Evaporation process continues, till the entire water changes into vapour form. So in an open vessel, the water-vapour equilibrium cannot be achieved.

\[\underset{\begin{smallmatrix} \\ \text{In initial} \\ \text{At}\,\,\text{equilibrium} \end{smallmatrix}}{\mathop{{}}}\,\underset{\begin{smallmatrix} 2\,\,mol \\ \left( 2-2\alpha \right)mol \end{smallmatrix}}{\mathop{2HI\left( g \right)}}\,\,\rightleftharpoons \underset{\begin{smallmatrix} 0\,\,mol \\ \alpha \,mol \end{smallmatrix}}{\mathop{{{H}_{2}}\left( g \right)}}\,+\underset{\begin{smallmatrix} 0\,\,mol \\ \alpha \,mol \end{smallmatrix}}{\mathop{{{I}_{2}}\left( g \right)}}\,\]
$$\eqalign{ & {\text{So, at equilibrium total moles}} \cr & = 2 - 2\alpha + \alpha + \alpha \cr & = 2 - 2\alpha + 2\alpha \cr & = 2 \cr} $$

Volume of ice is greater than that of water. The direction in which the reaction will proceed can be predicted by applying Le-Chatelier's principle
$${\text{Pressure}} \propto \frac{1}{{{\text{Volume}}}}$$
So equilibrium, will shift forward.

$${X_2}\left( g \right) + 4{Y_2} \rightleftharpoons 2X{Y_4}\left( g \right),$$      where $$\Delta H < 0$$   and $$\Delta n < 0\left[ {\Delta n = {n_P} - {n_R}} \right]$$
∴ The forward reaction is favoured at high pressure and low temperature. ( According to Le-Chatelier's principle )

$$\Delta {n_g} = {n_{g\left( p \right)}} - {n_{g\left( r \right)}}$$

If $${Q_c} = {K_c},$$  reaction is in equilibrium.

Since reaction is exothermic it is favoured at low temperature. There is a decrease in number of moles hence reaction is favoured at high pressure.

\[\underset{\begin{smallmatrix} \text{start} \\ \text{At equib}\text{.} \end{smallmatrix}}{\mathop{N{{H}_{4}}HS\left( s \right)}}\,\rightleftharpoons \underset{\begin{smallmatrix} \text{0}\text{.5 atm} \\ 0.54\,+\,x\,\,atm \end{smallmatrix}}{\mathop{N{{H}_{3}}\left( g \right)}}\,+\underset{\begin{smallmatrix} 0\text{ }atm \\ x\text{ }atm. \end{smallmatrix}}{\mathop{{{H}_{2}}S\left( g \right)}}\,\]
$$\eqalign{ & {\text{Then}}\,\,0.5 + x + x = 2x + 0.5 = 0.84\,{\text{(given)}} \cr & \Rightarrow x = 0.17\,{\text{atm}}{\text{.}} \cr & {p_{N{H_3}}} = 0.5 + 0.17 = 0.67\,{\text{atm}}\,;\,\,{p_{{H_2}S}} = 0.17\,{\text{atm}} \cr & K = {p_{N{H_3}}} \times {p_{{H_2}S}} = 0.67 \times 0.17\,{\text{at}}{{\text{m}}^2} = 0.1139 = 0.11 \cr} $$

$$\eqalign{ & {K_p} = \frac{{{p_{C{H_3}OH}}}}{{{p_{CO}} \times {p_{{H_2}}}}} \cr & = \frac{2}{{1 \times {{\left( {0.1} \right)}^2}}} \cr & = 200; \cr & {\text{For reverse reaction}} \cr & \frac{1}{{{K_p}}} = \frac{1}{{200}} = 5 \times {10^{ - 3}}at{m^2} \cr} $$

Key Idea The reaction for which the number of moles of gaseous products $$\left( {{n_p}} \right)$$  is not equal to the number of moles of gaseous reactants $$\left( {{n_R}} \right),$$  has different value of $${K_C}$$  and $${K_p}.$$
$$\eqalign{ & {\text{From the equation,}}\,{K_p} = {K_C} \times {\left( {RT} \right)^{\Delta \,{n_g}}} \cr & {\text{where, }}\left[ {\Delta {n_g}\,{\text{gaseous}} = {n_p} - {n_R}} \right] \cr & \left( {\text{a}} \right)\,{n_p} = {n_R} = 2,\,{\text{thus,}}\,{K_p} = {K_C} \cr & \left( {\text{b}} \right)\,{n_p} = {n_R} = 2,\,{\text{thus,}}\,{K_p} = {K_C} \cr & \left( {\text{c}} \right)\,{n_p} = {n_R} = 2,\,{\text{thus,}}\,{K_p} = {K_C} \cr & \left( {\text{d}} \right)\,{n_p} = 2,\,\,{n_R} = 1,\,{\text{thus,}}\,{K_p} \ne {K_C} \cr} $$