Question

9.2 gm of $N_2{O}_4\left(g\right)$ is taken in a closed one-litre vessel and heated till the following equilibrium is reached

$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$.

At equilibrium 50% of $N_2{O}_4\left(g\right)$ is dissociated. What is the equilibrium constant?

[in mole lit${}^{-1}$] [M.wt. of $N_2{O}_4$ is 92]

• A. 0.1
• B. 0.2
• C. 0.4
• D. 2

Answer 1

Answer: (B)

0.2

9.2 g of $N_2{O}_4\left(g\right)$ corresponds to $\frac{9.2}{92}=0.1$ moles.

Out of this 0.05 moles (50% of 0.1 moles) dissociate to form $2\times 0.05=0.1$ moles of $N{O}_2\left(g\right)$.

The equilibrium constant is $\frac{[N{O}_2{]}^2}{\left[N_2{O}_4\right]}=\frac{(0.1{)}^2}{0.05}=0.2$.

Hence, the correct option is $\text{B}$