Question

$20$% of $N_2{O}_4$ molecules are dissociated in a sample of gas at ${27}^{o}C$ and $760$ torr. Mixture has the density at equilibrium equal to:

• A. $1.48g/L$
• B. $1.84g/L$
• C. $2.25g/L$
• D. $3.12g/L$

Answer 1

The correct option is D $3.12g/L$

$N_2{O}_4\rightleftharpoons 2N{O}_4$
$1$ $0$
$1-\alpha$ $2\alpha$
$20$ dissociated
$\therefore \alpha =0.20$
$N_2{O}_4=1-\alpha$ $N_2=2\times 0.20$
$=0.80$ left $=0.40$
Total made $1-\alpha +2\alpha$
$=1+\alpha$
Max $=x_{N_2{O}_4},M_{N_2{O}_2}+x_{N{O}_2}{M}_{N{O}_2}$
$=\frac{1-\alpha }{1+\alpha }\times 92+\frac{2\alpha }{1+\alpha }\times 46$
$=\frac{0.80}{1.20}\times 92+\frac{0.46}{1.20}\times 46$
$=61.22+15.33$
$=76.66$
$PM=dRT$
$d=\frac{PM}{KT}=\frac{760\times 76.66}{62.36\times 300}$
$=3.11g/L$