Chemical Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

Exothermic reaction, decrease in number of moles, increase in concentration of reactants.

$$\eqalign{ & {I_2}\,\,\,\,\,\, \rightleftharpoons \,\,\,2{I^ - } \cr & 1 - x\,\,\,\,\,\,\,\,\,\,2x \cr & {K_c} = \frac{{{{\left( {2x} \right)}^2}}}{{\left( {1 - x} \right)}} = {10^{ - 6}} \cr & {\text{Soln}}{\text{. shows that}}\,\left( {1 - x} \right) > 2x \cr & \therefore \,\,\left[ {{I_2}\left( g \right)} \right] > \left[ {{I^ - }\left( g \right)} \right] \cr} $$

$${\text{When}}\,\,\Delta {n_g} = 0,{K_p} = {K_c}$$

$$N{H_4}COON{H_{2\left( s \right)}} \rightleftharpoons $$     $$\mathop {2N{H_{3\left( g \right)}}}\limits_{2p} + \mathop {C{O_{2\left( g \right)}}}\limits_p $$
When volume and temperature are constant, the number of moles of a gas is proportional to its partial pressure.
$$\eqalign{ & {\text{So,}}\,2p + p = 3 \cr & 3p = 3, \cr & \therefore p = 1\,atm \cr} $$
$${K_p} = {\left( {2p} \right)^2} \times p = 4{p^3}$$     $$ = 4 \times {\left( 1 \right)^3} = 4\,at{m^3}$$

$$\eqalign{ & {\text{At equilibrium the moles of}}\,C{l_2}\,{\text{must be}} \cr & = 0.15 \times 3 = 0.45 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,PC{l_5}\,\,\,\,\, \rightleftharpoons \,PC{l_3}\, + \,\,C{l_2} \cr & {\text{Eqm}}{\text{. Conc}}.\,\,\frac{{x - 0.45}}{3}\,\,\,\,\,\,\,\,\,\,\frac{{0.45}}{3}\,\,\,\,\,\,\,\,\frac{{0.45}}{3} \cr & {K_c} = \frac{{\left[ {PC{l_3}} \right]\left[ {C{l_2}} \right]}}{{\left[ {PC{l_5}} \right]}} \cr & \therefore \,\,0.04 = \frac{{0.15 \times 0.15}}{{\frac{{\left( {x - 0.45} \right)}}{3}}} \cr & \therefore \,\,x = 2.1\,moles \cr} $$

Due to exothermicity of reaction low or optimum temperature will be required. Since 3 moles are changing to 2 moles.
∴ High pressure will be required.

$$\eqalign{ & D = \frac{{PM}}{{RT}} \cr & = \frac{{1 \times 82.1}}{{0.0821 \times 400}} \cr & = 2.5\,g\,{L^{ - 1}} \cr & d = 2.0\,g\,{L^{ - 1}}{\text{(given)}} \cr & \alpha = \frac{{D - d}}{{d\left( {n - 1} \right)}} \cr & = \frac{{2.5 - 2}}{{2\left( {3 - 1} \right)}} \cr & = 0.125 \cr & = 12.5\% \cr} $$

Statement $$(a)$$ is correct and the rest statements are wrong. $${K_p}$$  depends only on temperature hence at constant temp. $${K_p}$$  will not change.

$$\eqalign{ & {N_2}\left( g \right) + {O_2}\left( g \right) \rightleftharpoons 2\,NO\left( g \right);\,{K_1}\,\,\,...{\text{(i)}} \cr & 2NO\left( g \right) + {O_2}\left( g \right) \rightleftharpoons 2N{O_2}\left( g \right);\,{K_2}\,\,\,...{\text{(ii)}} \cr & {\text{On adding Eqs}}{\text{. (i) and (ii)}} \cr} $$
$${{\text{N}}_2}\left( g \right) + 2{O_2}\left( g \right) \rightleftharpoons 2N{O_2}\left( g \right);$$      $$K = {K_1} \times {K_2}\,\,\,...{\text{(iii)}}$$
On dividing $${\text{(iii)}}$$  by $$\frac{1}{2}$$  and on reversing we get,
$$\eqalign{ & N{O_2}\left( g \right) \rightleftharpoons \frac{1}{2}{N_2}\left( g \right) + {O_2}\left( g \right); \cr & So,\,\,K = \frac{{{{\left( {{N_2}} \right)}^{\frac{1}{2}}}\left( {{O_2}} \right)}}{{\left( {N{O_2}} \right)}} \cr & \therefore \,\,\,\,\,\,K = {\left[ {\frac{1}{{{K_1}{K_2}}}} \right]^{\frac{1}{2}}} \cr} $$

Let the total mass of the mixture of $$CO$$  and $$C{O_2}$$  is $$100\,g,$$  then $$CO = 90.55\,g$$    and $$C{O_2} = 100 - 90.55 = 9.45\,g$$
$$\eqalign{ & {\text{Moles}}\,\,{\text{of}}\,\,CO = \frac{{90.55}}{{28}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3.234 \cr & {\text{Moles}}\,\,{\text{of}}\,\,C{O_2} = \frac{{9.45}}{{44}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.215 \cr & {\text{Mole fraction of}} \cr & CO = \frac{{3.234}}{{3.234 + 0.215}} \cr & \,\,\,\,\,\,\,\,\,\, = 0.938 \cr & {\text{Mole fraction of}} \cr & C{O_2} = \frac{{0.215}}{{3.234 + 0.215}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 0.062 \cr & \therefore {p_{CO}} = {\text{mole fraction}} \times {\text{total pressure}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{ 0}}{\text{.938}} \times {\text{1}}\,atm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{0}}{\text{.938}}\,atm \cr & {p_{C{O_2}}} = 0.062 \times 1\,atm \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 0.062\,atm \cr & {K_p}\,{\text{for the reaction,}} \cr & {C_{\left( s \right)}} + C{O_{2\left( g \right)}} \rightleftharpoons 2C{O_{\left( g \right)}} \cr & {K_p} = \frac{{\mathop {{p_{CO}}}\limits^2 }}{{{p_{C{O_2}}}}} \cr & \,\,\,\,\,\,\,\, = \frac{{{{\left( {0.938} \right)}^2}}}{{0.062}} \cr & \,\,\,\,\,\,\,\, = 14.19 \cr & {\text{Now,}}\,\Delta {n_g} = 2 - 1 = 1,{K_p} = {K_c}{\left( {RT} \right)^{\Delta {n_g}}} \cr & {\text{or}}\,\,{K_c} = \frac{{{K_p}}}{{RT}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{14.19}}{{0.0821 \times 1127}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.153 \cr} $$