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Question

For the reaction 2NH3 (g) N2(g) + 3H2(g), if the total equilibrium pressure is P and degree of dissociation is α, the equilibrium constant Kp is given by

27α4p216(1α2)2


A

True

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B

False

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Solution

The correct option is A

True


Initially let us assume n moles of NH3 are there.

2NH3(g)N2(g)+3H2(g)

Moles at t = 0 n 0 0

Moles at t=teq(1α) n(α2) n(3α2)

Total amount of the subtance = n (1 - α) + n α2+n3α2 = n (1 - α)

Partial pressure of NH3 , PNH3 = xNH3 P = n(1α)n(1+α) P = 1α1+αP

Partial pressure of N2, PN2 = xN2 P = n(α2)n(1+α) P = α2(1+α)P

Partial pressure of H2, PH2 = xH2 P = n(3α2)n(1+α) P = 3α2(1+α)P

the expression of Kp is

Kp = PN2p3H2P2NH3 = [ α2(1+α) P ] [3α2(1+α)P]3 [1+α1α1P]2 = 27α4P216(1α2)2


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