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 $K_c$ for the reaction, $NO\left(g\right)\rightleftharpoons \frac{1}{2}{N}_2\left(g\right)+O_2\left(g\right)$ at the same temperature is:

• A. $0.02$
• B. $2.5\times {10}^2$
• C. $4\times {10}^{-4}$
• D. $50$

Answer 1

Answer: (D)

$50$

$Explanation:$

According to the law of mass action, it is the proposition that the rate of the chemical reaction is directly proportional to the product of the activities or concentration of the reactants.

Equation 1:

$N_{2\left(g\right)}+O_{\left(g\right)}\rightleftharpoons 2N{O}_{\left(g\right)}$

The equilibrium constant $=K_c=\frac{{\left[NO\right]}^2}{\left[N_2\right]\left[O_2\right]}=4\times {10}^{-4}$

Equation 2:

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

$K^{{}^{\prime }}=\frac{{\left[N_2\right]}^{\frac{1}{2}}{\left[O_2\right]}^{\frac{1}{2}}}{\left[NO\right]}=\frac{1}{\sqrt{4\times {10}^{-4}}}=\frac{1}{2\times {10}^{-2}}=\frac{100}{2}=50$

$K^{{}^{\prime }}=50$

Hence the correct answer is option $D$.