1. Write the reaction and mole balance

The reaction is $X_2(g) \rightleftharpoons 2X(g)$

Initially:

• moles of $X_2 = 1$
• moles of $X = 0$

If $\beta$ moles of $X$ are formed, then since 1 mole of $X_2$ gives 2 moles of $X$, $X_2 \text{ consumed} = \frac{\beta}{2}$

So at any stage: $n_{X_2}=1-\frac{\beta}{2}, \qquad n_X=\beta$

Total moles: $n_{\text{tot}}=1-\frac{\beta}{2}+\beta=1+\frac{\beta}{2}$

2. Reaction quotient at the start

For the reaction, $Q_p=\frac{(P_X)^2}{P_{X_2}}$

Initially, no $X$ is present, so $P_X=0 \implies Q_p=0$

Given that $\Delta_r G^\circ >0$, we have $\Delta_r G^\circ = -RT\ln K_p$ Rightarrow $K_p<1$

But initially, $Q_p=0<K_p$ so $\Delta_r G = \Delta_r G^\circ + RT\ln Q_p <0$ (in the limit, since $\ln 0 \to -\infty$).

Hence, forward dissociation starts spontaneously. So Option B is correct.

3. Effect of pressure

The reaction is $X_2(g) \rightleftharpoons 2X(g)$ Here, gaseous moles increase from 1 to 2.

By Le Chatelier’s principle, decreasing total pressure favors the side with more moles, i.e. the products. So more $X$ is formed.

Hence, Option A is correct.

4. Check whether $K_c<1$

We know $\Delta_r G^\circ>0$, so $K_p<1$

Now, $K_p=K_c(RT)^{\Delta n}$ Here, $\Delta n=2-1=1$ Therefore, $K_p=K_c(RT)$ So, $K_c=\frac{K_p}{RT}$

Given: $R=0.083\ \text{L bar K}^{-1}\text{mol}^{-1}, \quad T=298\ \text{K}$ $RT=0.083\times 298\approx 24.7$

Thus, $K_c=\frac{K_p}{24.7}$ Since $K_p<1$, $K_c<\frac{1}{24.7}<1$ Hence, Option D is correct.

5. Check Option C: $\beta_{\text{equilibrium}}=0.7$

At equilibrium under total pressure $P=2$ bar:

Mole fractions are $y_X=\frac{\beta}{1+\beta/2}, \qquad y_{X_2}=\frac{1-\beta/2}{1+\beta/2}$

So partial pressures are $P_X=2\cdot \frac{\beta}{1+\beta/2}, \qquad P_{X_2}=2\cdot \frac{1-\beta/2}{1+\beta/2}$

Therefore, $K_p=\frac{P_X^2}{P_{X_2}}$ $K_p=\frac{\left(2\beta/(1+\beta/2)\right)^2}{2(1-\beta/2)/(1+\beta/2)}$ $K_p=\frac{2\beta^2}{(1+\beta/2)(1-\beta/2)}$ $K_p=\frac{2\beta^2}{1-\beta^2/4}$

Now test $\beta=0.7$: $K_p=\frac{2(0.7)^2}{1-(0.7)^2/4}$ $=\frac{0.98}{1-0.1225}$ $=\frac{0.98}{0.8775}\approx 1.12$

But from $\Delta_r G^\circ>0$, we know $K_p<1$ So $\beta=0.7$ would imply $K_p>1$, which is impossible.

Hence, Option C is incorrect.

6. Final conclusion

The incorrect statement is: $\boxed{\text{C}}$