For the equilibrium (s) ⇌ (g) + (g), equilibrium pressure is X bar at T K. The value of ΔrG° for the — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Write the Decomposition Equation and Identify Phases The thermal decomposition of solid ammonium hydrosulfide (\ce{NH4HS}) is represented by the heterogeneous equilibrium: \[\ce{NH4HS(s) <=> NH3(g) + H2S(g)}\] In this heterogeneous system: * The reactant \ce{NH4HS(s)} is in the solid phase. The active mass (activity) of a pure solid is constant and taken as unity (\(1\)): \[a_{\ce{NH4HS(s)}} = 1\] * Consequently, the solid reactant does not appear in the expression for \(K_p\). * The products, ammonia (\ce{NH3}) and hydrogen sulfide (\ce{H2S}), are both in the gaseous phase and contribute to the total equilibrium pressure. Step 2 - Determine the Partial Pressures of the Gaseous Products Let \(P_{\text{total}}\) be the total equilibrium pressure. We are given: \[P_{\text{total}} = X\text{ bar}\] From stoichiometry, decomposition of 1 mole of solid produces 1 mole of \ce{NH3(g)} and 1 mole of \ce{H2S(g)}: \[n_{\ce{NH3}} = n_{\ce{H2S}}\] By Dalton's Law, since mole fractions are equal, partial pressures are equal: \[p_{\ce{NH3}} + p_{\ce{H2S}} = X\text{ bar} \implies p_{\ce{NH3}} = p_{\ce{H2S}} = \frac{X}{2}\text{ bar}\] Step 3 - Express the Equilibrium Constant (\(K_p\)) \[K_p = p_{\ce{NH3}} \cdot p_{\ce{H2S}} = \left(\frac{X}{2}\right)\left(\frac{X}{2}\right) = \frac{X^2}{4}\text{ bar}^2\] Step 4 - Calculate \(\Delta_r G^\circ\) Using the thermodynamic relation: \[\Delta_r G^\circ = -RT \ln K_p = -RT \ln\left(\frac{X^2}{4}\right)\] Simplify using logarithm rules: \[\ln\left(\frac{X^2}{4}\right) = \ln\left[\left(\frac{X}{2}\right)^2\right] = 2\ln\left(\frac{X}{2}\right) = 2(\ln X - \ln 2)\] Therefore: \[\Delta_r G^\circ = \boxed{-2RT(\ln X - \ln 2)}\] Step 5 - Evaluate the Options * **Option (A) \(-RT \ln X\)**: Incorrect. Assumes \(K_p = X\), ignoring the two products and equal partial pressures. * **Option (B) \(-2RT \ln X\)**: Incorrect. Omits the division by 2, assuming \(K_p = X^2\). * **Option (C) \(-2RT(\ln X - \ln 2)\)**: Correct. Correctly uses \(p_{\ce{NH3}} = p_{\ce{H2S}} = X/2\) giving \(K_p = X^2/4\). * **Option (D) \(-2RT \ln(2X)\)**: Incorrect. Mathematical error in logarithm manipulation.