Due to absence of hydrolysation of $$FeC{l_3}$$  ackward reaction will not take place.

$$\eqalign{ & {\text{For}}\,\,Fe_{\left( {aq} \right)}^{3 + } + 3OH_{\left( {aq} \right)}^ - \rightleftharpoons Fe{\left( {OH} \right)_{3\left( s \right)}} \cr & {K_c} = \frac{1}{{\left[ {F{e^{3 + }}} \right]{{\left[ {O{H^ - }} \right]}^3}}} \cr} $$

On adding inert gas at constant volume the total pressure of the system is increased, but the partial pressure of each reactant and product remains the same. Hence no effect on the state of equilibrium.

Smaller value of equilibrium constant $$\left( { < {{10}^{ - 3}}} \right)$$   signifies the greater concentration of reactants as compared to that of products.

Equilibrium shifted in the backward direction. ( $$Le$$  Chatelier's principle )

$$\eqalign{ & NO \rightleftharpoons \frac{1}{2}{N_2} + \frac{1}{2}{O_2};{K_c} = 0.25 \cr & \frac{1}{2}{N_2} + \frac{1}{2}{O_2} \rightleftharpoons NO;{K_c} = \frac{1}{{0.25}} = 4 \cr} $$

The value of $${K_c}$$  for the reverse reaction will be $$\frac{1}{{8.2 \times {{10}^4}}}.$$

Any change in the concentration, pressure and temperature of the reaction results in change in the direction of equilibrium. This change in the direction of equilibrium is governed by Le-Chatelier's principle. According to this equilibrium shifts in the opposite direction to undo the change,
$${N_2}\left( g \right) + 3{H_2}\left( g \right) \rightleftharpoons 2N{H_3}\left( g \right) + Heat$$
(a) Increasing the concentration of $$N{H_3}\left( g \right):$$   On increasing the concentration of $$N{H_3}\left( g \right),$$   the equilibrium shifts in the backward direction where concentration of $$N{H_3}\left( g \right)$$  decreases.
(b) Decreasing the pressure: Since, $$p \propto n$$  ( number of moles ), therefore, equilibrium shifts in the backward direction where number of moles ate increasing.
(c) Decreasing the concentration of $${N_2}\left( g \right)$$  and $${H_2}\left( g \right):$$   Equilibrium shifts in the backward direction when concentration of $${N_2}\left( g \right)$$  and $${H_2}\left( g \right)$$  decreases.
(d) Increasing pressure and decreasing temperature: On increasing pressure, equilibrium shifts in the forward direction where number of moles decreases. It is an example of exothermic reaction therefore decreasing temperature favours the forward direction.

$$\eqalign{ & \,\,A\,\,\,\,\, + \,\,\,\,\,\,\,\,\,2B\,\,\,\,\,\,\,\,\, \rightleftharpoons \,\,\,\,2C + D \cr & \,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1.5a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,0 \cr & \left( {a - x} \right)\,\left( {1.5a - 2x} \right)\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,x \cr & {\text{Hence}}\,\,{K_C} = \frac{{\left( {2{x^2}} \right) \times x}}{{\left( {a - x} \right){{\left( {1.5a - 2x} \right)}^2}}} \cr & {\text{Given, at equilibrium}} \cr & \therefore \,\,\left( {a - x} \right) = \left( {1.5a - 2x} \right) \cr & \therefore \,\,a = 2x \cr & {\text{On solving}}\,\,{K_C} = 4 \cr} $$

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2S{O_3}\, \rightleftharpoons \,2S{O_2} + {O_2} \cr & {\text{Initial}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \cr & {\text{At equilibrium}}\,\,\,\,\,\,1 - 2\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,2\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha \cr & \therefore \,\,2\alpha \, = 0.6,\,\,\,\therefore \,\,\alpha \, = 0.3 \cr & \left[ {S{O_3}} \right] = 1 - 2\alpha \, = 1 - 0.6 = 0.4 \cr & \left[ {S{O_2}} \right] = 2\alpha \, = 0.6 \cr & \left[ {{O_2}} \right] = \alpha \, = 0.3 \cr & {K_{eq}} = \frac{{\alpha \, \times {{\left( {2\alpha \,} \right)}^2}}}{{{{\left( {1 - 2\alpha \,} \right)}^2}}} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{0.3 \times 0.6 \times 0.6}}{{0.4 \times 0.4}} \cr & \,\,\,\,\,\,\,\,\,\, = 0.675 \cr} $$