Question

The equilibrium constant for the reaction,
$N_2+2O_2\rightleftharpoons 2N{O}_2$ is $100$. The equilibrium constant for the reactions:
(a) $2N{O}_2\rightleftharpoons N_2+2O_2$ is $K_1$
(b) $N{O}_2\rightleftharpoons \frac{1}{2}{N}_2+O_2$ is $K_2$
Find the value of $K_1\times K_2$.

• A. $0.1$
• B. $0.01$
• C. $0.001$
• D. $1$

Answer 1

The correct option is C $0.001$

The equilibrium constant for the reaction
$N_2+2O_2\rightleftharpoons 2N{O}_2$
$K_C=\frac{[N{O}_2{]}^2}{\left[N_2\right][O_2{]}^2}=\mathrm{100...}\left(1\right)$
Equilibrium constant for reaction (a)
(a) $2N{O}_2\rightleftharpoons N_2+2O_2$
$K_1=\frac{\left[N_2\right][O_2{]}^2}{[N{O}_2{]}^2}...\left(2\right)$
On comparing equation $\left(1\right)$ and $\left(2\right)$, we can see that equation $\left(2\right)$ is reverse of equation $\left(1\right)$
Hence,
$K_1=\frac{1}{K_C}$
$K_1=\frac{1}{100}={10}^{-2}$
Equilibrium constant for reaction (b)
(b) $N{O}_2\rightleftharpoons \frac{1}{2}{N}_2+O_2$
$K_2=\frac{\left[N_2{]}^{1/2}\right[O_2]}{\left[N{O}_2\right]}...\left(3\right)$
On comparing equation $\left(2\right)$ and $\left(3\right)$, we get
$K_2=\sqrt{K_1}=\sqrt{{10}^{-2}}={10}^{-1}=0.1$
Hence,
$K_1\times K_2=0.01\times 0.1=0.001$