# 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

N

_{2}O

_{4}(g) <=> 2NO

_{2}

Initial 1 0

Equilibr. (1-x) 2X

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

p(N

_{2}O

_{4}) = (1-x)/(1+x)*P

Given X = 0.25 P = 1 atm.

p(N

_{2}O

_{4}) = 0.6 atm

p(NO

_{2}) = 0.4 atm

K

_{p}= p(NO

_{2})

^{2}/p(N

_{2}O

_{4}) = 0.267 atm

If degree of dissociation of N

_{2}O

_{4}at 0.1 atm is Y

p(N

_{2}O

_{4}) = (1-Y)/(1+Y)*0.1

p(NO

_{2}) = 2Y/(1+Y)*0.1

K

_{p}= ((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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