Question

Show that for reaction AB(g)$\rightleftharpoons$ A(g) + B(g), the total pressure at which AB is 50% dissociated is numerically equal to three times of $K_p$.

Answer 1

$AB\rightleftharpoons A+B$

Let 2P atm be initial pressure of AB. 50% of AB is dissociated. 50% of 2P is P.

Out of 2P atm of AB, P atm is dissociated and $2P-P=P$ atm remains. P atm of AB forms P atm of A and P atm of B. The total pressure is equal to sum of partial pressures of A, B and AB. It is $P+P+P=3P$.

The equilibrium constant $K_P=\frac{P_A\times P_B}{P_{AB}}$

$K_P=\frac{P\times P}{P}=P$

Hence, total pressure is equal to three times of $K_P$.