If reaction is multiplied by 2, the equilibrium constant becomes square of the previous value.
$$K = {7^2} = 49$$

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

$$\eqalign{ & {K_p} = {K_c}{\left( {RT} \right)^{\Delta n}}; \cr & \Delta n = 1 - \left( {1 + \frac{1}{2}} \right) \cr & = 1 - \frac{3}{2} \cr & = - \frac{1}{2} \cr & \therefore \,\,\,\frac{{{K_p}}}{{{K_c}}} = {\left( {RT} \right)^{ - \frac{1}{2}}} \cr} $$

$$\mathop {PC{l_5}\left( g \right)}\limits_{1 - x} \rightleftharpoons \mathop {PC{l_3}\left( g \right)}\limits_x + \mathop {C{l_2}\left( g \right)}\limits_x $$
Total moles after dissociation $$1 - x + x + x = 1 + x$$
$${p_{PC{l_3}}} = $$  mole fraction of $$PC{l_3}$$  × Total pressure
$$\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = }}\left( {\frac{x}{{1 + x}}} \right)P$$

By addition of $$S{O_2}$$  equilibrium will shift to $$RHS$$  which is exothermic. Hence temperature will increase.

Since, $$O{H^ - }$$  are generated from weak acid $$\left( {{H_2}O} \right),$$  and a weak acid ( like $$C{O_2}$$  ) should be used to remove it. Because if we add strong acid like $$(HCl)$$  it reverse the reaction. $$KOH$$  increases the concentration of $$O{H^ - },$$  thus again shifts the reaction in backward side.
$$C{O_2}$$  combines with $$O{H^ - }$$  to give carbonate which is easily removed.
$$S{O_2}$$  reacts with water to give strong acid, so it cannot be used.

$$\eqalign{ & {K_c} = \frac{{\left[ {PC{l_3}} \right]\left[ {C{l_2}} \right]}}{{\left[ {PC{l_5}} \right]}} \cr & \,\,\,\,\,\,\,\, = \frac{{2.1 \times 2.1}}{{1.9}} \cr & \,\,\,\,\,\,\,\, = 2.32 \cr} $$

\[\begin{matrix} {} & 2S{{O}_{2}} & + \\ \text{Initial}\,\,\text{moles} & 5 & {} \\ \text{At}\,\,\text{equilibrium} & 5-3 & {} \\ \end{matrix}\begin{matrix} {{O}_{2}} & \rightleftharpoons & 2S{{O}_{3}} \\ 5 & {} & 0 \\ 5-\frac{3}{2} & {} & 3\left( 5\times \frac{60}{100}=3 \right) \\ \end{matrix}\]

Total number of moles in the vessel = 2 + 3.5 + 3 = 8.5

$$\eqalign{ & {\text{For equation (i),}} \cr & NO\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \rightleftharpoons N{O_2}\left( g \right) \cr & {K_1} = \frac{{\left[ {N{O_2}} \right]}}{{\left[ {NO} \right]{{\left[ {{O_2}} \right]}^{\frac{1}{2}}}}}\,\,\,...{\text{(i)}} \cr & {\text{For equation (ii),}} \cr & 2N{O_2}\left( g \right) \rightleftharpoons 2NO\left( g \right) + {O_2}\left( g \right) \cr & {K_2} = \frac{{{{\left[ {NO} \right]}^2}\left[ {{O_2}} \right]}}{{{{\left[ {N{O_2}} \right]}^2}}}\,\,\,...{\text{(ii)}} \cr & {\text{Now, on reversing equation (i), we get,}} \cr & \frac{1}{{{K_1}}} = \frac{1}{{\frac{{\left[ {N{O_2}} \right]}}{{\left[ {NO} \right]{{\left[ {{O_2}} \right]}^{\frac{1}{2}}}}}}} \cr & \,\,\,\,\,\,\,\,\, = \frac{{\left[ {NO} \right]{{\left[ {{O_2}} \right]}^{\frac{1}{2}}}}}{{\left[ {N{O_2}} \right]}} \cr & {\left( {\frac{1}{{{K_1}}}} \right)^2} = {\left\{ {\frac{{\left[ {NO} \right]{{\left[ {{O_2}} \right]}^{\frac{1}{2}}}}}{{\left[ {N{O_2}} \right]}}} \right\}^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left[ {NO} \right]}^2}\left[ {{O_2}} \right]}}{{{{\left[ {N{O_2}} \right]}^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {K_2} \cr & \frac{1}{{K_1^2}} = {K_2} \cr} $$

$$\eqalign{ & {\text{Equilibrium constant,}} \cr & K = \frac{{{k_f}}}{{{k_b}}} \cr & \,\,\,\,\,\, = \frac{{{\text{Forward rate constant}}}}{{{\text{Backward rate constant}}}} \cr & K = \frac{{1.1 \times {{10}^{ - 2}}}}{{1.5 \times {{10}^{ - 3}}}} \cr & \,\,\,\,\,\, = \frac{{1.1 \times 10}}{{1.5}} \cr & \,\,\,\,\,\, = \frac{{11}}{{1.5}} \cr & \,\,\,\,\,\, = 7.33 \cr} $$