$$\eqalign{ & S{O_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \rightleftharpoons S{O_3}\left( g \right) \cr & {K_p} = {K_C}{\left( {RT} \right)^x} \cr} $$
where $$x = \Delta {n_g} = $$   number of gaseous moles in product - number of gaseous moles in reactant
$$\eqalign{ & = 1 - \left( {1 + \frac{1}{2}} \right) \cr & = 1 - \frac{3}{2} \cr & = - \frac{1}{2} \cr} $$

$$\eqalign{ & Mg{\left( {OH} \right)_2} \to \left[ {M{g^{2 + }}} \right] + 2\left[ {O{H^ - }} \right] \cr & {K_{sp}} = \left[ {Mg} \right]{\left[ {OH} \right]^2} \cr & = \left[ x \right]{\left[ {2x} \right]^2} \cr & = x.4{x^2} \cr & = 4{x^3}. \cr} $$

The concentration of reactants decreases and that of products increases with time. Rate of reaction increases with time.
At equilibrium, $${{\text{R}}_f} = {R_b}$$

$${K_c} = \frac{{0.2 \times 0.2}}{{0.1}} = 0.4$$

Vapours and liquid are at the same temperature.

$$\left(\text{I}\right)N_2+2O_2\stackrel{K_1}{\rightleftharpoons }2N{O}_2$$
$${K_1} = \frac{{{{\left[ {N{O_2}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{O_2}} \right]}^2}}}\,\,\,...\left( {\text{i}} \right)$$
$$\left(\text{II}\right)2N{O}_2\stackrel{K_2}{\rightleftharpoons }{N}_2+2O_2$$
$${K_2} = \frac{{\left[ {{N_2}} \right]{{\left[ {{O_2}} \right]}^2}}}{{{{\left[ {N{O_2}} \right]}^2}}}\,\,...\left( {{\text{ii}}} \right)$$
$$\left(\text{III}\right)N{O}_2\stackrel{K_3}{\rightleftharpoons }\frac{1}{2}{N}_2+O_2$$
$$\eqalign{ & {K_3} = \frac{{{{\left[ {{N_2}} \right]}^{\frac{1}{2}}}\left[ {{O_2}} \right]}}{{\left[ {N{O_2}} \right]}} \cr & \therefore \,\,{\left( {{K_3}} \right)^2} = \frac{{\left[ {{N_2}} \right]{{\left[ {{O_2}} \right]}^2}}}{{{{\left[ {N{O_2}} \right]}^2}}}\,\,...\left( {{\text{iii}}} \right) \cr & \therefore \,\,{\text{from equation (i), (ii) and (iii)}} \cr & {K_1} = \frac{1}{{{K_2}}} = \frac{1}{{{{\left( {{K_3}} \right)}^2}}} \cr} $$

\[\underset{\begin{smallmatrix} {{P}_{\text{initial}}}\,0.5\,atm \\ {{P}_{\text{final}}}\left( 0.5-x \right)\,atm \end{smallmatrix}}{\mathop{C{{O}_{2}}+{{C}_{\left( \text{graphite} \right)}}}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ 2x\,atm \end{smallmatrix}}{\mathop{2CO}}\,\]

$$\eqalign{ & {\text{Total }}P{\text{ at equilibrium }} \cr & {\text{ = }}\,{\text{0}}{\text{.5}} - x + 2x = 0.5 + x\,atm \cr & 0.8 = 0.5 + x \cr & \therefore \,\,x = 0.8 - 0.5 = 0.3\,atm \cr & {\text{Now}}\,\,{k_p} = \frac{{{{\left( {{P_{CO}}} \right)}^2}}}{{{P_{C{O_2}}}}} \cr & = \frac{{{{\left( {2'0.3} \right)}^2}}}{{\left( {0.5 - 0.3} \right)}} \cr & = \frac{{{{\left( {0.6} \right)}^2}}}{{\left( {0.2} \right)}} \cr & = 1.8\,atm \cr} $$

$$\eqalign{ & PC{l_{5\left( g \right)}} \rightleftharpoons PC{l_{3\left( g \right)}} + C{l_{2\left( g \right)}} \cr & {Q_c} = \frac{{0.5 \times 0.3}}{{0.6}} = 0.25 \cr} $$
$${K_c} = 0.2,$$  Since, $${Q_c} > {K_c}$$   reaction will proceed in backward direction.

$$\eqalign{ & \Delta {G^ \circ }\,\,{\text{is related to}}\,\,{K_{sp}}\,\,{\text{by the equation,}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {G^ \circ } = - 2.303RT\,\,{\text{log}}\,\,{K_{sp}} \cr & {\text{Given,}}\,\,\Delta {G^ \circ } = + 63.3\,kJ \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 63.3 \times {10^3}\,J \cr & {\text{Thus, substitute}}\,\,\Delta {G^ \circ } = 63.3 \times {10^3}\,\,J, \cr} $$
$$R = 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}}\,\,{\text{and}}$$       $$T = 298\,K\left[ {25 + 273\,K} \right]$$
$$\eqalign{ & {\text{from the above equation we get,}} \cr & 63.3 \times {10^3} = - 2.303 \times 8.314 \times 298\,{\text{log}}\,\,{K_{sp}} \cr & \therefore \,{\text{log}}\,\,{K_{sp}} = - 11.09 \cr & \Rightarrow {K_{sp}} = {\text{antilog}}\left( { - 11.09} \right) \cr & \,\,\,\,\,\,\,{K_{sp}} = 8.0 \times {10^{ - 12}} \cr} $$

The condition for the equilibrium is
$${\left( {\Delta G} \right)_{T,P}} = 0\,\,{\text{and}}\,\,{\left( {\Delta S} \right)_{U,V}} = 0$$