Question

If the equilibrium constant for the given reaction $NO\rightleftharpoons \frac{1}{2}{N}_2+\frac{1}{2}{O}_2$ is $0.25$, then the equilibrium constant for the reaction $\frac{1}{2}{N}_2+\frac{1}{2}{O}_2\rightleftharpoons NO$ will be:

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (D)

$4$

The equilibrium constant written for a reaction written in the reverse direction is the reciprocal of the one for the forward reaction.

$NO\rightleftharpoons \frac{1}{2}{N}_2+\frac{1}{2}{O}_2$

The equilibrium constant for the reaction is $K_c=0.25$

The equilibrium constant $K_c^{\prime }$ for the backward or reverse reaction is

$K_c^{\prime }=\frac{1}{K_c}=\frac{1}{0.25}$

$\therefore$ The equilibrium constant for the reverse reaction is $K_c^{\prime }=4$.