Question

Gaseous $N_2{O}_4$ dissociates into gaseous $N{O}_2$ according to the reaction $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$ at 300 K and 1 atm pressure, the degree of dissociation of $N_2{O}_4$ is 0.2. If one mole of $N_2{O}_4$ gas is contained in a vessel, then the density of the equilibrium mixture is :

• A. 6.22 g/L
• B. 3.11 g/L
• C. 4.56 g/L
• D. 1.56 g/L

Answer 1

Answer: (B)

$N_2{O}_4⟶2N{O}_2$

at $t=0$, moles of $N_2{O}_4=1$, moles of $N{O}_2=0$

at $t=equilibrium$, mole of $N_2{O}_4=1-a$, mole of $N{O}_2=2a$

$a$=degree of dissociation.

Molecular weight of mixture$=\frac{(1-a)\times \text{molar mass of}{N}_2{O}_4+2a\times \text{molar mass of}N{O}_2}{(1-a+2a)}$

$=\frac{(1-0.2)(28+64)+2\times 0.2\times (14+32)}{1+0.2}$

$M=76.66$

$P=1atm,T=300K,$

$d=PM/RT$

$=\frac{1\times 76.66}{0.082\times 300}=3.11gm/lit$ .