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Question

For the dissociation reaction, N2O42NO2(g), the equilibrium constant Kp is 0.120 atm at 298 K and total pressure of system is 2 atm. Calculate the degree of dissociation of N2O4.

A
0.586
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B
0.121
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C
0.150
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D
0.254
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Solution

The correct option is B 0.121
Let the degree of dissociation of N2O4 be α.
Let the final pressure be P.
N2O42NO2
initial moles 1 0
At equilibrium 1α 2α
So, total moles = 1+α

Kp = P2[NO2]P[N2O4]
PNO2 = 2α1+α.P
PN2O4 = 1α1+α.P
Put in the formual of Kp
Kp=(2α1+α)2.P21α1+α.P
0.120 = 4(α)21(α)2.P
On solving for α, we get α = 0.121

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