Question

Before equilibrium is set-up for the chemical reaction $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$, vapour density $d$ of the gaseous mixture was measured. If $D$ is the theoretical value of vapour density, variation of $\alpha$ with $D/d$ is given by the graph. What is the value of $D/d$ at point $A$?

• A. $0$
• B. $0.5$
• C. $1$
• D. $1.5$

Answer 1

The correct option is C $1$

Solution

Let $x$ be initial conc. of $N_2{O}_4$

$\alpha$ be the dissociation value

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

t=0 $x$ 0

t=t $x-x\alpha$ $2x\alpha$

$\frac{D}{d}=\frac{\text{moles at equilibrium}}{\text{initial moles}}$

Moles at equilibrium =$x-x\alpha +2x\alpha =x(1+\alpha )$

$\frac{D}{d}=\frac{x(1+\alpha )}{x}$

$\frac{D}{d}=1+\alpha$

At point A, $\alpha =0$

Hence $\frac{D}{d}=1$

The correct option is C