Two gases A and B in the molar ratio 1- 2 were admitted to any empty vessel and allowed to reach equilibrium at 400- C and 8 atm pressure as A - 2 B - 2 C- The mole fraction of C at equilibrium is 0-4 - Calculate K P atm -1 for the reaction-

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Standard XII

Chemistry

Introduction to Alcohols

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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Two gases A and B in the molar ratio 1:2 were admitted to any empty vessel and allowed to reach equilibrium at 400 oC and 8 atm pressure as A+2B⇌2C. The mole fraction of C at equilibrium is 0.4. Calculate KP (atm−1) for the reaction.

A + 2B ⇌ 2Ca 2aa−x 2a−2x 2x Total moles at equilibrium 3a-x Mole fraction of C =2x3a−x=0.4 ⇒2x=1.2a−0.4x ⇒x=1.2a2.4 ⇒x=a2 Now, A + 2B ⇌ 2Ca2 a a Total moles=5a2 PA=a25a2×8=85 atmPB=a5a2×8=165 atmPC=a5a2×8=165 atmKP=(Pc)2(PB)2PA=58KP=0.625 atm−1

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.