A sparingly soluble salt having general formula $A_x^{p+} B_y^{q-}$. Its molar solubility is $S \mathrm{~mol} \mathrm{~L}^{-1}$.

Then, $\quad A_x^{\rho+} B_y^{q-} \rightleftharpoons x A^{\rho+}(a q)+y B^{q-}(a q)$

$S$ moles of $A_x B_y$ dissolve to give $x$ moles of $A^{P+}$ and $y$ moles of $B^{q-}$.

Therefore, solubility product $\left(K_{\text {sp }}\right)=\left[A^{P_{+}}\right]^x\left[B^{q-}\right]^y$

$$\begin{aligned} & =[x S]^x[y S]^y \\ & =x^x y^y S^{x+y} \end{aligned}$$

The relation between $$\Delta G$$ and Q is

$$\begin{aligned} \Delta G & =\Delta G^{\ominus}+R T \ln Q \\ \Delta G & =\text { change in free energy as the reaction proceeds. } \\ \Delta G^{\ominus} & =\text { standard free energy } \\ Q & =\text { reaction quotient } \\ R & =\text { gas constant } \\ T & =\text { absolute temperature in } K \end{aligned}$$

$$\begin{aligned} \text{Since,}\quad \Delta G^{\ominus} & =-R T \ln K \\ \therefore\quad \Delta G & =-R T \ln K+R T \ln Q \\ \Delta G & =R T \ln \frac{Q}{K} \end{aligned}$$

If $Q

If $Q=K, \Delta G=0$ reaction is in equilibrium and there is no net reaction.

(b) $\begin{aligned} \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g}) & \rightleftharpoons \mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \\ K_{\mathrm{c}} & =\frac{\left[\mathrm{CH}_4\right]\left[\mathrm{H}_2 \mathrm{O}\right]}{[\mathrm{CO}]\left[\mathrm{H}_2\right]^3}\end{aligned}$

On increasing pressure, volume decreases. If we doubled the pressure, volume will be halved but the molar concentrations will be doubled. Then,

$Q_{\mathrm{C}}=\frac{2\left[\mathrm{CH}_4\right] \cdot 2\left[\mathrm{H}_2 \mathrm{O}\right]}{2[\mathrm{CO}]\left\{2\left[\mathrm{H}_2\right]\right\}^3}=\frac{1}{4} \frac{\left[\mathrm{CH}_4\right]\left[\mathrm{H}_2 \mathrm{O}\right]}{[\mathrm{CO}]\left[\mathrm{H}_2\right]^3}=\frac{1}{4} K_c$

Therefore, $Q_c$ is less than $K_c$, so $Q_c$ will tend to increase to re-establish equilibrium and the reaction will go in forward direction.