Question

$N_2$ + $O_2\leftrightharpoons 2NO$ ...... (1)
$NO\leftrightharpoons \frac{1}{2}{N}_2+\frac{1}{2}{O}_2$....... (2)

If $K_1$ and $K_2$ are the equilibrium constants for the reaction (1) and (2) respectively, then:

• A. $K_1={\left(\frac{1}{K_2}\right)}^2$
• B. $K_1=K_2^2$
• C. $K_1=\frac{1}{K_2}$
• D. $K_2=\frac{1}{K_1^2}$

Answer 1

The correct option is A $K_1={\left(\frac{1}{K_2}\right)}^2$

The reaction $NO\leftrightharpoons \frac{1}{2}{N}_2+\frac{1}{2}{O}_2$ is multipled by 2 and reversed to obtain the reaction $N_2+O_2\leftrightharpoons 2NO$.

Hence, the value of the equilibrium constant of the reaction $(1$) is the square of the reciprocal of the value of the equilibrium constant of the reaction $\left(2\right)$

Thus, $K_1={\left(\frac{1}{K_2}\right)}^2$.