Question

The value of the equilibrium constant for the reaction, $N_{2}+2 O_{2} \rightleftharpoons 2 N O_{2}$ is $100$ . The equilibrium constant for the reaction $N O_{2} \rightleftharpoons \frac{1}{2} N_{2}+O_{2}$ will be

• A. 100
• B. 0.010
• C. 0.10
• D. 1000

Answer 1

Answer: (C)

For the reaction, $N _{2}+2 O _{2} 2 NO _{2}$
$K_{c}=\frac{\left[N O_{2}\right]^{2}}{\left[N_{2}\right]\left[O_{2}\right]^{2}}=100, ...$ (i)
For the reaction, $NO _{2} \frac{1}{2} N _{2}+ O _{2}$
$K_{c}=\frac{\left[N_{2}\right]^{1 / 2}\left[O_{2}\right]}{\left[N O_{2}\right]} $
or $\left(K_{c}\right)^{2}=\frac{\left[N_{2}\right]\left[O_{2}\right]}{\left[N O_{2}\right]^{2}} $
$=\frac{1}{100}$ ( from equation (i) ]
$\therefore K_{c}=\sqrt{\frac{1}{100}}=\frac{1}{10}=0.10$