The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at 298 \(\mathrm{K}\) is $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad K=1.28 \times 10^{83}$$ a. Calculate \(8^{\circ}\) and \(\Delta G^{\circ}\) at 298 \(\mathrm{K}\) for the fuel cell reaction. b. Predict the signs of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the fuel cell reaction. c. As temperature increases, does the maximum amount of work obtained from the fuel cell reaction increase, decrease, or remain the same? Explain.

We use the relation between ΔG° and the equilibrium constant, K: ΔG° = -RT ln(K) where R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin (298 K), and K is the equilibrium constant (1.28 × 10^83). ΔG° = - (8.314 J/mol K) × (298 K) × ln(1.28 × 10^83) ΔG° = - (245943 J/mol) So, the value of ΔG° is -245943 J/mol.

We know that, at a constant temperature, ΔG° is the combination of ΔH° and ΔS°. ΔG° = ΔH° - TΔS° We have already calculated the value of ΔG°. Now, we need to analyze the reaction. This is an exothermic reaction because hydrogen and oxygen combine to produce water, which releases energy. Therefore, ΔH° should be negative. Since the reaction involves the conversion of gases to a liquid, the system becomes more ordered. This increase in order leads to a decrease in entropy, and thus, ΔS° should be negative. Therefore, the signs of ΔH° and ΔS° are both negative for the fuel cell reaction.

To understand how temperature affects the maximum amount of work obtained from the fuel cell reaction, we must analyze the effect of temperature on the equilibrium constant K and on ΔG°. Since the reaction is exothermic (ΔH° is negative), an increase in temperature will shift the equilibrium towards the reactants. This implies that the equilibrium constant K will decrease with increasing temperature. Recall that ΔG° is the maximum amount of work that can be extracted from the system. From the equation: ΔG° = ΔH° - TΔS° As the temperature increases, the term TΔS° will also increase. Since both ΔH° and ΔS° are negative, TΔS° will be a positive quantity with a larger magnitude as temperature increases. This causes ΔG° to become less negative as the temperature increases. Therefore, as the temperature increases, the maximum amount of work obtained from the fuel cell reaction will decrease.