Question

Calculate the equilibrium concentrations of N2O4 and NO2 at 25 ∘C in a vessel that contains an initial N2O4 concentration of 0.0654 M . The equilibrium constant Kc for the reaction N2O4(g)⇌2NO2(g) is 4.64×10−3 at 25 ∘C.

Answer 1

Answer: The equilibrium concentration of $N_2O_4\text{ and }NO_2$ are 0.0164 M and 0.0572 M

Explanation:

We are given:

Initial concentration of $N_2O_4$ = 0.0654 M

The given chemical equation follows:

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

Initial:         0.0654

At eqllm:   0.0654 - x           2x

The expression of $K_c$ for above reaction follows:

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

We are given:

$K_c=4.64\times 10^{-3}$

$[NO_2]_{eq}=2x$

$[N_2O_4]_{eq}=0.0654-x$

Putting values in above equation, we get:

$4.64\times 10^{-3}=\frac{(2x)^2}{(0.0654-x)}\\\\4x^2+(4.64\times 10^{-3})x-(0.303\times 10^{-3})=0$

Solving for the value of 'x', we get:

$x=0.0082,-0.0093$

Neglecting the negative value of 'x', as concentration cannot be negative.

Now,

$[NO_2]_{eq}=2x=2(0.0082)=0.0164M$

$[N_2O_4]_{eq}=0.0654-x=(0.0654-0.0082)=0.0572M$

Hence, the equilibrium concentration of $N_2O_4\text{ and }NO_2$ are 0.0164 M and 0.0572 M