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Question

The equilibrium constant Kp( in atm) for the reaction is 2.46 at 7 atm and 300 K.
A2(g)B2(g)+C2(g)
Calculate the average molar mass (in g/mol) of an equilibrium mixture.
Given: Molar mass of A2,B2and C2 are 70,49 and 21 g/mol respectively.
(Given R=0.082 L atm K1mol1,0.10.3

A
56.7
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B
45
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C
40
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D
37.5
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Solution

The correct option is A 56.7
A) Given Kp=2.46
R=0.082 L atm K1mol1
P=7 atm
T=300 K
Δng=Number of moles of Product-Number of moles of reactant =21=1
Kp=Kc(RT)ng
2.46=Kc(0.082×300)1
Kc=2.460.082×300=0.1
Let equilibrium number of moles of A,B and C be 1,x and x respectively.
Kc=[B2][C2][A2]
On substituting all values, we get
0.1=(x)×(x)1
X2=0.1
X=0.3
So. equilibrium moles of A, B and C are 1,0.3 and 0.3 respectively.

Mole fraction of A(χA)=nA2nA2+nB2+nC2

=11+0.3+0.3=11.6=0.6250.62

Similarly χB2=0.31.6=0.18750.19

nB2=nC2

χB2=χC2=0.19

Average molecular mass,
=MA2×χA2+MB2×χB2+MC2×χC2
=70×0.62+49×0.19+21×0.19
=56.7

Mavg=56.7g/mol
Correct option is (A).

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