Question
For the reaction 2HI (g) ⇋ H2(g) + I2 (g), the degree of dissociation α of HI (g) is related to the equilibrium constant (Kp) by the expression
Answer: Option D
:
D
2HI(g)⇋H2(g)+l2(g)Atequilibrium1−αα/2α/2
Total moles at equilibrium =1
PHI=(1−α)×P,PH2=α/2×PPl2=α/2×P
Kp=PH2×Pl2/(PHI)2
Substituting above values we get,
Kp=a2/[4×(1−α2)]
solving,α/1−α=2(Kp)12
α=2√Kp1+2√Kp
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:
D
2HI(g)⇋H2(g)+l2(g)Atequilibrium1−αα/2α/2
Total moles at equilibrium =1
PHI=(1−α)×P,PH2=α/2×PPl2=α/2×P
Kp=PH2×Pl2/(PHI)2
Substituting above values we get,
Kp=a2/[4×(1−α2)]
solving,α/1−α=2(Kp)12
α=2√Kp1+2√Kp
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