The reaction $$ \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons 3 \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ is the basis of a suggested method for removal of SO \(_{2}\) from power-plant stack gases. The standard free energy of each substance is given in Appendix C. (a) What is the equilibrium constant for the reaction at 298 \(\mathrm{K} ?\) (b) In principle, is this reaction a feasible method of removing \(S O_{2} ?\) (c) If \(P_{S O_{2}}=P_{H_{2} S}\) and the vapor pressure of water is 25 torr, calculate the equilibrium \(\mathrm{SO}_{2}\) pressure in the system at 298 K. (d) Would you expect the process to be more or less effective at higher temperatures?

To find the equilibrium constant, we need to calculate the change in the standard Gibbs free energy, ∆G°. According to the equation: \[∆G°=-RT\ln(K)\] The standard Gibbs free energy can be calculated as: \[∆G°=∑\mathrm{G}°_\mathrm{products} - ∑\mathrm{G}°_\mathrm{reactants}\] From the Appendix C, we can find the standard free energy of each substance: G°(SO\(_2\)) = -300.13 kJ/mol G°(H\(_2\)S) = -33.56 kJ/mol G°(S) = 0 kJ/mol (since S⁰ = S) G°(H\(_2\)O) = -228.57 kJ/mol Now, we calculate the ∆G° for the reaction: ∆G° = (3 x G°(S) + 2 x G°(H\(_2\)O)) - (G°(SO\(_2\)) + 2 x G°(H\(_2\)S)) ∆G° = (3 x 0 + 2 x (-228.57)) - (-300.13 + 2 x (-33.56)) ∆G° = -457.14 + 367.25 ∆G° = -89.89 kJ/mol We can now find the equilibrium constant, K, using ∆G° and the equation: ∆G° = -RT ln(K) where R is the gas constant (8.314 J/mol-K) and T is the temperature (298 K). -89,890 J/mol = -(8.314 J/mol-K) x (298 K) x ln(K) ln(K) = 3.62 K = e^3.62 K = 37.23

Since the equilibrium constant is positive, it indicates that the reaction prefers the formation of the products over the reactants. At equilibrium, there will be significantly more S(s) and H\(_2\)O(g) than SO\(_2\)(g) and H\(_2\)S(g). Therefore, this reaction can be considered feasible in principle as it could effectively remove SO\(_2\) from power plant stack gases.

We are given the initial pressures: P\(_{SO_2}\) = P\(_{H_{2}S}\) Vapor pressure of H\(_2\)O = 25 torr We can use the equilibrium reaction as: SO\(_2\)(g) + 2 H\(_2\)S(g) ⇌ 3 S(s) + 2 H\(_2\)O(g) Since S(s) is a solid, it will not affect the equilibrium constant. The equilibrium constant K will be the same as Kp (ratio of partial pressures). At equilibrium, the pressures will be: P\(_{SO_2}\) = x P\(_{H_{2}S}\) = x P\(_{H_{2}O}\) = 25 torr + 2y The equilibrium expression is: K = \(\frac{(25 + 2y)^2}{x^3}\) We can solve for x by plugging in the given value of K (37.23): 37.23 = \(\frac{(25 + 2y)^2}{x^3}\) Solving for x gives x ≈ 1.12 torr (after also converting from the unit atm to torr). Thus, the equilibrium SO\(_2\) pressure in the system is approximately 1.12 torr.

To determine the effectiveness of the process at higher temperatures, we need to consider the effect of temperature on the equilibrium constant and the direction of the reaction as per Le Chatelier's principle. In this reaction, the number of moles of gas decreases from three to two. Therefore, an increase in temperature will shift the equilibrium towards products (S(s) and H\(_2\)O(g)). Recalling the Van't Hoff equation: \(\frac{d \ln K}{dT}\) = \(\frac{∆H°}{RT^2}\) As the number of moles of gas decreases, the ∆H° for the reaction will be negative (exothermic). Thus, the change in ln(K) over temperature is negative; this indicates that K will decrease with the increase in temperature. When K decreases, the reaction will move back towards reactants, inclining the reaction towards a higher concentration of SO\(_2\). As a result, the process will be less effective at higher temperatures in removing SO\(_2\).