Chemical Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & 2N{O_{2\left( g \right)}} \rightleftharpoons {N_2}{O_{4\left( g \right)}} \cr & \Delta n = 1 - 2 = - 1;{K_p} = {K_c}{\left( {RT} \right)^{\Delta n}} \cr & \frac{{{K_p}}}{{{K_c}}} = {\left( {RT} \right)^{ - 1}} \cr} $$

As all the reactants and products are present in aqueous form in (D) so it is a reversible reaction. In others either solid or gas is generated which is insoluble or volatile and hence makes the reaction unidirectional.

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{N_2}{O_4}\left( g \right) \rightleftharpoons 2N{O_2}\left( g \right) \cr & t = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \cr & t = eq.\,\,\,\,\,\,\,\,\,1 - \alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\alpha \cr & {\text{Where}}\,\,\alpha = {\text{Degree of dissociation}}{\text{.}} \cr & Mol.\,wt.\,\,{\text{of mixture}} \cr & = \frac{{\left( {1 - \alpha } \right) \times {M_{{N_2}{O_4}}} + 2\alpha \times {M_{N{O_2}}}}}{{\left( {1 + \alpha } \right)}} \cr & = \frac{{\left( {1 - 0.2} \right)92 + 2 \times 0.2 \times 46}}{{\left( {1 + 0.2} \right)}} \cr & = 76.66 \cr & {\text{Now, As per ideal gas equation}} \cr & PV = nRT \cr & P{M_{{\text{mixture}}}} = dRT \cr & \therefore \,\,d = \frac{{P{M_{mix}}}}{{RT}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{1 \times 76.66}}{{0.821 \times 300}} \cr & \,\,\,\,\,\,\,\,\,\, = 3.11g/L \cr} $$

$$\eqalign{ & {\text{Consider reaction (i),}} \cr & {N_2}\left( g \right) + {O_2}\left( g \right) \rightleftharpoons 2NO\left( g \right)...{\text{(i)}} \cr & {K_1} = \frac{{{{\left[ {NO} \right]}^2}}}{{\left[ {{N_2}} \right]\left[ {{O_2}} \right]}} \cr & {\text{Now, consider reaction (ii),}} \cr & NO\left( g \right) \rightleftharpoons \frac{1}{2}{N_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right)...{\text{(ii)}} \cr & \,\,{K_2} = \frac{{{{\left[ {{N_2}} \right]}^{\frac{1}{2}}}{{\left[ {{O_2}} \right]}^{\frac{1}{2}}}}}{{\left[ {NO} \right]}} \cr & \frac{1}{{{K_2}}} = \frac{1}{{\frac{{{{\left[ {{N_2}} \right]}^{\frac{1}{2}}}{{\left[ {{O_2}} \right]}^{\frac{1}{2}}}}}{{\left[ {NO} \right]}}}} \cr & \,\,\,\,\,\,\,\,\, = \frac{{\left[ {NO} \right]}}{{{{\left[ {{N_2}} \right]}^{\frac{1}{2}}}{{\left[ {{O_2}} \right]}^{\frac{1}{2}}}}} \cr & {\left( {\frac{1}{{{K_2}}}} \right)^2} = {\left\{ {\frac{{\left[ {NO} \right]}}{{{{\left[ {{N_2}} \right]}^{\frac{1}{2}}}{{\left[ {{O_2}} \right]}^{\frac{1}{2}}}}}} \right\}^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left[ {NO} \right]}^2}}}{{\left[ {{N_2}} \right]\left[ {{O_2}} \right]}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {K_1} \cr} $$

\[\begin{matrix} {} & 2H{{I}_{\left( g \right)}} & \rightleftharpoons \\ \text{Initial}\,\,\text{conc}\text{.} & 1 & {} \\ \text{At}\,\,\text{equilibrium} & 0.5 & {} \\ \end{matrix}\begin{matrix} {} \\ {} \\ {} \\ \end{matrix}\begin{matrix} {{H}_{2\left( g \right)}} & + & {{I}_{2\left( g \right)}}; \\ 0 & {} & 0 \\ x & {} & x \\ \end{matrix}\begin{matrix} {{K}_{c}}=24.8 \\ {} \\ {} \\ \end{matrix}\]

$${K_c} = \frac{{\left[ {{H_2}} \right]\left[ {{I_2}} \right]}}{{{{\left[ {HI} \right]}^2}}}\,\,{\text{or}}\,\,\frac{1}{{24.8}} = $$      $$\frac{{x.x}}{{0.5 \times 0.5}}{\text{or}}\,\,{x^2} = \frac{{0.25}}{{24.8}} = 0.010$$
$$x = 0.10\,mol\,{L^{ - 1}}$$

$$\eqalign{ & \Delta {G^ \circ } = 2494.2J \cr & 2A \rightleftharpoons B + C. \cr & R = 8.314\,J/K/mol. \cr & e = 2.718 \cr} $$
$$\left[ A \right] = \frac{1}{2},\left[ B \right] = 2,\left[ C \right] = \frac{1}{2};$$      $$Q = \frac{{\left[ B \right]\left[ C \right]}}{{{{\left[ A \right]}^2}}} = \frac{{2 \times \frac{1}{2}}}{{{{\left( {\frac{1}{2}} \right)}^2}}} = 4$$
$$\Delta {G^ \circ } = - 2.303\,RT\,\log \,{K_c}.$$
$$\eqalign{ & 2494.2J = - 2.303 \times \left( {8.314\,J/K/mol} \right) \times \left( {300K} \right)\log {K_c} \cr & \Rightarrow \log \,{K_c} = - \frac{{2494.2J}}{{2.303 \times \times 8.314J/K/mol \times 300K}} \cr & \Rightarrow \log \,{K_c} = - 0.4341;\,{K_c} = 0.37;Q > {K_c}. \cr} $$

$$\eqalign{ & {K_p} = {K_c}{\left( {RT} \right)^{\Delta n}};\Delta n = 1 \cr & {K_p} = 167\,bar, \cr & {K_c} = \frac{{167\,bar}}{{0.0831\,L\,bar\,{K^{ - 1}}\,mo{l^{ - 1}} \times 1073\,K}} \cr & \,\,\,\,\,\,\,\, = 1.873\,mol\,{L^{ - 1}} \cr} $$

\[\begin{array}{*{20}{c}} {} \\ {{\text{Initial}}} \\ {{\text{At eqn}}{\text{.}}} \\ \begin{gathered} {\text{Partial }} \hfill \\ {\text{pressure}} \hfill \\ \end{gathered} \end{array}\begin{array}{*{20}{c}} {PC{l_5}} \\ 1 \\ {1 - \alpha } \\ { = \frac{{1 - \alpha }}{{1 + \alpha }}.P} \end{array}\begin{array}{*{20}{c}} \rightleftharpoons \\ {} \\ {} \\ {} \end{array}\begin{array}{*{20}{c}} {PC{l_3}} \\ 0 \\ \alpha \\ {\frac{\alpha }{{1 + \alpha }}.P} \end{array}\begin{array}{*{20}{c}} + \\ {} \\ {} \\ {} \end{array}\begin{array}{*{20}{c}} {C{l_2}} \\ 0 \\ \alpha \\ {\frac{\alpha }{{1 + \alpha }}.P} \end{array}\]
$$\eqalign{ & {\text{Total mole}} = 1 - \alpha + \alpha + \alpha = 1 + \alpha \cr & {K_P} = \frac{{\frac{\alpha }{{1 + \alpha }}P.\frac{\alpha }{{1 + \alpha }}.P}}{{\frac{{1 - \alpha }}{{1 + \alpha }}P}} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{{\alpha ^2}P}}{{1 - {\alpha ^2}}} \cr} $$

$$\eqalign{ & {K_c} = \frac{{{{\left[ {NO} \right]}^2}}}{{\left[ {{N_2}} \right]\left[ {{O_2}} \right]}} \cr & = 4 \times {10^{ - 4}} \cr & K{'_c} = \frac{{{{\left[ {{N_2}} \right]}^{\frac{1}{2}}}{{\left[ {{O_2}} \right]}^{\frac{1}{2}}}}}{{\left[ {NO} \right]}} \cr & = \frac{1}{{\sqrt {{K_c}} }} \cr & = \frac{1}{{\sqrt {4 \times {{10}^{ - 4}}} }} \cr & = 50 \cr} $$

$$XeF{e_6}\left( g \right) + {H_2}O\left( g \right) \rightleftharpoons $$     $$XeO{F_4}\left( g \right) + 2HF\left( g \right)$$
$${K_1} = \frac{{\left[ {XeO{F_4}} \right]{{\left[ {HF} \right]}^2}}}{{\left[ {Xe{F_6}} \right]\left[ {{H_2}O} \right]}}\,\,\,...{\text{(i)}}$$
$$Xe{O_4}\left( g \right) + Xe{F_6}\left( g \right) \rightleftharpoons $$     $$XeO{F_4}\left( g \right) + Xe{O_3}{F_2}\left( g \right)$$
$${K_2} = \frac{{\left[ {XeO{F_4}} \right]\left[ {Xe{O_3}{F_2}} \right]}}{{\left[ {Xe{O_4}} \right]\left[ {Xe{F_6}} \right]}}\,\,\,...{\text{(ii)}}$$
$${\text{For the reaction,}}$$
$$Xe{O_4}\left( g \right) + 2HF\left( g \right) \rightleftharpoons $$     $$Xe{O_3}{F_2}\left( g \right) + {H_2}O\left( g \right)$$
$$\eqalign{ & K = \frac{{\left[ {Xe{O_3}{F_2}} \right]\left[ {{H_2}O} \right]}}{{\left[ {Xe{O_4}} \right]{{\left[ {HF} \right]}^2}}}\,\,\,...{\text{(iii)}} \cr & {\text{By dividing eq}}{\text{. (ii) by (i) we get,}} \cr & K = \frac{{{K_2}}}{{{K_1}}} \cr} $$