Chemical Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

The reaction given is an exothermic reaction thus accordingly to Lechatalier’s principle lowering of temperature, addition of $${F_2}$$  and or $$C{l_2}$$  favour the for ward direction and hence the production of $$Cl{F_3}$$ .

Endothermic reaction is favoured at high temperature.

$$\eqalign{ & {K_c} = \frac{{{k_f}}}{{{k_r}}},{k_f} = {K_c} \cdot {k_r} \cr & = \left( {57.0} \right)\left( {1.16 \times {{10}^{ - 3}}\,{M^{ - 1}}\,{s^{ - 1}}} \right) \cr & = 6.61 \times {10^{ - 2}}\,{M^{ - 1}}\,{s^{ - 1}} \cr} $$

$$2HI \rightleftharpoons {H_2} + {I_2}.$$    It is 22% decomposed ,
$$\therefore \,\,\frac{{3.20 \times 22}}{{100}} = 0.704$$
$$(3.2–0.704)$$   is equal to $$HI$$  present at equilibrium which is = 2.496

$$\eqalign{ & {\text{Moles of}}\,{N_2} = \frac{{28}}{{28}} = 1, \cr & {\text{Moles of}}\,{H_2} = \frac{6}{2} = 3 \cr & {\text{Moles of}}\,{H_2}S{O_4}\,{\text{required}} = \frac{{500 \times 1}}{{1000}} = 0.5 \cr & {\text{Moles of}}\,N{H_3}\,{\text{neutralised by}}\,{H_2}S{O_4} = 1.0 \cr & \left( {2N{H_3} + {H_2}S{O_4} \to {{\left( {N{H_4}} \right)}_2}S{O_4}} \right) \cr} $$

Hence 1 $$mol$$  of $$N{H_3}$$  by the reaction between $${N_2}$$  and $${H_2}.$$

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{N_2}\,\,\,\,\,\,\, + \,\,\,\,\,\,3{H_2}\,\,\,\,\,\,\,\,\,\, \rightleftharpoons 2\,N{H_3} \cr & {\text{initial}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \cr & {\text{at eqm}}\,\,\,\,1 - 0.5\,\,\,\,\,\,\,3 - 0.5 \times 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \cr & {K_c} = \frac{{1 \times 1}}{{0.5 \times {{\left( {1.5} \right)}^3}}} = 0.592 \cr} $$

Equilibrium can be attained by either side of the reactions of equilibrium.

In this reaction the ratio of number of moles of reactants to products is same i.e. 2 : 2, hence change in volume will not alter the number of moles.

\[\underset{\begin{smallmatrix} \\ \text{Moles}\,\,\text{at}\,\,\text{eqm}\text{.} \end{smallmatrix}}{\mathop{{}}}\,\underset{2-x}{\mathop{{{A}_{2\left( g \right)}}}}\,+\underset{4-x}{\mathop{{{B}_{2\left( g \right)}}}}\,\rightleftharpoons \underset{2x}{\mathop{2A{{B}_{\left( g \right)}}}}\,\]
$$\eqalign{ & {K_c} = \frac{{4{x^2}}}{{\left( {2 - x} \right)\left( {4 - x} \right)}} \Rightarrow x = 1.33\,mole \cr & \left[ {AB} \right] = \frac{{2 \times 1.33}}{4} = 0.66\,M \cr} $$

For the reaction,
$$\eqalign{ & C{H_4}\left( g \right) + 2{O_2}\left( g \right) \rightleftharpoons C{O_2}\left( g \right) + 2{H_2}O\left( l \right) \cr & \Delta {H_r} = - 170.8\,k\,J\,mo{l^{ - 1}} \cr} $$
This equilibrium is an example of heterogeneous chemical equilibrium. Hence, for it
$${K_c} = \frac{{\left[ {C{O_2}} \right]}}{{\left[ {C{H_4}} \right]{{\left[ {{O_2}} \right]}^2}}}\,\,\,...{\text{(i)}}$$
( equilibrium constant on the basis of concentration )
and   $${K_p} = \frac{{{p_{C{O_2}}}}}{{{p_{C{H_4}}} \times {P_{O_2^2}}}}\,\,...{\text{(ii)}}$$
( equilibrium constant according to partial pressure )
Thus, in this concentration of $$C{O_2}\left( g \right)$$   and $${H_2}O\left( l \right)$$  are not equal at equilibrium.
The equilibrium constant $$\left( {{K_p}} \right) = \frac{{\left[ {C{O_2}} \right]}}{{\left[ {C{H_4}} \right]\left[ {{O_2}} \right]}}$$     is not correct expression.
On adding $$C{H_4}\left( g \right)$$   or $${O_2}\left( g \right)$$  at equilibrium, $${K_C}$$  will be decreased according to expression (i) but $${K_C}$$  remains constant at constant temperature for a reaction, so for maintaining the constant value of $${K_C},$$  the concentration of $$C{O_2}$$  will increased in same order. Hence, on addition of $$C{H_4}$$  or $${O_2}$$  equilibrium will cause to the right.
Combustion reaction is an example of exothermic reaction.

The given reaction will be exothermic in nature due to the formation of three $$X{\text{ - }}Y$$  bonds from the gaseous atoms. The reaction is also accompanied with the decrease in the gaseous species $$\left( {{\text{i}}{\text{.e}}.\,\,\Delta n\,{\text{is}}\,{\text{negative}}} \right).$$
Hence, the reaction will be affected by both temperature and pressure. The use of catalyst does not affect the equilibrium concentrations of the species in the chemical reaction.