Assertion (A) An aqueous solution of ammonium acetate can act as a buffer.
Reason (R) Acetic acid is a weak acid and $\mathrm{NH}_4 \mathrm{OH}$ is a weak base.
Assertion (A) In the dissociation of $\mathrm{PCl}_5$ at constant pressure and temperature the addition of helium at equilibrium increases the dissociation of $\mathrm{PCl}_5$.
Reason (R) Helium removes $\mathrm{Cl}_2$ from the field of action.
How can you predict the following stages of a reaction by comparing the value of $K_c$ and $Q_c$ ?
(i) Net reaction proceeds in the forward direction.
(ii) Net reaction proceeds in the backward direction.
(iii) No net reaction occurs.
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.
On the basis of Le-Chatelier principle explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction.
$$\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) \Delta \mathrm{H}=-92.38 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
What will be the effect of addition of argon to the above reaction mixture at constant volume?
$\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g) ; \Delta H=-92.38 \mathrm{~kJ} \mathrm{~mol}^{-1}$
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 $\mathrm{A}_x^{\mathrm{p}+} \mathrm{B}_y^{\mathrm{q}-}$ and molar solubility S is in equilibrium with its saturated solution. Derive a relationship between the solubility and solubility product for such salt.
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 $\left(K_{\text {sp }}\right)=\left[A^{P_{+}}\right]^x\left[B^{q-}\right]^y$
$$\begin{aligned} & =[x S]^x[y S]^y \\ & =x^x y^y S^{x+y} \end{aligned}$$