Question

The equilibrium constant for the given reaction is $100$.
$N_2(g)+2O_2(g)\rightleftharpoons 2N{O}_2(g)$
What is the equilibrium constant for the reaction given below?
$N{O}_2(g)\rightleftharpoons \frac{1}{2}{N}_2(g)+O_2(g)$

• A. 10
• B. 1
• C. 0.1
• D. 0.01

Answer 1

Answer: (C)

$N_2+2O_2\rightleftharpoons 2N{O}_2$
$K_1=\frac{{\left[N{O}_2\right]}^2}{\left[N_2\right]{\left[O_2\right]}^2}$
or $100=\frac{{\left[N{O}_2\right]}^2}{\left[N_2\right]{\left[O_2\right]}^2}\dots$. (i)
Again, $\left[N{O}_2\right]\rightleftharpoons \frac{1}{2}{N}_2+O_2$
$K_2=\frac{{\left[N_2\right]}^{1/2}\left[O_2\right]}{\left[N{O}_2\right]}$
or $K_2^2=\frac{\left[N_2\right]{\left[O_2\right]}^2}{{\left[N{O}_2\right]}^2}\dots$ (ii)
On multiplying Eqs. (i) and (ii), we get
$100\times K_2^2=1$
or $K_2^2=\frac{1}{100}$
$K_2=\frac{1}{10}=0.1$