Hydrogen sulfide can be removed from natural gas by the reaction $$2 \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{~S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ Calculate \(\Delta G^{\circ}\) and \(K\) (at \(298 \mathrm{~K}\) ) for this reaction. Would this reaction be favored at a high or low temperature?

To find \(\Delta G^{\circ}\) for the reaction, we use the following relationship for reactions involving gases: $$\Delta G^{\circ} = \sum_{products}(\Delta G^{\circ}_{prod} . \nu_{prod}) - \sum_{reactants}(\Delta G^{\circ}_{reac} . \nu_{reac})$$ where \(\Delta G^{\circ}_{prod}\) and \(\Delta G^{\circ}_{reac}\) are the standard Gibbs free energy of formation values for the products and reactants, respectively, and \(\nu_{prod}\) and \(\nu_{reac}\) are the stoichiometric coefficients for each species in the balanced chemical equation. We can look up the standard Gibbs free energy of formation values for each species from a standard thermodynamics table, which gives: \(\Delta G^{\circ}_{H_{2}S(g)} = -33.6 \,\text{kJ}\,\text{mol}^{-1}\) \(\Delta G^{\circ}_{SO_{2}(g)} = -300.1 \,\text{kJ}\,\text{mol}^{-1}\) \(\Delta G^{\circ}_{S(s)} = 0 \,\text{kJ}\,\text{mol}^{-1}\) (Since the element in its most stable state has \(\Delta G^{\circ} = 0\)) \(\Delta G^{\circ}_{H_{2}O(g)} = -228.6 \,\text{kJ}\,\text{mol}^{-1}\) Now, we can substitute these values into the equation for \(\Delta G^{\circ}\): $$\Delta G^{\circ} = (3\Delta G^{\circ}_{S(s)} + 2\Delta G^{\circ}_{H_{2}O(g)}) - (2\Delta G^{\circ}_{H_{2}S(g)} + \Delta G^{\circ}_{SO_{2}(g)})$$

Now, we can plug in the values for the standard Gibbs free energy of formation for each species and solve for \(\Delta G^{\circ}\): $$\Delta G^{\circ} = (3(0) + 2(-228.6)) - (2(-33.6) + (-300.1))$$ $$\Delta G^{\circ} = (-457.2) - (-367.3)$$ $$\Delta G^{\circ} = -89.9 \, \text{kJ}\, \text{mol}^{-1}$$

We have the following relationship between the standard Gibbs free energy change, the gas constant \(R\), the temperature \(T\), and the equilibrium constant \(K\): $$\Delta G^{\circ} = -RT\ln K$$ We are asked to find \(K\) at \(T = 298 \,\text{K}\), and we know the gas constant \(R = 8.314 \,\text{J} \, \text{mol}^{-1} \, \text{K}^{-1}\). Rearranging the equation to solve for \(K\) and converting the units of \(\Delta G^{\circ}\) to J/mol to match the units of \(R\), we get: $$K = e^{\frac{-\Delta G^{\circ}}{RT}}$$ $$K = e^{\frac{-(-89.9 \, \times \, 1000)}{(8.314)(298)}}$$

Now, using the exponential function, we can calculate the equilibrium constant \(K\): $$K = e^{\frac{89.9 \, \times \, 1000}{8.314 \times 298}}$$ $$K \approx 2.03 \times 10^6$$

To determine whether the reaction is favored at high or low temperatures, we need to examine the van 't Hoff equation, which relates the equilibrium constant \(K\) to the temperature \(T\), the standard enthalpy change \(\Delta H^{\circ}\), and the gas constant \(R\): $$\frac{d \ln K}{dT} = \frac{\Delta H^{\circ}}{RT^2}$$ As the sign of \(\frac{d\ln K}{dT}\) depends only on the sign of \(\Delta H^{\circ}\), we can determine the temperature dependence of the reaction based on the standard enthalpy change alone. Using the relationship between \(\Delta G^{\circ}\), \(\Delta H^{\circ}\), and the standard entropy change \(\Delta S^{\circ}\): $$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$$ We see that the \(\Delta G^{\circ}\) is negative for the given reaction and entropy change \(\Delta S^{\circ}\) is also usually negative for exothermic reactions with fewer moles of gas in the product side. Therefore, the reaction is exothermic (\(\Delta H^{\circ} < 0\)). By examining the van 't Hoff equation, we can deduce that if \(\Delta H^{\circ} < 0\), the equilibrium constant \(K\) will decrease with increasing temperature. Hence, the reaction is favored at low temperatures. In conclusion, the standard Gibbs free energy change for the given reaction is \(\Delta G^{\circ} = -89.9 \,\text{kJ}\,\text{mol}^{-1}\), and the equilibrium constant at 298 K is \(K \approx 2.03 \times 10^6\). The reaction is favored at low temperatures.