Question

At equilibrium, $N _{2} O _{4(g)} \rightleftharpoons 2 NO _{2(g)}$ the observed molecular weight of $N _{2} O _{4}$ is $80\, g\, mol ^{-1}$ at $350\, K$. The percentage dissociation of $N _{2} O _{4(g)}$ at $350 \,K$ is

• A. $10 \%$
• B. $15 \%$
• C. $20 \%$
• D. $18 \%$

Answer 1

Answer: (B)

$ N _{2} O _{4(g)} \rightleftharpoons 2 NO _{2(g)}$

Degree of dissociation may be calculated as

$x=\frac{ N -m}{(n-1) m} [\therefore n=2$ (number of moles produced by $1 mol )]$

$=\frac{92-80}{(2-1) 80}=\frac{12}{80}=0.15$

Percentage dissociation $=0.15 \times 100=15 \%$