For a reversible reaction, if the concentrations of the reactants are doubled, the equilibrium constant will be
A.
one-fourth
B.
halved
C.
doubled
D.
the same
Answer :
the same
Solution :
$$\eqalign{
& {\text{Consider a hypothetical change,}} \cr
& A + B \rightleftharpoons C + D \cr
& {\text{For this reaction,}} \cr
& {K_{eq}} = \frac{{\left[ C \right]\left[ D \right]}}{{\left[ A \right]\left[ B \right]}} \cr} $$
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.
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