Thermodynamic and Equilibrium Analysis of Gaseous Dissociation Reaction

To determine which statement is INCORRECT, we analyze the given thermodynamic and chemical equilibrium properties of the reaction:

$X_2(g) \rightleftharpoons 2X(g)$

1. Equilibrium Constant Expression ($K_p$)

Let the initial moles of $X_2$ be $1\text{ mol}$ and $X$ be $0\text{ mol}$. At equilibrium, the moles of $X$ produced is $\beta_{\text{equilibrium}}$.

• Moles of $X_2$ at equilibrium = $1 - \frac{\beta_{\text{equilibrium}}}{2}$
• Moles of $X$ at equilibrium = $\beta_{\text{equilibrium}}$
• Total moles at equilibrium = $\left(1 - \frac{\beta_{\text{equilibrium}}}{2}\right) + \beta_{\text{equilibrium}} = 1 + \frac{\beta_{\text{equilibrium}}}{2} = \frac{2 + \beta_{\text{equilibrium}}}{2}$

Given the total pressure $P = 2\text{ bar}$, the partial pressures are: $P_{X_2} = \left(\frac{1 - \beta_{\text{equilibrium}}/2}{1 + \beta_{\text{equilibrium}}/2}\right) P = \left(\frac{2 - \beta_{\text{equilibrium}}}{2 + \beta_{\text{equilibrium}}}\right) (2\text{ bar})$

$P_X = \left(\frac{\beta_{\text{equilibrium}}}{1 + \beta_{\text{equilibrium}}/2}\right) P = \left(\frac{2\beta_{\text{equilibrium}}}{2 + \beta_{\text{equilibrium}}}\right) (2\text{ bar})$

The equilibrium constant $K_p$ is: $K_p = \frac{(P_X)^2}{P_{X_2}} = \frac{\left[\frac{4\beta_{\text{equilibrium}}}{2 + \beta_{\text{equilibrium}}}\right]^2}{\frac{2(2 - \beta_{\text{equilibrium}})}{2 + \beta_{\text{equilibrium}}}} = \frac{8\beta_{\text{equilibrium}}^2}{4 - \beta_{\text{equilibrium}}^2}$

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2. Evaluation of Options

• Statement (A): For the given reaction, $\Delta n_g = 2 - 1 = +1 > 0$. According to Le Chatelier's Principle, a decrease in total pressure causes the equilibrium to shift towards the side with a greater number of gas moles (the forward direction). This leads to the formation of a greater quantity of gaseous $X$. Thus, Statement (A) is CORRECT.

• Statement (B): At the beginning of the reaction, no $X$ is present ($P_X = 0$), which implies $Q_p = 0$. The reaction Gibbs free energy is given by: $\Delta_r G = \Delta_r G^\circ + RT \ln Q_p$ As $Q_p \to 0$, $\ln Q_p \to -\infty$, causing $\Delta_r G \to -\infty$. Since $\Delta_r G < 0$, the reaction occurs spontaneously in the forward direction initially. Thus, Statement (B) is CORRECT.

• Statement (C): We are given that $\Delta_r G^\circ > 0$. From thermodynamics: $\Delta_r G^\circ = -RT \ln K_p^\circ > 0 \implies \ln K_p^\circ < 0 \implies K_p < 1$ Substituting the expression for $K_p$: $\frac{8\beta_{\text{equilibrium}}^2}{4 - \beta_{\text{equilibrium}}^2} < 1$ $8\beta_{\text{equilibrium}}^2 < 4 - \beta_{\text{equilibrium}}^2$ $9\beta_{\text{equilibrium}}^2 < 4 \implies \beta_{\text{equilibrium}} < \frac{2}{3} \approx 0.67$ Since $\beta_{\text{equilibrium}}$ must be strictly less than $0.67$, it cannot equal $0.7$. Thus, Statement (C) is INCORRECT.

• Statement (D): The relationship between $K_p$ and $K_c$ is: $K_p = K_c (RT)^{\Delta n_g}$ Since $\Delta n_g = 1$, $R = 0.083\text{ L bar K}^{-1}\text{ mol}^{-1}$, and $T = 298\text{ K}$: $RT = 0.083 \times 298 = 24.734\text{ L bar mol}^{-1}$ $K_c = \frac{K_p}{RT} = \frac{K_p}{24.734}$ Since $K_p < 1$, we get: $K_c < \frac{1}{24.734} < 1$ Thus, Statement (D) is CORRECT.

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Conclusion

The incorrect statement is C.