Prediction of the following stages of a reaction by comparing the value of $K_c$ and $Q_c$ are

(i) If $Q_c< K_c$, the reaction will proceed in the direction of the products (forward reaction).

(ii) If $Q_c>K_c$, the reaction will proceed in the direction of reactants (reverse reaction).

(iii) If $Q_c=K_c$, the reaction mixture is already at equilibrium.

$\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g) ; \Delta H=-92.38 \mathrm{~kJ} \mathrm{~mol}^{-1}$

It is an exothermic process as $\Delta H$ is negative.

Effect of temperature According to Le-Chatelier's principle, low temperature is favourable for high yield of ammonia, but practically very low temperatures slow down the reaction. So, optimum temperature 700 K is favourable in attainment of equilibrium.

Effect of pressure Similarly, high pressure about 200 atm is favourable for high yield of ammonia. On increasing pressure, reaction goes in the forward direction because the number of moles decreases in the forward direction.

Addition of argon At constant volume addition of argon does not affect the equilibrium because it does not change the partial pressures of the reactants or products involved in the reaction and the equilibrium remains undisturbed.

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.