Question

$N_2{O}_4$ is 25% dissociated at ${37}^{\circ }$C and one atmosphere pressure. Calculate the percentage dissociation at 0.1 atmosphere and ${37}^{\circ }$C.

• A. 26 %
• B. 42 %
• C. 56 %
• D. 65 %

Answer 1

The correct option is D

65 %

$N_2{O}_4$ $\leftrightharpoons$ $2N{O}_4$

1 0 before dissociation

1 - $\alpha$ 2 $\alpha$ after dissociation

Total moles after dissociation (n) = 1 - $\alpha$ + 2$\alpha$ = (1 + $\alpha$)

where $\alpha$ is the degree of dissociation.

$K_p$ = $\frac{{nN{H}_2}^2}{n{N}_2{O}_4}$ $\times$ ${\left(\frac{P}{\sum }\right)}^{△n}$

= $\frac{{2\alpha }^2}{1-\alpha }$ ${\left[\frac{P}{1+\alpha }\right]}^1$

or $K_P$ = $\frac{4{\alpha }^{2}p}{1-{\alpha }^2}$

$K_P$= $\frac{4{0.25}^2\times 1}{1-{\left(0.25\right)}^2}$

($\alpha$ = 25 % , given = 0.25)

$\frac{0.25}{0.9375}$ = 0.267

Again at 0.1 atm.

$K_p$ = $\frac{4{\alpha }^{2}p}{1-{\alpha }^2}$

or 0.267 = $\frac{4{\alpha }^2\times 0.1}{1-{\alpha }^2}$ ; $\alpha$ = $\pm$ 0.65 or 65%