Question

A certain mass of a gas A is kept in a closed container where it undergoes dimerisation, according to the reaction:
$2A\left(g\right)\to A_2\left(g\right)$
Assuming that the temperature is constant, it was found that the partial pressure of $A_2$ gas after time t was one-fifth of the initial pressure in the container. If each molecule of gas A weighs ${10}^{-22}g$ then select the correct statement(s):

• A. The ratio of initial total pressure to the total pressure at time t is 3 : 2
• B. The mole fraction of $A_2$ in the vessel after time t is 0.20
• C. The percentage dimerisation of A at time t is 40%
• D. Average molecular mass of the mixture at time t is 75 amu

Answer 1

Answer: (C), (D)

The correct options are
C The percentage dimerisation of A at time t is 40%
D Average molecular mass of the mixture at time t is 75 amu

$2A\to A_2$
t = 0 $n$
t = t $n-x$ $\frac{x}{2}$

Therefore
P $\propto$ moles
$P_i\propto n$
$P_f\propto (n-\frac{x}{2})$
$\frac{P_i}{n}=\frac{P_f}{n-\frac{x}{2}}$ ...(1)

Also, from Dalton's law
$p_{A_2}={\chi }_{A_2}\times P_1$
$p_{A_2}=\left(\frac{\frac{x}{2}}{(n-\frac{x}{2})}\right)\times P_1$ ...(2)
Also,
$p_{A_2}=\frac{1}{5}\times P_1$ ...(3)
From equation (1), (2) and (3)
$x=\frac{2n}{5}$
$\frac{P_i}{P_f}=\frac{n}{n-\frac{x}{2}}=\frac{n}{n-\frac{n}{5}}=\frac{5}{4}$

${\chi }_{A_2}=\left(\frac{\frac{x}{2}}{(n-\frac{x}{2})}\right)=\frac{\frac{n}{5}}{\frac{4n}{5}}=0.25$

% dimerisation = $\frac{x}{n}\times 100=\frac{\frac{2n}{5}}{n}\times 100=40$%

Molecular mass of A = ${10}^{-22}\times 6\times {10}^{23}=60amu$
Therefore
$M_{avg}=\frac{n_A\times M_A+n_{A_2}\times M_{A_2}}{n_A+n_{A_2}}$
$M_{avg}=\frac{\frac{3n}{5}\times 60+\frac{n}{5}\times 120}{\frac{3n}{5}+\frac{n}{5}}=75amu$