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  4. At 60° and 1 atm, N2O4 is 50 % dissociated into NO2 then Kp is

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Q. At $60^{\circ}$ and $1$ atm, $N_2O_4$ is $50\%$ dissociated into $NO_2$ then $K_p$ is

AIIMSAIIMS 2012Equilibrium

Solution:

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$N_2O_4$ is $50\%$ dissociated, so $\alpha=\frac{1}{2}$
$K_{p}=\frac{p^{2}NO_{2}}{pN_{2}O_{4}}=\frac{\left(2\times\frac{1}{2}\right)^2}{\left(1-\frac{1}{2}\right)}=2$ atm