Question

9.2 grams of $N_2{O}_4\left(g\right)$ is taken in a closed one litre vessel and heated till the following equilibrium is reached, $N_2{O}_4\left(g\right)\rightleftharpoons {2NO}_2\left(g\right)$. At equilibrium, 50% $N_2{O}_4\left(g\right)$ is dissociated. What is the equilibrium constant? (in $M$) (molecular weight of $N_2{O}_4=92$)

• A. 0.1
• B. 0.4
• C. 0.3
• D. 0.2

Answer 1

The correct option is D 0.2

Given : $\alpha =50\%=0.5$

$m=9.2g$; $M=92g/mol$

$c=\frac{m}{M}\times \frac{1}{V}$

$=\frac{9.2}{92}\times \frac{1}{1}=0.1M$

$[N_2{O}_4{]}_{eq}=\frac{1}{2}(0.1)=0.05mol{L}^1$

$[N{O}_2{]}_{eq}=2\times 0.05=0.1mol{L}^1$

Equilibrium constant $K_e=\frac{[N{O}_2{]}^2}{\left[N_2{O}_4\right]}$

$K_e=\frac{(0.1{)}^2}{\left(0.05\right)}=0.2mol{L}^{-1}$

Answer D) 0.2