Question

In the following equilibrium $N_2{O}_{4\left(g\right)}\rightleftharpoons {2NO}_{2\left(g\right)}$ ${NO}_{2\left(g\right)}$ is $501$ of the total volume. hence degree of dissociation and vant hoff factor (i) respectively are:

• A. $0.5,1.5$
• B. $0.25,1.25$
• C. $0.33,1.33$
• D. $0.66,1.66$

Answer 1

The correct option is B $0.25,1.25$

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

${NO}_2$ is $50$% of the total volume in equilibrium set $49$.

$\therefore$ The degree of dissociation is $0.25$

$\Rightarrow$ The mole fraction of $N_2{O}_4$ is $\frac{1}{3}$ & for ${NO}_2$ is $\frac{2}{3}$.

$\Rightarrow$ $\frac{2}{3}$ gives $50$%

$\therefore$ $\frac{1}{3}$ would give how much percentage?

$\frac{1}{3}\times 50\times \frac{3}{2}=25$%

So, degree of dissociation is $0.25$, Vanthoff factor is $1.25$.