Question

The dissociation of $N_2{O}_4$ takes place as per the equation $N_2{O}_4\left(g\right)\iff 2N{O}_2\left(g\right)$. $N_2{O}_4$ is 20% dissociated while, the equilibrium pressure of mixture is 600 mm of Hg. Calculate $K_p$ assuming the volume to be constant.

• A. 50
• B. 100
• C. 166.8
• D. 600

Answer 1

The correct option is B 100

$x=0.2$ and total pressure $=600mmofHg$

Let initially the concentration of $N_2{O}_4$ be x, then at equilibrium its concentration will be 1-x.

And initially the concentration of $N{O}_2$ is 0. Thus, at equilibrium its concentration will be 2x.

Thus total concentration at equilibrium will be: 1-x+2x = 1+x
$\therefore P_{N{O}_2}=\frac{2x}{1+x}{P}_{total}=200$
$P_{N_2{O}_4}=\frac{1-x}{1+x}{P}_{total}=\frac{2}{3}{P}_{total}=400$
$K_p=\frac{(P_{N{O}_2}{)}^2}{P_{N_2{O}_4}}=100$