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Q. For the reaction $A_{2}(g) + 2B_{2} \rightleftharpoons 2C_{2}(g)$
the partial pressure of $A_{2}$ and $B_{2}$ at equilibrium are $0.80\, atm$ and $0.40\, atm$ respectively. The pressure of the system is $2.80\, atm$. The equilibrium constant $K_p$ will be

Equilibrium

Solution:

$A_{2}(g )+2B_{2}(g)\rightleftharpoons 2C_{2}(g)$
$P_{A_{2}} = 0.80$ atm
$P_{B_{2}} = 0.4$ atm
Total pressure o f the system $= 2.8$ atm
$\therefore P_{c_{2}} = 2.8 -0.8 -0.4 =1.6$
$K_{p} =\frac {P^{2}_{C_2}}{P_{A_2} \times P^{3}_{B_2}} =\frac {(1.6)^{2}}{0.8 \times (0.4)^{2}}=20$