Question

The initial concentration of N2O4 is 0.45M The system is heated and allowed to reach equilibrium. At equilibrium concentration of NO2 is 0.30 M. Calculate the equilibrium constant for the reaction: N2O4 \Longleftrightarrow ⟺ 2NO2

Answer 1

Answer : The value of equilibrium constant $K_c$ for the following reaction will be, 2.4

Explanation :  Given,

Initial concentration of $N_2O_4$ = 0.45 M

Equilibrium concentration of $NO_2$ = 0.30 M

$K_c$ is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The given balanced equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

As we know that the concentrations of pure solids are constant that means they do not change. Thus, they are not included in the equilibrium expression.

Actual concentration of $NO_2$ at equilibrium = $2\times 0.30M=0.6M$

Concentration of $N_2O_4$ at equilibrium = $[N_2O_4]-[NO_2]=0.45-0.30=0.15M$

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

Now put all the given values in this expression, we get :

$K_c=\frac{(0.6)^2}{(0.15)}$

$K_c=2.4$

Therefore, the value of equilibrium constant $K_c$ for the following reaction will be, 2.4