Question

The vapour density of $N_2{O}_4$ and $N{O}_2$ mixture at a certain temperature is $30$.

Calculate the percentage dissociation of $N_2{O}_4$ at this temperature. $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right).$

• A. $53.33$%
• B. $58.56$%
• C. $62.43$%
• D. $67.67$%

Answer 1

Answer: (A)

$53.33$%

$\begin{array}{ccc}{N}_2{O}_4& \leftrightarrows & 2N{O}_2\\ 1& & 0\\ 1-x& & 2x\end{array}$
Initial vapour density = 46.
vapour density at this temp. = 30.

So, $\frac{(1-x)\times 92+2x\times 46}{(1-x)+2x}\times \frac{1}{2}=30$

$⟹\frac{92}{1+x}\times \frac{1}{2}=30$

$⟹\frac{46}{1+x}=30$

$⟹\frac{46}{30}=1+x$

So, $x=\frac{46}{30}-1$

So % dissociation $=0.53\times 100=53\%$