Question

For the following reaction:
$N{O}_{(}g)+O_3(g)\rightleftharpoons N{O}_{2\left(g\right)}+O_{2\left(g\right)}$
The value of $K_C$ is $8.2\times {10}^4$. What will be the value of $K_C$ for the reverse reaction?

• A. $8.2\times {10}^4$
• B. $\frac{1}{8.2\times {10}^4}$
• C. $(8.2\times {10}^4{)}^2$
• D. $\sqrt{8.2\times {10}^4}$

Answer 1

The correct option is B $\frac{1}{8.2\times {10}^4}$

The equilibrium constant written for a reaction written in the reverse direction is the reciprocal of the one for the forward reaction.
$K_c^{\prime }=\frac{1}{K_c}$
$NO+O_3\rightleftharpoons N{O}_2+O_2$
Here $K_c$ for the above reaction is given as $8.2\times {10}^4$, then the $K_c^{\prime }$ for the reverse reaction is $\frac{1}{8.2\times {10}^4}$