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Q. At $600\, K$ & $200$ bar, a $1: 3$ (molar ratio) mixture of $A _{2}$ & $B _{2}$ react to form an equilibrium mixture containing $\chi_{ AB _{3}}=0.60$ Assuming ideal gas behaviour, calculate $K _{ p }$ for the reaction:
$A _{2}( g )+3 B _{2}( g ) \rightleftharpoons 2 AB _{3}( g )$

Equilibrium

Solution:

image

$(\because$ total no. of moles $= n (1- x )+3 n (1- x )+2 nx = n (4-2 x))$

Also, $\chi_{ AB _{3}}=\frac{2 x }{4-2 x }=0.6$

$x=0.75$

So, $\chi_{A_{2}}=0.1 \& X_{B_{2}}=0.3$

Use the expression, $K_{p}=\frac{\left(p A B_{3}\right)^{2}}{\left(p A_{2}\right)\left(p B_{2}\right)^{3}}$

$=3.33 \times 10^{-3}$