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Question:

The equilibrium constant Kp for this reaction at 298K, in terms of βequilibrium, is:

4β²equilibrium4−β²equilibrium

8β²equilibrium2−βequilibrium

4β²equilibrium2−βequilibrium

8β²equilibrium4−β²equilibrium

Solution:

X₂(g)⇌2X(g)
Initial mole 1 0
t=teq (1−α) 2α
Given 2α=βequilibrium α=βequilibrium/2
Total mole at equilibrium=(1+α)=(1+(βeq/2))
Px2= [(1−βeq/2)/(1+βeq/2)]Ptotal = [(2−βeq)/(2+βeq)]Ptotal
Px(g)= [(βeq/(1+βeq/2))]Ptotal = [(2βeq)/(2+βeq)]Ptotal
So,Kp=(Px)²/(Px2)=[((2βeq)/(2+βeq))Ptotal]²/[((2−βeq)/(2+βeq))Ptotal]
Kp=4β²eq/(4−β²eq)×Ptotal = (8β²eq/(4−β²eq))
So, answer is=B.