Let’s consider the reaction: \[ \mathrm{Cl}_{2(g)} \rightleftarrows 2 \mathrm{Cl}_{(g)} \nonumber \]
We can divide the exercise into two parts. In the first one, we will deal with calculating the equilibrium constant at \(T=2\,298\;\mathrm{K}\) from the data at \(T=298\;\mathrm{K}\). In the second one, we will calculate the change in mole fraction when the pressure is increased from \(P^{-\kern-6pt{\ominus}\kern-6pt-}=1\;\text{bar}\) to \(P_f=2.5 \;\text{bar}\).
Let’s begin the first part by calculating \(\Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}\) and \(\Delta_{\text{rxn}} H^{-\kern-6pt{\ominus}\kern-6pt-}\) from: \[ \begin{aligned} \Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}&= 2 \Delta_{\text{f}} G^{-\kern-6pt{\ominus}\kern-6pt-}_{\text{Cl}_{(g)}} - \Delta_{\text{f}} G^{-\kern-6pt{\ominus}\kern-6pt-}_{\text{Cl}_{2(g)}} \\ \Delta_{\text{rxn}} H^{-\kern-6pt{\ominus}\kern-6pt-}&= 2 \Delta_{\text{f}} H^{-\kern-6pt{\ominus}\kern-6pt-}_{\text{Cl}_{(g)}} - \Delta_{\text{f}} H^{-\kern-6pt{\ominus}\kern-6pt-}_{\text{Cl}_{2(g)}}, \end{aligned} \nonumber \] and since \(\text{Cl}_{2(g)}\) is an element in its most stable form at \(T=298\;\mathrm{K}\), its standard enthalpy and Gibbs free energy of formation are \(\Delta_{\text{f}} H^{-\kern-6pt{\ominus}\kern-6pt-}_{\text{Cl}_{2(g)}} = \Delta_{\text{f}} G^{-\kern-6pt{\ominus}\kern-6pt-}_{\text{Cl}_{2(g)}} = 0\). Therefore:\(^3\) \[ \begin{aligned} \Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}&= 2 \cdot 105.3 - 0 = 210.6 \;\text{kJ/mol} \\ \Delta_{\text{rxn}} H^{-\kern-6pt{\ominus}\kern-6pt-}&= 2 \cdot 121.3 - 0 = 242.6\;\text{kJ/mol}. \end{aligned} \nonumber \] Using Equation 10.1.8 to calculate \(K_P (P^{-\kern-6pt{\ominus}\kern-6pt-},298\;\text{K})\), we obtain:\(^4\) \[ \ln [ K_P (P^{-\kern-6pt{\ominus}\kern-6pt-},298\;\text{K}) ] = \dfrac{ - \Delta_{\text{rxn}} G^{-\kern-6pt{\ominus}\kern-6pt-}}{RT} = \dfrac{-210.6\times10^3}{8.31 \cdot 298} = - 85.0. \nonumber \] We can now use the integrated van ’t Hoff equation, Equation 10.2.5, to calculate \(K_P\) at \(T=2\,298\;\text{K}\): \[ \begin{aligned} \ln [K_P (P^{-\kern-6pt{\ominus}\kern-6pt-},&2\,298\;\text{K})] = \ln [K_P (P^{-\kern-6pt{\ominus}\kern-6pt-},298\;\text{K})] \;+ \\ &-\dfrac{\Delta_{\text{rxn}} H^{-\kern-6pt{\ominus}\kern-6pt-}}{R} \left(\dfrac{1}{2\,298}-\dfrac{1}{298} \right), \end{aligned} \nonumber \] which becomes: \[ \begin{aligned} \ln [K_P (P^{-\kern-6pt{\ominus}\kern-6pt-},&2\,298\;\text{K})] = - 85.0 \;+\\&-\dfrac{242.6\times 10^{3}}{8.31} \left(\dfrac{1}{2\,298}-\dfrac{1}{298} \right) = 0.262\;, \end{aligned} \nonumber \] which corresponds to: \[ K_P (P^{-\kern-6pt{\ominus}\kern-6pt-},2\,298\;\text{K}) = \exp (0.262)=1.30. \nonumber \]
Let’s now move to the second part of the exercise, where we increase the pressure from \(1\;\text{bar}\) to \(2.5\;\text{bar}\) at constant \(T=2\,298\;\text{K}\). We start by writing the definition of \(K_P\) and \(K_y\): \[ K_P=\dfrac{P_\mathrm{Cl_{(g)}}^2}{P_{\mathrm{Cl}_{2(g)}}} \qquad \qquad K_y=\dfrac{y_\mathrm{Cl_{(g)}}^2}{y_{\mathrm{Cl}_{2(g)}}}, \nonumber \] and using Equation 10.1.13: \[ \begin{aligned} \Delta \nu &= 2 - 1 = 1 \\ K_P &= K_y \cdot \dfrac{P}{P^{-\kern-6pt{\ominus}\kern-6pt-}} \quad \xrightarrow \qquad K_y=K_P \left( \dfrac{P}{P^{-\kern-6pt{\ominus}\kern-6pt-}} \right)^{-1}, \end{aligned} \nonumber \] we can calculate the initial \(K_y\) at \(P_i=P^{-\kern-6pt{\ominus}\kern-6pt-}\), using: \[ K_y (P^{-\kern-6pt{\ominus}\kern-6pt-},2\,298\;\text{K}) = 1.30 =\dfrac{1.30}{1}. \nonumber \] and calculate the initial concentration of \(\mathrm{Cl}_{(g)}\) and \(\mathrm{Cl}_{(g)}\) at \(P^{-\kern-6pt{\ominus}\kern-6pt-}\), recalling that \(y_{\mathrm{Cl}_{2(g)}}=1-y_{\mathrm{Cl}_{(g)}}:\) \[ K_y (P_i,2\,298\;\text{K})=\dfrac{\left(y^i_{\mathrm{Cl}_{(g)}}\right)^2}{y^i_{\mathrm{Cl}_{(g)}}} = 1.30. \nonumber \] Solving the quadratic equation, we obtain one negative answer—which is unphysical—,\(^5\) and: \[ y_{\mathrm{Cl}_{(g)}}^i= 0.662 \quad \xrightarrow \qquad y_{\mathrm{Cl}_{2(g)}}^i=1-0.662 = 0.338. \nonumber \] At the end of the process, \(P_f=2.5\;\text{bar}\), and we obtain: \[ K_y (P_f,2\,298\;\text{K}) = 0.520 = K_P \dfrac{P^{-\kern-6pt{\ominus}\kern-6pt-}}{P_f} = \dfrac{1.30}{2.5}, \nonumber \] and, using the same technique used before to solve the quadratic equation: \[ K_y (P_f,2\,298\;\text{K})=\dfrac{\left(y^f_{\mathrm{Cl}_{(g)}}\right)^2}{y^f_{\mathrm{Cl}_{(g)}}} = 0.520, \nonumber \] gives: \[ y_{\mathrm{Cl}_{(g)}}^f=0.507 \quad \xrightarrow \qquad y_{\mathrm{Cl}_{2(g)}}^i=1-0.507 = 0.493. \nonumber \] To summarize, when we increase the pressure from \(1\;\text{bar}\) to \(2.5\;\text{bar}\) at \(T=2\,298\;\text{K}\), the equilibrium constant in terms of the mole fraction decreases from \(K_y(P^{-\kern-6pt{\ominus}\kern-6pt-},2\,298\;\text{K})=1.30\) to \(K_y(P_f=2.5\;\text{bar},2\,298\;\text{K})=0.520\). This reduction is causing a shift of the equilibrium towards the reactants, with the concentration of \(\text{Cl}_{2(g)}\) increasing from \(y_{\text{Cl}_{2(g)}}^i = 0.338\) to \(y_{\text{Cl}_{2(g)}}^f = 0.493\) and the concentration of \(\text{Cl}_{(g)}\) decreasing from \(y_{\text{Cl}_{2(g)}}^i = 0.662\) to \(y_{\text{Cl}_{(g)}}^f = 0.507\).
