Question

$N_2{O}_4$ is 25% dissociated at ${37}^{o}C$ and one atmosphere pressure. Calculate (i) $K_p$ and (ii) the percentage dissociation at $0.1$ atmosphere and ${37}^{o}C$.

Answer 1

$N_2{O}_4\leftrightharpoons 2N{O}_2$

Initial $1$ $0$

Eqb. $1-x$ $2x$

total Moles$=(1-x)+\left(2x\right)=1+x$

$P_{N_2{O}_4}=\left(\frac{1-x}{1+x}\right)xp$

$x=0.25$ and $p=1$ $atm$

$P_{N_2{O}_4}=\frac{1-0.25}{1+0.25}\times 1=0.6atm$

$P_{N{O}_2}=$ $\left(\frac{2x}{1+x}\right)P=\left(\frac{2\times 0.25}{1+0.25}\right)1=0.4atm$

$K_p=\left[\frac{{\left(P_{{NO}_2}\right)}^2}{P_{N_2{O}_4}}\right]=\frac{(0.4{)}^2}{0.6}=0.267atm$

Now, Let degree of dissociation at $0.1$ $atm$ be $g$

$P_{N_2{O}_4}=\left(\frac{1-y}{1+y}\right)P=\frac{1-y}{1+y}\left(0.1\right)$

$P_{{NO}_2}=\left(\frac{2y}{1+y}\right)P=\frac{2y}{1+y}\left(0.1\right)$

$K_p=\left[\frac{{\left(P_{{NO}_2}\right)}^2}{P_{N_2{O}_4}}\right]=\frac{{\left(\frac{2y}{1+y}\right)}^2({0.1)}^2}{\left(\frac{1-y}{1+y}\right)\left(0.1\right)}=0.267atm$

$0.667y^2=0.276$ $atm$

$y=0.632$

$\%y=63.2\%$