Question

Two equations are simultaneously existing in a vessel at ${25}^{o}C$
$NO\left(g\right)+N{O}_2\left(g\right)\rightleftharpoons N_2{O}_3\left(g\right);K_{P1}$ (say)
If initially only $NO$ and $N{O}_2$ are present in a $3:5$ mole ration and the total pressure at equilibrium is 5.5 atm with the pressure of $N{O}_2$ at $0.5atm,$ calculate $K_{P1}$.

Answer 1

The given reaction is :-

$NO\left(g\right)+{NO}_2\left(g\right)\rightleftharpoons N_2{O}_3\left(g\right)$

Initial Pressure : $3P$ $SP$ $O$

At Equilibrium : $(3P-\rho )$ $(SP-\rho )$ $\rho$

So, Total pressure at equilibrium $=3P-\rho +5P-\rho +\rho$

$=\theta P-\rho =5.5$ atm (given)

$\Rightarrow \theta P-\rho =5.5atm\to \left(1\right)$

& pressure of ${NO}_2$ at equilibrium $=0.5$ atm

$\Rightarrow 5P-\rho =0.5atm\to \left(2\right)$

$\left(1\right)-\left(2\right)\Rightarrow 3P=5atm\Rightarrow P=\frac{5}{3}atm$

& $\rho =(5P-0.5)atm=5\times \frac{5}{3}-0.5=\frac{25}{3}-\frac{1}{2}$

$=\frac{50-3}{6}=\frac{47}{6}atm$

Now, $K{\rho }_1=\frac{\rho N_2{O}_3}{pNO.{\rho NO}_2}$

$=\frac{P}{(3P-\rho ).(5P-\rho )}$

$=\frac{1}{(3\frac{P}{\rho }-1).(5\frac{P}{\rho }-1)}$ $(\because \frac{P}{\rho }=\frac{5}{3}\times \frac{6}{47}=\frac{10}{47})$

$=\frac{1}{(3\times \frac{10}{47}-1).(5\times \frac{10}{47}-1)}$

$=\frac{47\times 47}{-17\times 3}=-43.31$