Question

At $46$, $K_p$ for the reaction ${N_{2}O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$ is $0.667$ atm. Compute the percent dissociation of ${N_{2}O}_4$ at $46$ at total pressure of $380$ Torr.

Answer 1

$N_2{O}_4⟶{2NO}_2$

Pressure at

$t=0$ $0$

at equilibrium

$x-a$ $2a$

$(x-a)+2a=x+a=\frac{380}{760}=\frac{1}{2}$

$\Rightarrow K_p=\frac{{\left(2a\right)}^2}{(x-a)}=\frac{{\left(4a\right)}^2}{(\frac{1}{2}-2a)}=0.667$

$\Rightarrow {8a}^2=0.667(1-4a)=0.667-2.667a$

$\Rightarrow {8a}^2=0.667(1-4a)=0.667-2.667a$

$\Rightarrow {8a}^2+2.67a-0.667=0$

$a={(-2.67+({2.67}^2+4\times 8\times 0.667))}^{\frac{1}{2}}/2\times 8$ atm

percent dissociation $=\frac{a}{(\frac{1}{2}-a)}\times 100=50$.