Question

A vessel contains $N_2{O}_4$ and ${NO}_2$ is $2:3$ molar ratio at $10$ atm under equilibrium, now $K_p$ for $N_2{O}_4\rightleftharpoons 2{NO}_2$ is:

• A. $9{atm}^{-1}$
• B. $4.5{atm}^{-1}$
• C. $4.5{atm}^2$
• D. $10atm$

Answer 1

Answer: (A)

$9{atm}^{-1}$

We know $PV=nRT$

at equilibrium $P\propto n$ {$V$ is same as both in same vessel }

as $\frac{P_1}{P_2}=\frac{n_{1}RT/V}{n=RT/V}$ ($P_1$ = pressure of $N_2{O}_4$)

$\frac{P_1}{P_2}=\frac{n_1}{n_2}=\frac{2x}{3x}$ ($P_2$= pressure of $N{O}_2$ )

$P_{total}=P_1+P_2$

$10=2x+3x=5x$

$x=\frac{10}{5}=2a+m$

in reaction $N_2{O}_4\rightleftharpoons 2N{O}_2$

$K_p=\frac{(P^{N{O}_2}{)}^2}{\left(P{N}_2{O}_4\right)}=\frac{(3x{)}^2}{2x}=\frac{9x}{2}$

$k_p=\frac{9\times 2}{2}=9$