Question

The reversible reaction between hydrogen chloride gas and one mole of oxygen gas produces steam and chlorine gas: $$ 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=0.79 $$ Predict the direction in which the system will move to reach equilibrium if one starts with $$ \text { (a) } P_{\mathrm{H}_{2} \mathrm{O}}=P_{\mathrm{HCl}}=P_{\mathrm{O}_{2}}=0.20 \mathrm{~atm} $$ (b) \(P_{\mathrm{HCl}}=0.30 \mathrm{~atm}, P_{\mathrm{H}_{2} \mathrm{O}}=0.35 \mathrm{~atm}, P_{\mathrm{Cl}_{2}}=0.2 \mathrm{~atm}\) \(P_{\mathrm{O}_{2}}=0.15 \mathrm{~atm}\)

Short answer

Answer: For both initial sets of partial pressures, the reaction will proceed in the reverse direction to reach equilibrium.

Step-by-step solution

Write the expression for the reaction quotient Q

For a given reaction, the reaction quotient, Q, is defined as the ratio of the product of the concentrations (or pressures) of the products raised to their stoichiometric coefficients to the product of the concentrations (or pressures) of the reactants raised to their stoichiometric coefficients. In this case, we can write the expression for Q as: $$ Q = \frac{P_{Cl_2}^2 P_{H_2O}^2}{P_{HCl}^4 P_{O_2}} $$

Calculate Q and compare with K for (a)

Plug in the given pressure values into the Q expression for (a): $$ Q = \frac{(0.20)^2 (0.20)^2}{(0.20)^4 (0.20)} = 1 $$ Now, compare Q with the provided equilibrium constant, K = 0.79. Since Q > K, the reaction will proceed in the reverse direction to reach equilibrium.

Calculate Q and compare with K for (b)

Plug in the given pressure values into the Q expression for (b): $$ Q = \frac{(0.2)^2 (0.35)^2}{(0.3)^4 (0.15)} \approx 3.24 $$ Now, compare Q with the provided equilibrium constant, K = 0.79. Since Q > K, the reaction will also proceed in the reverse direction to reach equilibrium in case (b). In conclusion, for both initial sets of partial pressures, the reaction will proceed in the reverse direction to reach equilibrium.