Question

In the reaction,
$N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$ if $D$ and $d$ are the vapour densitites at initial stage and at equilibrium then what will the value of $\frac{D}{d}$ at point $A$ in the given graph?

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

Answer 1

The correct option is C $1$

In the reaction,
$N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$ if $D$ and $d$ are the vapour densities at initial stage and at equilibrium.
The value of $\frac{D}{d}$ at point $A$ in the given graph will be $1$.
$\alpha =\frac{D-d}{(n-1)d}$
For the given reaction $n=2$ and $\alpha =0$
$0=\frac{D-d}{(2-1)d}$
$0=D-d$
$D=d$
$\frac{D}{d}=1$