Chemical Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & {\text{Given,}}\,\,{N_2} + 3{H_2} \rightleftharpoons 2N{H_3},\,\,{K_1}...{\text{(i)}} \cr & {N_2} + {O_2} \rightleftharpoons 2NO,\,\,{K_2}\,\,...{\text{(ii)}} \cr & {H_2} + \frac{1}{2}{O_2} \to {H_2}O,\,\,{K_3}\,\,...{\text{(iii)}} \cr & {\text{To calculate,}} \cr} $$
$$2N{H}_3+\frac{5}{2}{O}_2\stackrel{K}{\rightleftharpoons }2NO+3H_{2}O,$$
$$K = ?\,\,\,\,...{\text{(iv)}}$$
On reversing the equation (i) and multiplying the equation (iii) by 3, we get
$$\eqalign{ & 2N{H_3} \rightleftharpoons {N_2} + 3{H_2},\frac{1}{{{K_1}}}\,\,...{\text{(v)}} \cr & 3{H_2} + \frac{3}{2}{O_2} \to 3{H_2}O,\,\,K_3^3\,\,...{\text{(vi)}} \cr} $$
Now, add equation. (ii), (v) and (vi), we get the resultant equation. (iv).
$$2N{H}_3+\frac{5}{2}{O}_2\stackrel{K}{\rightleftharpoons }2NO+3H_{2}O$$
$$\therefore \,\,K = \frac{{{K_2}K_3^3}}{{{K_1}}}$$

$$ - \Delta {G^ \circ } = RT\,\ln \,{K_c}$$

$$PC{l_{5\left( g \right)}} \rightleftharpoons PC{l_{3\left( g \right)}} + C{l_{2\left( g \right)}}$$
If concentration of $$PC{l_5}$$  is increased, the forward reaction is favoured.

\[\begin{matrix} {} \\ \text{Initial}\,\,\text{conc}\text{.} \\ \text{At}\,\,\text{equilibrium} \\ \text{Conc}\text{.} \\ \end{matrix}\begin{matrix} N{{O}_{\left( g \right)}} \\ 1 \\ 0.5 \\ \frac{0.5}{10} \\ \end{matrix}\begin{matrix} + \\ {} \\ {} \\ {} \\ \end{matrix}\begin{matrix} {{O}_{3\left( g \right)}} \\ 1 \\ 0.5 \\ \frac{0.5}{10} \\ \end{matrix}\begin{matrix} \rightleftharpoons \\ {} \\ {} \\ {} \\ \end{matrix}\begin{matrix} N{{O}_{2\left( g \right)}} \\ 0 \\ 0.5 \\ \frac{0.5}{10} \\ \end{matrix}\begin{matrix} + \\ {} \\ {} \\ {} \\ \end{matrix}\begin{matrix} {{O}_{2\left( g \right)}} \\ 0 \\ 0.5 \\ \frac{0.5}{10} \\ \end{matrix}\]

$${K_c} = \frac{{0.05 \times 0.05}}{{0.05 \times 0.05}} = 1$$

It is an exothermic reaction, hence the forward reaction is favoured at low temperature which means colourless $${N_2}{O_4}$$  will be formed resulting in decrease in intensity of brown colour. High temperature favours backward reaction resulting in formation of $$N{O_2},$$  thus intensifying brown colour.

$${K_c} = \frac{{{k_f}}}{{{k_r}}} = 57.0\,\,{\text{or}}\,{k_f} = 57.0\,{k_r}$$

$$\eqalign{ & {H_2}\left( g \right) + {I_2}\left( g \right) \rightleftharpoons 2HI\left( g \right) \cr & {K_p} = {K_c}{\left( {RT} \right)^{\Delta n}}; \cr & \Delta n = 2 - 2 = 0;\,\,\,\therefore \,\,{K_p} = {K_c} \cr} $$

\[\begin{matrix} {} & {{S}_{8\left( g \right)}} & \rightleftharpoons & 4{{S}_{2\left( g \right)}} \\ \text{Initial}\,\,\text{pressure} & 1 & {} & 0 \\ \text{At}\,\,\text{equilibrium} & 1-\frac{25}{100} & {} & 4\times \frac{25}{100} \\ {} & =0.75 & {} & =1 \\ \end{matrix}\]
$$\eqalign{ & {K_p} = \frac{{{{\left( {{P_{{S_2}}}} \right)}^4}}}{{{P_{{S_8}}}}} \cr & \,\,\,\,\,\,\,\,\, = \frac{{{{\left( 1 \right)}^4}}}{{0.75}} \cr & \,\,\,\,\,\,\,\,\, = 1.33\,at{m^3} \cr} $$

$$\eqalign{ & {\text{For the reaction,}} \cr & {N_2}\left( g \right) + 3{H_2}\left( g \right) \rightleftharpoons 2N{H_3}\left( g \right) \cr & Q{\text{(Quotient) = }}\frac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}}, \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {n_g} = 2 - 4 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 2 \cr} $$
At equilibrium $$Q$$ is equal to $${K_c}$$  but for the progress of reaction towards right side, $$Q > {K_c}$$

As the reaction is endothermic, the increase in temperature will favour the forward reaction. More $$PC{l_5}$$  will dissociate to form $$PC{l_3}$$  and $$C{l_2}.$$