Question

The equilibrium constant ($K_c$) for the reaction $N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right)$ at temperature $T$ is $4\times {10}^{-4}$. The value of $K_P$ for the reaction $NO\left(g\right)\rightleftharpoons \frac{1}{2}{N}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$ at the same temperature is:

• A. $0.02$
• B. $2.5\times {10}^2$
• C. $4\times {10}^{-4}$
• D. $50.0$

Answer 1

Answer: (D)

$50.0$

Using the relation

$K_p={K_C\left(RT\right)}^{\Delta n}$

for the reaction:

$N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right)$

as $\Delta n=2-(1+1)=0K_p=K_C=\frac{p_{NO}^2}{p_{O_2}{p}_{N_2}}$

$K_p=4\times {10}^{-4}$

for the reaction $NO\left(g\right)\rightleftharpoons \frac{1}{2}{N}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$

${K^{{}^{\prime }}}_p=\frac{p_{O_2}^{1/2}{p}_{N_2}^{1/2}}{p_{NO}}={\left(\frac{p_{O_2}{p}_{N_2}}{p_{NO}^2}\right)}^{1/2}{K^{{}^{\prime }}}_p=\frac{1}{{\left(K_p\right)}^{1/2}}=\frac{1}{{(4\times {10}^{-4})}^{1/2}}{K^{{}^{\prime }}}_p=\frac{1}{2\times {10}^{-2}}=50$

Therefore option D is correct.