$$0.6\,mole$$ of $$PC{l_5},0.3\,mole$$ of $$PC{l_3}$$ and $$0.5\,mole$$ of $$C{l_2}$$ are taken in a $$1\,L$$ flask to obtain the following equilibrium : $$PC{l_{5\left( g \right)}} \rightleftharpoons PC{l_{3\left( g \right)}} + C{l_{2\left( g \right)}}$$
If the equilibrium constant $${K_c}$$ for the reaction is 0.2, predict the direction of the reaction.
A.
Forward direction
B.
Backward direction
C.
Direction of the reaction cannot be predicted
D.
Reaction does not move in any direction.
Answer :
Backward direction
Solution :
$$\eqalign{
& PC{l_{5\left( g \right)}} \rightleftharpoons PC{l_{3\left( g \right)}} + C{l_{2\left( g \right)}} \cr
& {Q_c} = \frac{{0.5 \times 0.3}}{{0.6}} = 0.25 \cr} $$
$${K_c} = 0.2,$$ Since, $${Q_c} > {K_c}$$ reaction will proceed in backward direction.
Releted MCQ Question on Physical Chemistry >> Chemical Equilibrium
Releted Question 1
For the reaction : $${H_2}\left( g \right) + {I_2}\left( g \right) \rightleftharpoons 2HI\left( g \right)$$ the equilibrium constant $${K_p}$$ changes with