mathrm{C}} & = & \Delta\bar{H}^\... ...athrm{kJ\,K^{-1}mol^{-1}})\\ & = & -241.223\,\mathrm{kJ/mol}. \end{eqnarray*}$$

The odd notation $\Delta\bar{G}^\circ_{0^\circ\mathrm{C}}$ indicates that this $\Delta\bar{G}$ corresponds to standard conditions (reactants and products at unit activity), except for the temperature. The rest of the calculation is identical to that in question 2a:

$$\begin{eqnarray*}% latex2html id marker 103 \Delta\bar{G} & = & -241.223\,\mathr... ...\mathrm{kJ/mol}}{2.02\,\mathrm{g/mol}} = -119\,\mathrm{MJ/kg}. \end{eqnarray*}$$

The calculation at $-20^\circ\mathrm{C}$ is identical:

$$\begin{eqnarray*}% latex2html id marker 122 \Delta\bar{G}^\circ_{-20^\circ\mathr... ...\mathrm{kJ/mol}}{2.02\,\mathrm{g/mol}} = -120\,\mathrm{MJ/kg}. \end{eqnarray*}$$

The maximum work is 119MJ/kg at 0 and 120MJ/kg at $-20^\circ\mathrm{C}$. A heating system is therefore not required from the point of view of efficiency, but would be required to prevent the aqueous medium from freezing.

3.

$$% latex2html id marker 243 \begin{array}{rcl} \Delta\bar{G} ... ... \Delta\bar{S} & = & \Delta\bar{S}^\circ - R\ln Q. \end{array}$$