Question

Vapour density of the equilibrium mixture of $N{O}_2$ and $N_2{O}_4$ is found to be $38.33$. For the equilibrium:

$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$.

Calculate:

a. Abnormal molecular weight.

b. Degree of dissociation.

c. Percentage of $N{O}_2$ in the mixture.

d. $K_P$ for the reaction if total pressure is 2 atm.

• A. a. 76.66, b. 20%, c. 33%, d. 0.33
• B. a. 78.66, b. 20%, c. 33%, d. 0.66
• C. a. 78.66, b. 20%, c. 66%, d. 0.33
• D. a. 76.66, b. 40%, c. 33%, d. 0.33

Answer 1

Answer: (A)

a. 76.66, b. 20%, c. 33%, d. 0.33

Vapor Density is 38.33

Reaction $N_2{O}_4\rightleftarrows 2N{O}_2$

Initial moles $1$ $0$

Moles at Eq $1-x$ $2x$

(A) Abnormal molecular weight

We know Vapor density$=$ Molecular weight/2

$\Rightarrow$Molecular weight = V.D.$\times$2

$=38.33\times 2=76.66$g/mol

(B) Degree of dissociation ($x$)

V.D. at starting is given by $\frac{M_{N_2{O}_4}}{2}$$=\frac{96}{2}=46$

$x=\frac{V.D{.}_{starting}-V.D{.}_{Eq}}{V.D{.}_{Eq}}$

$=\frac{46-38.33}{38.33}=0.200$

Or %$x$$=0.200\times 100\%=20\%$

(C) Percentage of $N{O}_2$ in the mixture

Initial moles $1$

Final moles $1-x+2x=1+x$

$n\left(N{O}_2\right)=2x$

$\%N{O}_2=\frac{2x}{1+x}\times 100$

$=\frac{2\times 0.20}{1+0.20}\times 100=\frac{0.4}{1.20}\times 100$

$=33\%$

(D) $K_p$ for the reaction if total pressure is $2atm$

$p_{N_2{O}_4}=P\left(\frac{1-x}{1+x}\right)=2\left(\frac{1-0.2}{1+0.2}\right)=1.33$

$p_{N{O}_2}=P\left(\frac{2x}{1+x}\right)=2\left(\frac{2\times 0.2}{1.2}\right)=0.66$

$k_p=\frac{{\left(p_{N{O}_2}\right)}^2}{{\left(p_{N_2{O}_4}\right)}^1}=\frac{{\left(0.66\right)}^2}{1.33}=0.33$