Chemical Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\mathop {\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]_{\left( {aq} \right)}^{3 + }}\limits_{{\text{Pink}}} + 4Cl_{\left( {aq} \right)}^ - \rightleftharpoons $$       $$\mathop {\left[ {CoC{l_4}} \right]_{\left( {aq} \right)}^{2 - }}\limits_{{\text{Blue}}} + 6{H_2}{O_{\left( l \right)}}$$
On cooling the mixture, the reaction proceeds in backward direction and on heating the mixture, the reaction proceeds in forward direction, hence, the reaction is endothermic i.e., $$\Delta H > 0.$$

$$\eqalign{ & A{B_2} \rightleftharpoons {A^{ + 2}} + 2{B^ - } \cr & \left[ A \right] = 1.0 \times {10^{ - 5}},\left[ B \right] = \left[ {2.0 \times {{10}^{ - 5}}} \right], \cr & {K_{sp}} = {\left[ B \right]^2}\left[ A \right] \cr & = {\left[ {2 \times {{10}^{ - 5}}} \right]^2}\left[ {1.0 \times {{10}^{ - 5}}} \right] \cr & = 4 \times {10^{ - 15}} \cr} $$

$$\eqalign{ & S{O_3}\left( g \right) \rightleftharpoons S{O_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \cr & {K_c} = \frac{{\left[ {S{O_2}} \right]{{\left[ {{O_2}} \right]}^{\frac{1}{2}}}}}{{\left[ {S{O_3}} \right]}} = 4.9 \times {10^{ - 2}}; \cr} $$
On taking the square of the above reaction
$$\eqalign{ & \frac{{{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]}}{{{{\left[ {S{O_3}} \right]}^2}}} = 24.01 \times {10^{ - 4}} \cr & {\text{now}}\,K{'_C}\,{\text{for 2S}}{{\text{O}}_2}\left( g \right) + {O_2}\left( g \right) \rightleftharpoons 2S{O_3} \cr & = \frac{{{{\left[ {S{O_3}} \right]}^2}}}{{{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]}} = \frac{1}{{24.01 \times {{10}^{ - 4}}}} = 416 \cr} $$

Value of $${K_c}$$  predicts the extent of the reaction.

When we add heat to the equilibrium between solid and liquid, then the equilibrium shifts towards liquid and hence, the amount of solid decrease and amount of liquid increase.

At initial stage of reaction, concentration of each product will increase and hence $$Q$$ will increase.

At constant temperature $${{K_p}}$$  pressure constant. With change of pressure, $$x$$ will change in such a way that $${{K_p}}$$  remains a constant.

\[\begin{matrix} {} & W & + \\ \text{Initial} & 1 & {} \\ \text{At eqn}\text{.} & 1-a & {} \\ \end{matrix}\begin{matrix} X & \rightleftharpoons & Y \\ 1 & {} & 0 \\ 1-a & {} & a \\ \end{matrix}\,\,\,\,\,\,\begin{matrix} + \\ {} \\ {} \\ \end{matrix}\,\,\,\,\,\,\begin{matrix} Z \\ 0 \\ a \\ \end{matrix}\]
$$\eqalign{ & {K_{eq}} = 9 = \frac{{{\alpha ^2}}}{{{{\left( {1 - \alpha } \right)}^2}}} \cr & \therefore \,\,\alpha = 0.75 \cr & {\text{Moles of}}\,\,Y = 0.75 \cr} $$

In a reversible reaction, catalyst speeds up both the forward and backward reactions to the same extent, so (C) is wrong. At equilibrium,
$$\eqalign{ & \Delta G = {G_{{\text{poducts}}}} - {G_{{\text{reactants}}}} = 0 \cr & \Rightarrow 2{G_{N{H_3}}} - \left( {{G_{{N_2}}} + 3{G_{{H_2}}}} \right) = 0 \cr & or\,\,2{G_{N{H_3}}} = {G_{{N_2}}} + 3{G_{{H_2}}} \cr} $$

$$\eqalign{ & {\text{Applying}}\,\,\Delta {G^ \circ } = - 2.303\,RT \times \log K \cr & = - 2.303 \times 2 \times 298 \times \log \,200 \cr & = - 3158.4\,cal \cr & = - 3.158\,kcal \cr} $$