The efflorescent salts loss water to atmosphere
$$\eqalign{ & {K_p} = p_{{H_2}O}^2 = 1.086 \times {10^{ - 4}}; \cr & {p_{{H_2}O}} = 1.042 \times {10^{ - 2}}atm = 7.92\,mm \cr} $$
If $${H_2}O$$  pressure at $${25^ \circ }C$$  is less than $$7.92\,mm.$$
The reaction
$$CuS{O_4}.5{H_2}O\left( s \right) \to CuS{O_4}.3{H_2}O\left( s \right) + {H_2}O\left( g \right)$$
will not proceed in $$RHS.$$

$$\eqalign{ & \Delta {G^ \circ }_{NO\left( g \right)} = 86.6k\,J/mol = 86600J/mol;\,{G^ \circ }_{N{O_2}\left( g \right)} = x\,J/mol \cr & T = 298,\,{K_p} = 1.6 \times {10^{12}} \cr & \Delta {G^ \circ } = - RT\,\ln \,{K_p} \cr & {\text{Given equation,}} \cr & 2NO\left( g \right) + {O_2}\left( g \right) \rightleftharpoons 2N{O_2}\left( g \right) \cr & \therefore \,\,2\Delta {G^ \circ }_{N{O_2}} - 2\Delta {G^ \circ }_{NO} = - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right) \cr & 2\Delta {G^ \circ }_{N{O_2}} - 2 \times 86600 = - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right) \cr & 2\Delta {G^ \circ }_{N{O_2}} = 2 \times 86600 - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right) \cr & \Delta {G^ \circ }_{N{O_2}} = \frac{1}{2}\left[ {2 \times 86600 - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)} \right] \cr & = 0.5\left[ {2 \times 86600 - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)} \right] \cr} $$

If the volume is kept constant and an inert gas is added that does not take part in the reaction, the equilibrium remains undisturbed.

$$PC{l_{5\left( g \right)}} \rightleftharpoons PC{l_{3\left( g \right)}} + C{l_{2\left( g \right)}};$$      $${K_c} = 8.3 \times {10^{ - 3}}$$
$${\text{For}}\,\,PC{l_{3\left( g \right)}} + C{l_{2\left( g \right)}} \rightleftharpoons PC{l_{5\left( g \right)}};$$       $${K_c} = \frac{1}{{8.3 \times {{10}^{ - 3}}}}$$
$${K_c} = \frac{1}{{8.3}} \times {10^3} = 120.48$$

$$\eqalign{ & \mathop a\limits_x + \mathop b\limits_x \rightleftharpoons \mathop c\limits_{2x} + \mathop d\limits_{2x} \cr & {K_c} = \frac{{2x.2x}}{{x.x}} = \frac{{4{x^2}}}{{{x^2}}} \Rightarrow {K_c} = 4 \cr} $$

In heterogeneous system, $${K_C}$$  and $${K_p}$$  are not depend upon the concentration or pressure of solid substance. Hence, at equilibrium their concentration or pressure are assumed as one.
$$\eqalign{ & MgC{O_3}\left( s \right) \rightleftharpoons MgO\left( s \right) + C{O_2}\left( g \right) \cr & \therefore \,\,\,{K_p} = {p_{C{O_2}}} \cr} $$

$$\eqalign{ & {P_{total}} = 3P \cr & \Rightarrow \,\,P = \frac{{0.318}}{3} = 0.106 \cr & \therefore \,\,{K_p} = 4{P^3} = 4.76 \times {10^{ - 3}} \cr} $$

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{N_2}{O_4}\left( g \right) \rightleftharpoons 2N{O_2}\left( g \right) \cr & {\text{Initial moles}}\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \cr & {\text{Moles at equil}}{\text{.}}\,\,\left( {1 - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\alpha \cr & \left( {\alpha = {\text{degree of dissociation}}} \right) \cr & {\text{Total number of moles at equil}}{\text{.}} \cr & = \left( {1 - \alpha } \right) + 2\alpha \cr & = \left( {1 + \alpha } \right) \cr & {p_{{N_2}{O_4}}} = \frac{{\left( {1 - \alpha } \right)}}{{\left( {1 + \alpha } \right)}} \times P \cr & {p_{N{O_2}}} = \frac{{2\alpha }}{{\left( {1 + \alpha } \right)}} \times P \cr & {K_P} = \frac{{{{\left( {{p_{N{O_2}}}} \right)}^2}}}{{{p_{{N_2}{O_4}}}}} \cr & = \frac{{{{\left( {\frac{{2\alpha }}{{\left( {1 + \alpha } \right)}} \times P} \right)}^2}}}{{\left( {\frac{{1 - \alpha }}{{1 + \alpha }}} \right) \times P}} \cr & = \frac{{4{\alpha ^2}P}}{{1 - {\alpha ^2}}} \cr & {\text{Given,}}\,\,{K_P} = 2,\,P = 0.5\,atm \cr & \therefore \,\,{K_P} = \frac{{4{\alpha ^2}P}}{{1 - {\alpha ^2}}} \cr & = \frac{{4{\alpha ^2} \times 0.5}}{{1 - {\alpha ^2}}} \cr & \alpha = 0.707 \approx 0.71 \cr & \therefore \,\,{\text{Percentage dissociation}} \cr & = 0.71 \times 100 = 71 \cr} $$

$$\eqalign{ & {\text{For a reaction,}} \cr & \mathop A\limits_{{\text{Reactant}}} \rightleftharpoons \mathop B\limits_{{\text{Product}}} \cr & K = \frac{{{{\left[ B \right]}_{eq}}}}{{{{\left[ A \right]}_{eq}}}} \cr & 1.6 \times {10^{12}} = \frac{{{{\left[ B \right]}_{eq}}}}{{{{\left[ A \right]}_{eq}}}} \cr & \therefore \,\,{\left[ B \right]_{eq}} > > {\left[ A \right]_{eq}} \cr} $$
So, mostly the product will be present in the equilibrium mixture.

\[\begin{matrix} {} & N{{H}_{4}}H{{S}_{\left( s \right)}} & \rightleftharpoons & N{{H}_{3\left( g \right)}} \\ \text{Initial}\,\,\text{moles} & 1 & {} & 0 \\ \text{At}\,\,\text{equilibrium} & \left( 1-x \right) & {} & x \\ {} & {} & {} & {} \\ \end{matrix}\begin{matrix} + \\ {} \\ {} \\ {} \\ \end{matrix}\begin{matrix} {{H}_{2}}{{S}_{\left( g \right)}} \\ 0 \\ x \\ {} \\ \end{matrix}\]
Total gaseous moles at equilibrium $$ = x + x = 2x$$
We know,$${\text{ }}{K_p} = {p_{N{H_3}}} \times {p_{{H_2}S}}$$
But, partial pressure $$(p)$$  = mole fraction × total pressure $$(P)$$
$$\eqalign{ & {K_p} = \left( {\frac{x}{{2x}} \times P} \right)\left( {\frac{x}{{2x}} \times P} \right) \cr & \,\,\,\,\,\,\,\,\, = {\left( {\frac{P}{2}} \right)^2} \cr & \,\,\,\,\,\,\,\,\, = {\left( {\frac{{1.12}}{2}} \right)^2} \cr & \,\,\,\,\,\,\,\,\, = 0.3136 \cr} $$