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Question

Consider a reaction: A(g)+B(g)C(g)+D(g)


A(g),B(g) and C(g) are taken in a container at 1 bar partial pressure each and adequate amount of liquid 'D' is added.

From the data given below. Calculate a+b+c+d

Given :
ΔG0fA(g)=30 kJ\ mole1; ΔG0fB(g)= 20 kJ mole1;

ΔG0fC(g)= 50 kJ mole1; ΔG0fD(g)= 100 kJ mole1;

Vapour pressure of D(l) at 300 K=16 bar [All data at 300 K]
where
a = Equilibrium constant of the given reaction
b = Twice the partial pressure of A at equilibrium
c = Twice the partial pressure of B at equilibrium
d = Twice the partial pressure of C at equilibrium

A
6
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B
5
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C
7
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D
8
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Solution

The correct option is A 6
A(g)+B(g)C(g)+D

=ΔG0reaction=(ΔG0fD)(ΔG0fA+ΔG0fB)

=(50kJ+100kJ)(30kJ+20kJ)

ΔG0=0

ΔG0=RT/Inkeq

keq=1

D(I)D(g) [Partial pressure of D gas = Vapour pressure]
A(g)+B(g)C(g) +D(g)

1 bar 1 bar 1 bar 1/6 bar
(1-x) bar (1-x) bar (1+x) bar (1/6) bar

Keq=1=16×[1+X](1X)2

1+x22x=16+x6

x22xx6+56=0

x213x6+56=0

6x213x+5=0

x=12

PA=12PB=12PC=32

where,
PA= partial pressure of A at equilibrium
PA= partial pressure of B at equilibrium
PA= partial pressure of C at equilibrium

Given that,
a = Equilibrium constant of the given reaction
b = Twice the partial pressure of A at equilibrium
c = Twice the partial pressure of B at equilibrium
d = Twice the partial pressure of C at equilibrium

a=1 b=1 c=1 d=3

a+b+c+d= 1+1+1+3 =6

Hence, the correct option is A

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