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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 expression

A
α=1+2KP2KP
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B
α=1+2KP2
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C
α=2KP1+2KP
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D
α=2KP1+2KP
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Solution

The correct option is D α=2KP1+2KP
Let the inital moles be 1 so, at equilibrium

2HI(g)H2(g)+I2(g)1αα2α2

total moles at equilibirum =1
we know partial pressure is given as
= (mole fraction) × (total pressure)

hence KP is given as:
KP=(α2×PT)2(1α)2P2T
where PT represents the total pressure.

Simplifying the values,
α1α=2KP
α=2KP1+2KP



Theory:

Playing with α, k and x :
Case 1 :
An(g) nA(g)
Before dissociation, a
(mole)
At t=teq ax nx
(mole)
Unknowns after dissociation aaα n aα

Case 2 :
2HI(g)H2(g) + I2(g)
Before dissociation
moles1

At t=teq
moles1xx2x2

Moles after dissociation 1αα2α2

Case 3 :

2HI(g)H2(g) + I2(g)
At t=0 (moles)
2
At t=teq (moles)
2xx2x2
Moles dissociated in terms of α
22ααα


Case 4:

2AB2(g)2AB(g) + B2(g)
At t=0 (moles)
c
At t=teq (moles )
ccαcαcα2

Case 5:

N2(g)+3H2(g)2NH3(g)
At t=0 (moles)
a3a
At t=teq (moles)
ax3a3x2x

Case 6:
N2(g)+3H2(g)2NH3(g)
At t=0 (moles) 15
At t=teq (moles)
1x53x2x


Case 7:
2HI(g)H2(g) + I2(g)
At t=0 (moles)
21
At t=teq (moles)
2xx21+x2
Moles dissociated in terms of α
22αα1+α

Case 8:
N2(g)+3H2(g)2NH3(g)
At t=0 (moles)
152
At t=teq (moles)
1x53x2+2x


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