for the reaction: (g) + (s) ⇌ 2(g) is 6.4 × 10^-4 atm. On close observation, it is found that the pa — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Express the equilibrium constant for the reaction with solid iodine Consider the first equilibrium reaction involving solid iodine: \[\ce{H2(g) + I2(s) <=> 2HI(g)}\] Since the reactant \ce{I2(s)} is in the solid phase, its active mass (or activity) is constant and defined as unity (\(a_{\ce{I2(s)}} = 1\)). Consequently, it does not appear in the expression for the pressure-based equilibrium constant (\(K_{p(\text{solid})}\)): \[K_{p(\text{solid})} = \frac{p_{\ce{HI}}^2}{p_{\ce{H2}}}\] We are given the value: \[K_{p(\text{solid})} = 6.4 \times 10^{-4}\text{ atm}\] Step 2 - Express the equilibrium constant for the reaction with gaseous iodine Consider the second equilibrium reaction, which is fully homogeneous in the gas phase: \[\ce{H2(g) + I2(g) <=> 2HI(g)}\] The pressure-based equilibrium constant for this homogeneous gaseous reaction (\(K_{p(\text{gas})}\)) is: \[K_{p(\text{gas})} = \frac{p_{\ce{HI}}^2}{p_{\ce{H2}} \cdot p_{\ce{I2}}}\] Step 3 - Relate the two equilibrium constants We can find the relationship between the two constants by rearranging the expression for \(K_{p(\text{gas})}\): \[K_{p(\text{gas})} = \frac{1}{p_{\ce{I2}}} \left( \frac{p_{\ce{HI}}^2}{p_{\ce{H2}}} \right)\] Substituting the expression of \(K_{p(\text{solid})} = \frac{p_{\ce{HI}}^2}{p_{\ce{H2}}}\) from Step 1: \[K_{p(\text{gas})} = \frac{K_{p(\text{solid})}}{p_{\ce{I2}}}\] Step 4 - Substitute the given values and calculate the final answer We are given: * \(K_{p(\text{solid})} = 6.4 \times 10^{-4}\text{ atm}\) * Gaseous iodine partial pressure at the same temperature, \(p_{\ce{I2}} = 1.6 \times 10^{-4}\text{ atm}\) Substituting these values into our relation: \[K_{p(\text{gas})} = \frac{6.4 \times 10^{-4}\text{ atm}}{1.6 \times 10^{-4}\text{ atm}}\] \[K_{p(\text{gas})} = \boxed{4}\] This corresponds to Option (B).