Question

For the reaction, $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right),△H=57.49{\text{kJ mol}}^{-1}$.
With the increase in temperature, the vapour density of the equilibrium mixture:

• A. Increases
• B. Decreases
• C. Remains the same
• D. Can not be predicted

Answer 1

The correct option is B Decreases

Since $△H$ for the reaction is positive,
$\Rightarrow$ the reaction is an endothermic reaction.
With the increase in temperature, the reaction proceeds forward.
As the reaction proceeds forward,
$\Rightarrow$ the number of moles increases.
We know that, $\text{molecular mass}=2\times \text{Vapour density}\therefore \text{M}\propto \text{V.D.}$

$\text{No. of moles}=\frac{\text{Weight}}{\text{Molecular mass}}\therefore \text{No. of mole}\propto \frac{1}{\text{Molecular mass}}$
So, $\text{No. of moles}\propto \frac{1}{\text{Vapour Density}}$

Therefore, with the increase in number of moles, vapour density decreases.