Chemical Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

TIPS/Formulae :
At constant temperature $${K_p}$$  or $${K_c}$$  remains constant.
For the equilibria :
$$\eqalign{ & {N_2}{O_4}\left( g \right) \rightleftharpoons 2N{O_2}\left( g \right) \cr & {\text{NOTE: }}{K_p}{\text{ = }}{K_c}{\text{ because here }}\Delta {\text{n = 1}} \cr} $$
$$\left[ {{K_P} = {K_C} \times {{\left( {RT} \right)}^{\Delta n}}} \right]$$     Since temperature is constant so$${K_c}$$  or $${K_p}$$  will remain constant. Further since volume is halved, the pressure will be doubled so $$\alpha $$ will decrease so as to maintain the constancy of $${K_c}$$  or $${K_p}$$ .
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{N_2}{O_4} \rightleftharpoons 2N{O_2} \cr & {\text{initial}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\, \cr & {\text{at equilibrium}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 - \alpha } \right)\,\,\,\,\,\,\,\,\,2\alpha \,\, \cr} $$
$$\eqalign{ & \therefore \,\,{\text{Total mole}} = 1 - \alpha + 2\alpha = 1 + \alpha \cr & {\text{Let total pressure = P}} \cr & \therefore \,\,p{N_2}{O_4} = \frac{{1 - \alpha }}{{1 + \alpha }}.p\,\,;\,pN{O_2} = \frac{{2\alpha }}{{1 - \alpha }}p \cr & {K_p} = \frac{{{{\left( {pN{O_2}} \right)}^2}}}{{p{N_2}{O_4}}} = \frac{{4{\alpha ^2} \times p}}{{\left( {1 - \alpha } \right)\left( {1 + \alpha } \right)}} = \frac{{4{\alpha ^2}p}}{{1 - {\alpha ^2}}} \cr & {\text{Since}}\,{K_p} = {\text{constant, so}}\,\alpha \propto \frac{1}{{\sqrt p }} \cr} $$
So when volume is halved, pressure gets doubled and thus $$\alpha $$ will decrease.

$$\mathop {Fe_{\left( {aq} \right)}^{3 + }}\limits_{{\text{Yellow}}} + \mathop {SCN_{\left( {aq} \right)}^ - }\limits_{{\text{Colourless}}} \rightleftharpoons \mathop {\left[ {Fe\left( {SCN} \right)} \right]_{\left( {aq} \right)}^{2 + }}\limits_{{\text{Deep red}}} $$
When oxalic acid is added, it reacts with $$F{e^{3 + }}$$  ions to form stable complex ion $${\left[ {Fe{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{3 - }},$$   thus decreasing the cone. of free $$Fe_{\left( {aq} \right)}^{3 + }.$$  Now, according to Le Chatelier's principle, the reaction will shift in backward direction to increase the cone. of free $$Fe_{\left( {aq} \right)}^{3 + }.$$  Thus, conc. of $${\left[ {Fe\left( {SCN} \right)} \right]^{2 + }}$$   decreases, so the intensity of red colour decreases.

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,HI \rightleftharpoons \,\,\,\,\,\,\,\frac{1}{2}{H_2} + \frac{1}{2}{I_2} \cr & {\text{at}}\,t = 0\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \cr & {\text{at eq}}{\text{.}}\,\,\,\,\,\,\,\,\,\,\left( {1 - 0.5} \right)\,\,\,\,\,\left( {0.5} \right)\,\,\,\,\,\,\left( {0.5} \right) \cr & \left( {\because \,\,\alpha = \frac{{50}}{{100}} = 0.5} \right) \cr & {\text{Now}}\,\,{K_C} = \frac{{{{\left( {0.5} \right)}^{\frac{1}{2}}}{{\left( {0.5} \right)}^{\frac{1}{2}}}}}{{0.5}} = 1 \cr} $$

$${K_c}$$  and $${K_p}$$  depend upon values of $$a,b, c$$  and $$d.$$

Since $${T_2}$$  has similar chemical properties as $${H_2}$$  so upon mixing, $$N{H_2}T$$  and $$NH{T_2}$$  are formed.

$$\eqalign{ & {\text{Consider a hypothetical change,}} \cr & A + B \rightleftharpoons C + D \cr & {\text{For this reaction,}} \cr & {K_{eq}} = \frac{{\left[ C \right]\left[ D \right]}}{{\left[ A \right]\left[ B \right]}} \cr} $$
For the above reaction if concentration of reactants are doubled then the rate of forward reaction increases for a short time but after sometime equilibrium will established. So, concentration has no effect on equilibrium constant. It remains unchanged after increasing the concentration of reactants.

$$\eqalign{ & {K_P} = {K_C}{\left( {RT} \right)^{\Delta {n_g}}} \cr & {\text{For the reaction}} \cr & CO\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \rightleftharpoons C{O_2}\left( g \right) \cr & \Delta {n_g} = 1 - \left( {1 + \frac{1}{2}} \right) = - \frac{1}{2} \cr & \therefore \,\,{K_P} = \frac{{{K_C}}}{{\sqrt {RT} }};\,\,\frac{{{K_P}}}{{{K_C}}} = \frac{1}{{\sqrt {RT} }} \cr} $$

Let the initial moles of $$X$$ be $$'a’$$ and that of $$Z$$ be $$'b’$$ then for the given reactions, we have
\[\underset{\begin{smallmatrix} \text{Initial} \\ \text{At equi}\text{.} \\ \text{(moles) } \end{smallmatrix}}{\mathop{{}}}\,\underset{\begin{smallmatrix} \text{a moles} \\ \text{a}\,\left( \text{1-}\alpha \right)\text{ } \end{smallmatrix}}{\mathop{X}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ 2\text{a}\alpha \end{smallmatrix}}{\mathop{2Y}}\,\]
$$\eqalign{ & {\text{Total no}}{\text{. of moles}} \cr & = \,{\text{a}}\left( {1 - \alpha } \right) + 2{\text{a}}\alpha \cr & = {\text{a}} - {\text{a}}\alpha + 2{\text{a}}\alpha = {\text{a}}\left( {1 + \alpha } \right) \cr & {\text{Now,}}\,\,\,{K_{{P_1}}} = \frac{{{{\left( {{n_y}} \right)}^2}}}{{{n_X}}} \times {\left( {\frac{{{P_{{T_1}}}}}{{\sum n }}} \right)^{\Delta n}} \cr & {\text{or}},\,\,\,\,\,\,\,\,{K_{{P_1}}} = \frac{{{{\left( {2{\text{a}}\alpha } \right)}^2}.{P_{{T_1}}}}}{{\left[ {{\text{a}}\left( {1 - \alpha } \right)} \right]\left[ {{\text{a}}\left( {1 + \alpha } \right)} \right]}} \cr} $$
\[\underset{\begin{smallmatrix} \text{Initial} \\ \text{At equi}\text{.} \\ \text{(moles) } \end{smallmatrix}}{\mathop{{}}}\,\underset{\begin{smallmatrix} b\text{ moles} \\ b\,\left( \text{1-}\alpha \right)\text{ } \end{smallmatrix}}{\mathop{X}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ b\alpha \end{smallmatrix}}{\mathop{P}}\,+\underset{\begin{smallmatrix} 0 \\ b\alpha \end{smallmatrix}}{\mathop{Q}}\,\]
$$\eqalign{ & {\text{Total no}}{\text{. of moles }} \cr & {\text{ = }}\,\,b\left( {1 - \alpha } \right) + b\alpha + b\alpha \cr & = b - b\alpha + b\alpha + b\alpha \cr & = b\left( {1 + \alpha } \right) \cr & {\text{Now}}\,\,{K_{{P_2}}} = \frac{{{n_Q} \times {n_P}}}{{{n_z}}} \times {\left[ {\frac{{{P_{{T_2}}}}}{{\sum\nolimits_n {} }}} \right]^{{\Delta _n}}} \cr & {\text{or}}\,\,\,{K_{{P_2}}} = \frac{{\left( {b\alpha } \right)\left( {b\alpha } \right).{P_{{T_2}}}}}{{\left[ {b\left( {1 - \alpha } \right)} \right]\left[ {b\left( {1 + \alpha } \right)} \right]}} \cr & {\text{or}}\,\,\,\frac{{{K_{{P_1}}}}}{{{K_{{P_2}}}}} = \frac{{4{\alpha ^2}.{P_{{T_1}}}}}{{\left( {1 - {\alpha ^2}} \right)}} \times \frac{{{{\left( {1 - \alpha } \right)}^2}}}{{{P_{{T_2}}}.{\alpha ^2}}} = \frac{{4{P_{{T_1}}}}}{{{P_{{T_2}}}}} \cr & {\text{or}}\,\frac{{{P_{{T_1}}}}}{{{P_{{T_2}}}}} = \frac{1}{9}\,\,\left[ {\because \,\,\,\,\frac{{{K_{{P_1}}}}}{{{K_{{P_2}}}}} = \frac{1}{9}\,{\text{given}}} \right] \cr & {\text{or}}\,\,\frac{{{P_{{T_1}}}}}{{{P_{{T_2}}}}} = \frac{1}{{36}}\,\,or\,\,1:36 \cr & {\text{i}}{\text{.e}}{\text{., (A) is the correct answer}}{\text{.}} \cr} $$

$$\eqalign{ & \mathop X\limits_{0.06} + \mathop {2Y}\limits_{0.12} \rightleftharpoons \mathop Z\limits_{0.216} \cr & {K_c} = \frac{{\left[ Z \right]}}{{\left[ X \right]{{\left[ Y \right]}^2}}} \cr & \,\,\,\,\,\,\,\, = \frac{{0.216}}{{0.06 \times 0.12 \times 0.12}} \cr & \,\,\,\,\,\,\,\, = 250 \cr} $$

$$2XY \rightleftharpoons {X_2} + {Y_2};{K_c} = 81$$
$$XY \rightleftharpoons \frac{1}{2}{X_2} + \frac{1}{2}{Y_2};K{'_c} = ?;$$      $$K{'_c} = \sqrt {{K_c}} = \sqrt {81} = 9$$