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Answer to Question #23578 in Physical Chemistry for gbvignesh

Question #23578
N2O4 IN 25% DISSOCIATION AT 37C AND ONE atm calculate kp and percentage dissociation at 0.1 atm and at 37c
Expert's answer

N2O4 (g) <=> 2NO2

Initial 1 0
Equilibr. (1-x) 2X

Total moles = (1-x) + 2X = (1+x)

p(N2O4) = (1-x)/(1+x)*P

Given X = 0.25 P = 1 atm.
p(N2O4) = 0.6 atm
p(NO2) = 0.4 atm

Kp = p(NO2)2/p(N2O4) = 0.267 atm

If degree of dissociation of N2O4 at 0.1 atm is Y
p(N2O4) = (1-Y)/(1+Y)*0.1

p(NO2) = 2Y/(1+Y)*0.1

Kp = ((2Y+ Y)^2 * (0.1)^2 ) / ((1-Y)/(1+Y) *0.1) = 0.667Y^2

0.267 atm = 0.667Y^2

Y=0.632 (63.2%)

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