If $K_p$ for the reaction $N_2{O}_4\rightleftharpoons 2N{O}_2$ is 0.66 then what is the equilibrium pressure of $N_2{O}_4$?

[Total pressure at equilibrium is 0.5 atm.]

The correct option is C 0.168
Given reaction is:

$N_2{O}_4\leftrightharpoons 2N{O}_2$

At t=0 1 0 Initial mole

At equilibrium 1-x 2x Moles at equilibrium

Moles fraction $\frac{1-x}{1+x}$ $\frac{2x}{1+x}$

at equlibrium

As the total pressure at equilibrium$=0.5atm$

So the partial pressure of $N_2{O}_4=\left(\frac{1-x}{1+x}\right)\times 0.5atm$

The partial pressure of $N{O}_4=\left(\frac{2x}{1+x}\right)\times 0.5atm$

Now, as the $K_p=\frac{\left[p\right(N{O}_2){]}^2}{\left[p\right(N_2{O}_4\left)\right]}$

$K_p=\frac{{\left(\frac{2x}{1+x}\right)}^2\times {0.5}^2}{\left(\frac{1-x}{1+x}\right)\times 0.5}$

$0.66=\frac{4x^2\times 0.5}{1-x^2}$

$0.66-0.66x^2=2x^2$

$x=\sqrt{\frac{0.66}{2.66}}=0.5$

Now put the value in partial pressure equation of $N_2{O}_4=\frac{1-0.5}{1+0.5}\times 0.5$

$=\frac{0.5\times 0.5}{1.5}$

$=0.168$