Question
A mixture of $$1.57\,mol$$ of $${N_2},1.92\,mol$$ of $${H_2}$$ and $$8.13\,mol$$ of $$N{H_3}$$ is introduced into $$20\,L$$ reaction vessel at $$500\,K.$$ At this temperature, the equilibrium constant, $${K_c}$$ for the reaction, $${N_{2\left( g \right)}} + 3{H_{2\left( g \right)}} \rightleftharpoons 2N{H_{3\left( g \right)}}$$ is $$1.7 \times {10^2}.$$ What is the direction of the net reaction?
A.
Forward
B.
Backward
C.
At equilibrium
D.
Data is insufficient
Answer :
Backward
Solution :
The reaction is $${N_{2\left( g \right)}} + 3{H_{2\left( g \right)}} \rightleftharpoons 2N{H_{3\left( g \right)}}$$
$$\eqalign{
& {Q_c} = \frac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}} \cr
& \,\,\,\,\,\,\,\, = \frac{{{{\left( {\frac{{8.13}}{{20}}\,mol\,{L^{ - 1}}} \right)}^2}}}{{\left( {\frac{{1.57}}{{20}}\,mol\,{L^{ - 1}}} \right){{\left( {\frac{{1.92}}{{20}}\,mol\,{L^{ - 1}}} \right)}^3}}} \cr
& \,\,\,\,\,\,\,\, = 2.38 \times {10^3} \cr} $$
As $${Q_c} \ne {K_c},$$ the reaction mixture is not in equilibrium.
As $${Q_c} > {K_c},$$ the net reaction will be in the backward direction.