Chemical Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

When the reaction is half completed, $$\left[ A \right] = \left[ B \right].$$
$${\text{Thus,}}\,K = \frac{{\left[ B \right]}}{{\left[ A \right]}} = 1$$
$$\therefore \Delta {G^ \circ } = - RT\,\ln \,K$$    $$ = - RT\,\ln \left( 1 \right) = 0$$

$$K = {K_2} \times \frac{1}{{{K_1}}} = \frac{{{K_2}}}{{{K_1}}}.$$

Plan As wecan see the reaction for which we have to find out equilibrium constant is different only in stoichiometric coefficient as compared to the given reaction. Hence, we can find equilibrium constant for the required reaction with the help of mentioned equilibrium constant in the problem.
Given, equilibrium constant for the reaction,
$${N_2}\left( g \right) + {O_2}\left( g \right) \rightleftharpoons 2NO\left( g \right)\,{\text{is}}\,K$$
i.e.   $$K = \frac{{{{\left[ {NO} \right]}^2}}}{{\left[ {{N_2}} \right]\left[ {{O_2}} \right]}}\,\,\,...(i)$$
Let equilibrium constant for the reaction,
$$\frac{1}{2}{N_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \rightleftharpoons NO\left( g \right)\,{\text{is}}\,K'$$
i.e.   $$K' = \frac{{\left[ {NO} \right]}}{{{{\left[ {{N_2}} \right]}^{\frac{1}{2}}}{{\left[ {{O_2}} \right]}^{\frac{1}{2}}}}}$$
On squaring both sides
$$K{'^2} = \frac{{{{\left[ {NO} \right]}^2}}}{{\left[ {{N_2}} \right]\left[ {{O_2}} \right]}}\,\,\,...(ii)$$
On comparing Eqs. (i) and (ii), we get
$$K = K{'^2}$$
or   $$K' = \sqrt K $$

$$\eqalign{ & {\text{For the reaction:}} \cr & {\text{PC}}{{\text{l}}_{5\left( g \right)}} \rightleftharpoons PC{l_{3\left( g \right)}} + C{l_{2\left( g \right)}} \cr & {K_c} = \frac{{\left[ {PC{l_3}} \right]\left[ {C{l_2}} \right]}}{{\left[ {PC{l_5}} \right]}} \cr & \,\,\,\,\,\,\,\,\, = \frac{{{{\left( {1.2 \times {{10}^{ - 3}}} \right)}^2}}}{{0.8 \times {{10}^{ - 3}}}} \cr & \,\,\,\,\,\,\,\,\, = 1.8 \times {10^{ - 3}}\,mol\,{L^{ - 1}} \cr} $$

$$\eqalign{ & {K_p} = {K_c}{\left( {RT} \right)^{\Delta n}};\,\,{\text{Here}}\,\Delta n = 1 - 2 = - 1 \cr & \therefore \,\,\frac{{{K_p}}}{{{K_c}}} = \frac{1}{{RT}} \cr} $$

$$Fe{\left( {OH} \right)_3}\left( s \right) \rightleftharpoons $$     $$F{e^{3 + }}\left( {aq} \right) + 3O{H^ - }\left( {aq} \right)$$
$$K = \frac{{\left[ {F{e^{3 + }}} \right]{{\left[ {O{H^ - }} \right]}^3}}}{{\left[ {Fe{{\left( {OH} \right)}_3}} \right]}}\,\,\,...{\text{(i)}}$$
To maintain equilibrium constant, let the concentration of $${F{e^{3 + }}}$$  is increased $$x$$  times, on decreasing the concentration of $${O{H^ - }}$$  by $$\frac{1}{4}$$  times
$$\eqalign{ & K = \frac{{\left[ {xF{e^{3 + }}} \right]{{\left[ {\frac{1}{4} \times O{H^ - }} \right]}^3}}}{{\left[ {Fe{{\left( {OH} \right)}_3}} \right]}}\,\,\,...{\text{(ii)}} \cr & {\text{By dividing eq}}{\text{. (ii) by (i) we get}} \cr & \frac{1}{{64}} \times x = 1\,\, \Rightarrow \,\,x = 64\,{\text{times}} \cr} $$

$$\eqalign{ & {N_{2\left( g \right)}} + {O_{2\left( g \right)}} \rightleftharpoons 2N{O_{\left( g \right)}};{K_c} = 0.1 \cr & \Delta n = 0,{\text{hence,}}\,{K_p} = {K_c} \cr} $$

For the reaction : $$N{H_4}C{l_{\left( s \right)}} \rightleftharpoons N{H_{3\left( g \right)}} + HC{l_{\left( g \right)}}$$
$$\eqalign{ & \Delta n = {n_{\left( {{\text{gaseous products}}} \right)}} - {n_{\left( {{\text{gaseous reactants}}} \right)}} \cr & \,\,\,\,\,\,\,\,\, = 2 - 0 \cr & \,\,\,\,\,\,\,\,\, = 2 \cr} $$

Only temperature affects the equilibrium constant. Since here $$\Delta H = 2 - 2 = 0,$$    so there is no change in K_p when total pressure changes.

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2HX\left( g \right) \rightleftharpoons \,\,\,\,{H_2}\left( g \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,{X_2}\left( g \right) \cr & {\text{At eqm}}{\text{.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{1}}{\text{.2}} \times {\text{1}}{{\text{0}}^{ - 3}}M\,\,\,\,\,\,\,\,\,1.2 \times {10^{ - 4}}M \cr & {K_{eq}} = \frac{{\left[ {{H_2}} \right]\left[ {{X_2}} \right]}}{{{{\left[ {HX} \right]}^2}}} \cr & {10^{ - 5}} = \frac{{1.2 \times {{10}^{ - 3}} \times 1.2 \times {{10}^{ - 4}}}}{{{{\left[ {HX} \right]}^2}}} \cr & \left[ {HX} \right] = \sqrt {\frac{{1.2 \times 1.2 \times {{10}^{ - 7}}}}{{{{10}^{ - 5}}}}} \cr & = 1.2 \times {10^{ - 1}} \cr & = 12 \times {10^{ - 2}}M \cr} $$