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Derivation of Kp and Kc

Two gases A a...

Question

Two gases $A$ and $B$ in the molar ratio $1 : 2$ were admitted to any empty vessel and allowed to reach equilibrium at $400^{o} C$ and $8 a t m$ pressure as $A + 2 B ⇌ 2 C$ . The mole fraction of $C$ at equilibrium is $0.4$ . Calculate $K_{P} (a t m^{- 1})$ for the reaction.

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Solution

$A + 2 B ⇌ 2 C a 2 a a - x 2 a - 2 x 2 x$
$Total moles at equilibrium 3a-x$
$Mole fraction of C =2x3a−x=0.4$
$\Rightarrow 2 x = 1.2 a - 0.4 x$
$\Rightarrow x = \frac{1.2 a}{2.4}$
$\Rightarrow x = \frac{a}{2}$
Now,
$A + 2 B ⇌ 2 C \frac{a}{2} a a$
$Total moles = \frac{5 a}{2}$
$P_{A} = \frac{\frac{a}{2}}{\frac{5 a}{2}} \times 8 = \frac{8}{5} a t m P_{B} = \frac{a}{\frac{5 a}{2}} \times 8 = \frac{16}{5} a t m P_{C} = \frac{a}{\frac{5 a}{2}} \times 8 = \frac{16}{5} a t m K_{P} = \frac{(P_{c})^{2}}{(P_{B})^{2} P_{A}} = \frac{5}{8} K_{P} = 0.625 a t m^{- 1}$

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