Calculate the equilibrium compositions due to the decomposition of 1 mole of nit

Question

Calculate the equilibrium compositions due to the decomposition of 1 mole of nitrogen tetroxide at 25C and 1 bar in each of the following cases.

N2O4(g) <==> 2NO2(g)   

N2O4  : Hf (formation) = 9160 joules/mole and Gf (formation) = 97540 joules/mole at 298 K

NO2 :   Hf (formation) = 33180 joules/mole and Gf (formation) = 51310 joules/mole at 298 K of

a.) Use the reaction equilibrium method with extent of reaction , and let the initial state be pure N2O4.

b.) Use the reaction equilibrium method with extent of reaction , and let the initial state be 1 mole inert Argon in addition to 1 mole of N2O4.

Explanation / Answer

dG = Gprod - Greact = 2*G-NO2 - G-N2O4

dG = 2*51310 - 97540 = 5,080 J/mol

K

dG = -RT*ln(K)

K = exp(-dG/(RT))

K = exp(-5080/(8.314*298)) = 0.128684

so

K = [NO2]^2 / [N2O4]

a)

use extent for initial N2O4 =

PV = nRT

n/V = P/(RT) = 1/(0.083*298) = 0.04043 mol/L

so..

[N2O4] = 0.04043 initially

for extent:

[N2O4] = 0.04043 - x

[NO2= 0+2x

then

0.128684 = (2x)^2 /(0.04043 - x)

0.128684*0.04043 - 0.128684x = 4x^2

0.0052-0.128684x - 4x^2 = 0

x = 0.0233

then

[N2O4] = 0.04043 - 0.0233 = 0.01713

[NO2= 0+2x = 2*0.0233*2 = 0.0466

b)

for 1 mol of Ar and 1 mol of N2O4, this is then

halved concentration

n/V = P/(RT) = (1/2)/(0.083*298) = 0.04043/2 = 0.020215 mol/L

so

0.128684 = (2x)^2 /(0.020215 - x)

0.020215*0.04043 - 0.020215x = 4x^2

4x^2 + 0.020215x - 0.000817292 = 0

x = 0.0119

[N2O4] = 0.04043 - 0.0119 = 0.02853

[NO2= 0+2*0.0119 = 0.0238

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