Question

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

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

Answer 1

The correct option is B $50$

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

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

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

$N{O}_{\left(g\right)}\rightleftharpoons \frac{1}{2}{N}_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}$

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

$=\frac{1}{2\times {10}^{-2}}=\frac{100}{2}=50$

Hence the correct answer is option B.