Question

The vapour density of ${N_2}{O_4}$ at a certain temperature is 30. What is the percentage dissociation of ${N_2}{O_4}$ at this temperature?
A. 53.3%
B. 76.6%
C. 26.7%
D. None of the above

Answer 1

Hint: The total mass in the reaction remains constant. Thus, the ratio of total moles of gas at any two temperatures is equal to the inverse ratio of the molar mass of the mixture at the respective temperatures.

Complete step by step answer:
Vapour density is defined as the density of a vapour in respect of that of hydrogen. It is a relative term. It can also be defined as the mass of a certain volume of a substance divided by mass of the same volume of hydrogen.
As we know that the molar mass of a substance is approximately twice its vapour density. Mathematically:
Molecular weight = 2 $\times$ Vapour density
So, the molar weight of the mixture at a certain temperature = 2 $\times$ 30
= 60 g
Now, ${N_2}{O_4}$ decomposes according to the reaction:
${N_2}{O_4} \to 2N{O_2}$
Let the initial number of moles of ${N_2}{O_4}$ be ‘n’. At the particular temperature, let its degree of dissociation be . So, the remaining moles of ${N_2}{O_4}$ = n(1 - ) and the number of moles of $N{O_2}$ formed = 2n. So, the total moles of the mixture at the same temperature = n(1 + ).
Now, as the total mass in the reaction remains constant, thus
$\dfrac{{{n_{initial}}}}{{{n_{final}}}} = \dfrac{{{M_{final}}}}{{{M_{initial}}}}$
$\Rightarrow \dfrac{n}{{n(1 + \alpha )}} = \dfrac{{60}}{{92}}$
$\Rightarrow \dfrac{1}{{1 + \alpha }} = \dfrac{{60}}{{92}}$
$\Rightarrow \alpha = \dfrac{{32}}{{60}} = 0.533$

Hence, the percentage dissociation of ${N_2}{O_4}$ = 53.3%
Therefore, the correct answer is (A) 53.3%

Note: Remember that the no of moles in a reaction at any time is inversely proportional to the molecular mass of the mixture only if the mass remains constant. This is not applicable in chemical reactions.