Question

The equilibrium constant for the given reaction is $100$.
$N_2(g) + 2O_2(g) \rightleftharpoons 2NO_2(g) $
What is the equilibrium constant for the reaction given below?
$NO_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}+2 O _{2} \rightleftharpoons 2 NO _{2} $
$ K _{1}=\frac{\left[ NO _{2}\right]^{2}}{\left[ N _{2}\right]\left[ O _{2}\right]^{2}}$
or $100=\frac{\left[ NO _{2}\right]^{2}}{\left[ N _{2}\right]\left[ O _{2}\right]^{2}} \ldots$. (i)
Again, $\left[ NO _{2}\right] \rightleftharpoons \frac{1}{2} N _{2}+ O _{2}$
$ K _{2}=\frac{\left[ N _{2}\right]^{1 / 2}\left[ O _{2}\right]}{\left[ NO _{2}\right]}$
or $ K _{2}^{2}=\frac{\left[ N _{2}\right]\left[ O _{2}\right]^{2}}{\left[ NO _{2}\right]^{2}} \ldots$ (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$