Chemical Equilibrium

Reaction $(c)$ can be obtained by adding reactions $(a)$ and $(b)$ therefore ${K_3} = {K_1}.{K_2}$

Hence $(c)$ is the correct answer.

Let the initial moles of $X$ be $'a'$

and that of $Z$ be $'b'$ the for the given reactions,

we have $X\,\,\,\,\,\,\,\,\,\rightleftharpoons\,\,\,\,\,\,\,\,\,2Y$

$\eqalign{ & Initial\,\,\,\,\,\,\,\,\,\,\,\,a\,\,moles\,\,\,\,\,\,\,0 \cr & At\,\,equi.\,\,\,\,\,\,\,a\left( {1 - \alpha } \right)\,\,\,\,\,\,\,2a\alpha \cr & (moles) \cr} $

Total no. of moles $ = a\left( {1 - \alpha } \right) + 2a\alpha $

$ = a - a\alpha + 2a\alpha $ $ = a\left( {1 + \alpha } \right)$

Now, $\,\,\,{K_{{p_1}}} = {{{{\left( {{n_y}} \right)}^2}} \over {{n_x}}} \times {\left( {{{{P_{{T_1}}}} \over {\sum n }}} \right)^{\Delta n}}$

or, $\,\,\,{K_{{p_1}}} = {{{{\left( {2a\alpha } \right)}^2}.{P_{{T_1}}}} \over {\left[ {a\left( {1 - \alpha } \right)} \right]\left[ {a\left( {1 + \alpha } \right)} \right]}}$

$\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $Z\,\,\,\,\,\,\,\,\,\rightleftharpoons\,\,\,\,\,\,\,\,\,P+Q$

$\eqalign{ & Initial\,\,\,\,\,\,\,\,\,\,\,b\,\,moles\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,0 \cr & Ar\,\,equi.\,\,\,\,\,\,b\left( {1 - \alpha } \right)\,\,\,\,\,\,\,\,b\alpha \,\,\,\,\,\,\,b\alpha \cr & (moles) \cr} $

Total no. of moles

$ = b\left( {1 - \alpha } \right) + b\alpha + b\alpha $

$ = b - b\alpha + b\alpha + b\alpha $

$ = b\left( {1 + \alpha } \right)$

Now $\,\,\,{K_{{P_2}}} = {{{n_Q} \times {n_P}} \over {{n_z}}} \times {\left[ {{{{P_{{T_2}}}} \over {\sum\nolimits_n \, }}} \right]^{\Delta n}}$

or$\,\,\,{K_{{P_2}}} = {{\left( {b\alpha } \right)\left( {b\alpha } \right).{P_{{T_2}}}} \over {\left[ {b\left( {1 - \alpha } \right)} \right]\left[ {b\left( {1 + \alpha } \right)} \right]}}$

or$\,\,\,$ $\,\,\,{{{K_{{P_1}}}} \over {{K_{P2}}}} = {{4{\alpha ^2}.{P_{{T_1}}}} \over {\left( {1 - {\alpha ^2}} \right)}} \times {{{{\left( {1 - \alpha } \right)}^2}} \over {{P_{{T_2}}}.{\alpha ^2}}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = {{4{P_{{T_1}}}} \over {{P_{{T_2}}}}}$
or$\,\,\,{{{P_{{T_1}}}} \over {{P_{T2}}}} = {1 \over 9}\,\,\,\,\,$ [ as $\,\,\,{{{K_{{P_1}}}} \over {{K_{{P_1}}}}} = {1 \over 9}\,\,$ given ]

or$\,\,\,{{{P_{{T_1}}}} \over {{P_{T2}}}} = {1 \over {36}}$

or$\,\,\,1:36$

$S{O_3}\left( g \right)\,\rightleftharpoons\,S{O_2}\left( g \right)\,\, + \,\,{1 \over 2}{O_2}\left( g \right)$

${K_c} = {{\left[ {S{O_2}} \right]{{\left[ {{O_2}} \right]}^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over {\left[ {S{O_3}} \right]}}$

$ = 4.9 \times {10^{ - 2}};$

On taking the square of the above reaction

${{{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]} \over {{{\left[ {S{O_3}} \right]}^2}}}$

$ = 24.01 \times {10^{ - 4}}$

now $K{'_C}$

for $\,\,2S{O_2}\left( g \right)\,\, + \,\,{O_2}\left( g \right)\,\rightleftharpoons\,2S{O_3}$

$ = {{{{\left[ {S{O_3}} \right]}^2}} \over {{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]}}$

$ = {1 \over {24.01 \times {{10}^{ - 4}}}}$

$ = 416$

$\mathop {PC{l_5}\left( g \right)}\limits_{1 - x} \mathop {\,\rightleftharpoons\,PC{l_3}\left( g \right)}\limits_x \,\, + \,\,\mathop {C{l_2}\left( g \right)}\limits_x $

Total moles after dissociation

$1 - x + x + x = 1 + x$

${P_{PC{l_3}}} = $ mole fraction of $PCl{}_3 \times $ Total pressure

$ = \left( {{x \over {1 + x}}} \right)P$

For the reaction : -

$2N{O_2}\left( g \right)\rightleftharpoons2NO\left( g \right) + {O_2}\left( g \right)$

Given ${K_c} = 1.8 \times {10^{ - 6}}\,\,$ at $\,\,{184^ \circ }C$

$R=0.0831$ $\,\,kJ/mol.k$

${K_p} = 1.8 \times {10^{ - 6}} \times 0.0831 \times 457$

$ = 6.836 \times {10^{ - 6}}$

$\left[ {} \right.$ as $\,\,\,\,\,{184^ \circ }C = \left( {273 + 184} \right) = 457\,k,\,\,$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. {\Delta n = \left( {2 + 1, - 1} \right) = 1} \right]$

Hence it is clear that ${K_p} > {K_c}$

The reaction given is an exothermic reaction thus accordingly to Lechatalier's principle lowering of temperature, addition of ${F_2}$ and or $C{l_2}$ favour the for ward direction and hence the production of $Cl{F_3}.$

$\mathop {N{H_4}HS\left( s \right)\,\rightleftharpoons\,}\limits_{\matrix{ {start} \cr {At\,\,equib.} \cr } } \,\,\mathop {N{H_3}\left( g \right)}\limits_{\matrix{ {0.5\,\,atm} \cr {0.5 + x\,\,atm} \cr } } \,\, + \mathop {{H_2}S\left( g \right)}\limits_{\matrix{ {0\,\,atm} \cr {x\,\,atm} \cr } } $

Then $0.5 + x + x = 2x + 0.5 = 0.84\,\,$ (Given)

$ \Rightarrow x = 0.17\,\,atm.$

${P_{N{H_3}}} = 0.5 + 0.17 = 0.67\,atm;$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ ${P_H}_{_2S} = 0.17\,atm$

$K = {P_{N{H_3}}} \times {P_{{H_2}S}}$

$ = 0.67 \times 0.17\,at{m^2}$

$ = 0.1139 = 0.11$

For ${P_4}\left( s \right) + 5{O_2}\left( g \right)\,\rightleftharpoons\,{P_4}{O_{10}}\left( 8 \right)$

${K_c} = {1 \over {{{\left( {{O_2}} \right)}^5}}}.$ The solids have concentration unity

${K_c} = {{{{\left[ {NO} \right]}^2}} \over {\left[ {{N_2}} \right]\left[ {{O_2}} \right]}} = 4 \times {10^{ - 4}}$

$K{'_c} = {{{{\left[ {{N_2}} \right]}^{1/2}}{{\left[ {{Q_2}} \right]}^{1/2}}} \over {\left[ {NO} \right]}}$

$ = {1 \over {\sqrt {{K_c}} }}$

$ = {1 \over {\sqrt {4 \times {{10}^{ - 4}}} }}$

$ = 50$

${K_p} = {K_c}{\left( {RT} \right)^{\Delta n}};$

Here $\Delta n = 1 - 2 = - 1$

$\therefore$ ${{{k_p}} \over {{K_c}}} = {1 \over {RT}}$

${K_c} = {{{{\left[ {N{O_2}} \right]}^2}} \over {\left[ {{N_2}{O_4}} \right]}} = {{\left[ {1.2 \times {{10}^{ - 2}}} \right]^2} \over {\left[ {4.8 \times {{10}^{ - 2}}} \right]}}$

$ = 3 \times {10^{ - 3}}\,mol/L$

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

$\therefore$ High pressure will be required.

In this reaction the ratio of number of moles of reactants to products is same $i.e.\,\,2:2,$ hence change in volume will not alter the number of moles.

${K_p} = {K_c}{\left( {RT} \right)^{\Delta n}};$

$\Delta n = 1 - \left( {1 + {1 \over 2}} \right)$

$ = 1 - {3 \over 2} = - {1 \over 2}.$

$\therefore$ $\,\,\,\,{{{K_p}} \over {{K_c}}} = {\left( {RT} \right)^{ - 1/2}}$