Question

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

$N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$

At equilibrium, $50$% $N_2{O}_4\left(g\right)$ is dissociated. What is the equilibrium constant? (in mol litre${}^{-1}$)

[Molecular weight of $N_2{O}_4=92$]

• A. $0.1$
• B. $0.4$
• C. $0.2$
• D. $2$

Answer 1

The correct option is C $0.2$

Given

Mass of $N_2{O}_4$ = 9.2g

Solution

$N_{2}O4\left(g\right)=2N{O}_2\left(g\right)$

Initial moles of $N_2{O}_4$ are $\frac{9.2}{92}$ =0.1 moles

50% get dissociated at equilibrium

Therefore At equilibrium, Moles of $N_2{O}_4$ dissociated are 0.05mole

In reaction, every Mole of $N_2{O}_4$ give 2 moles of $N{O}_2$.

Therefore At equilibrium no. of moles of $N{O}_2$ formed = 0.05* 2

$Kc=\frac{[N{O}_2{]}^2}{N_2{O}_4}$

$Kc=\frac{(0.10{)}^2}{0.05}$

Kc=0.2 mol/L

The correct option is C