Question

At ${60}^{\circ }$ and $1atm$ of $N_2{O}_4$ is $50$% dissociated into $N{O}_2$ then $K_p$ is:

• A. $1.33atm$
• B. $2atm$
• C. $2.67atm$
• D. $3atm$

Answer 1

The correct option is A $1.33atm$

$N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$ $50$% dissociated

At $P_0$ $0$ $\alpha =1/2$

$t=0$

At $P_0(1-\alpha )$ $2P_0\alpha$

$t_{eq}$

Given total $P$ at equilibrium 1 atm

$P_0(1+\alpha )=1$

$P_0\left(3/2\right)=1\Rightarrow P_0=2/3$

$K_P=\frac{{\left(2P_0\alpha \right)}^2}{P_0(1-\alpha )}\Rightarrow \frac{4P_0{\alpha }^2}{(1-\alpha )}$

$\frac{4\times \frac{1}{4}\times P_0}{1/2}\Rightarrow 2P_0=4/3atm$