Consider the reaction $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ where \(\Delta H^{\circ}=-103.8 \mathrm{kJ} / \mathrm{mol} .\) In a particular experiment, equal moles of \(\mathrm{H}_{2}(g)\) at 1.00 atm and \(\mathrm{Br}_{2}(g)\) at 1.00 atm were mixed in a 1.00-L flask at \(25^{\circ} \mathrm{C}\) and allowed to reach equilibrium. Then the molecules of \(\mathrm{H}_{2}\) at equilibrium were counted using a very sensitive technique, and \(1.10 \times 10^{13}\) molecules were found. For this reaction, calculate the values of \(K, \Delta G^{\circ},\) and \(\Delta S^{\circ} .\)

The initial pressures of H2 and Br2 are both 1.00 atm, which are equal. Given that the volume of the container is 1.00 L, we can use the ideal gas law (\(PV = nRT\)) to find the initial moles of the gases: For H2: $$n_{H_{2}} = \frac{P_{H_{2}}V}{RT} = \frac{1.00 \,\text{atm} \cdot 1.00\, \text{L}}{0.0821\, \frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot\text{K}} \cdot (25+273.15)\,\text{K}}\approx 0.0409\, \text{mol}$$ Similarly, for Br2: $$n_{Br_{2}} = \frac{P_{Br_{2}}V}{RT} = \frac{1.00 \,\text{atm} \cdot 1.00\, \text{L}}{0.0821\, \frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot\text{K}} \cdot (25+273.15)\,\text{K}}\approx 0.0409\, \text{mol}$$

Given the number of H2 molecules at equilibrium (1.10 x 10^13), we can convert it to moles using Avogadro's number (\(6.022\times10^{23}\, \text{molecules/mol}\)): $$n_{H_{2\,eq}} = \frac{1.10 \times 10^{13}\, \text{molecules}}{6.022\times10^{23}\, \text{molecules/mol}} \approx 1.83\times10^{-11}\, \text{mol}$$ We can find the change in moles of H2 and Br2, and the moles of HBr formed: $$\Delta n_{H_{2}} = \Delta n_{Br_{2}} = 0.0409\, \text{mol} - 1.83\times10^{-11}\, \text{mol} \approx 0.0409\, \text{mol}$$ $$n_{HBr} = 2 \times (0.0409\, \text{mol} - \Delta n_{H_{2}}) = 2 \times 1.83\times10^{-11}\, \text{mol} = 3.66\times10^{-11}\, \text{mol}$$ Now we can calculate the concentrations at equilibrium using the volume of the flask (1.00 L): $$[\mathrm{H}_{2}]_{\text{eq}} \approx \frac{1.83\times10^{-11}\, \text{mol}}{1.00\, \text{L}} = 1.83\times10^{-11}\, \text{M}$$ $$[\mathrm{Br}_{2}]_{\text{eq}} \approx \frac{0.0409\, \text{mol}}{1.00\, \text{L}} = 0.0409\, \text{M}$$ $$[\mathrm{HBr}]_{\text{eq}} \approx \frac{3.66\times10^{-11}\, \text{mol}}{1.00\, \text{L}} = 3.66\times10^{-11}\, \text{M}$$

We can find the equilibrium constant using the equilibrium concentrations. $$K = \frac{[\mathrm{HBr}]_{\text{eq}}^{2}}{[\mathrm{H}_{2}]_{\text{eq}} \cdot [\mathrm{Br}_{2}]_{\text{eq}}} = \frac{(3.66\times10^{-11}\, \text{M})^2}{(1.83\times10^{-11}\, \text{M}) \cdot (0.0409\, \text{M})}\approx 1.62 \times 10^{-9}$$

We can use the following relationship to find \(\Delta G^{\circ}\): $$\Delta G^{\circ} = -RT\ln{K} = -8.314\,\frac{\text{J}}{\text{mol}\cdot\text{K}}\cdot298.15\,\text{K}\ln{(1.62 \times 10^{-9})} \approx -59.2\,\frac{\text{kJ}}{\text{mol}}$$

Finally, we can calculate the standard entropy change by rearranging the Gibbs-Helmholtz equation: $$\Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T} = \frac{(-103.8\,\mathrm{kJ/mol}-(-59.2\,\mathrm{kJ/mol}))}{298.15\,\mathrm{K}} \approx 149.5\,\frac{\text{J}}{\text{mol}\cdot\text{K}}$$ Thus, for this reaction at given conditions, we have the equilibrium constant \(K \approx 1.62 \times 10^{-9}\), \(\Delta G^{\circ} \approx -59.2\,\frac{\text{kJ}}{\text{mol}}\), and \(\Delta S^{\circ} \approx 149.5\,\frac{\text{J}}{\text{mol}\cdot\text{K}}\).