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  : ΔHf (formation) = 9160 joules/mole and ΔGf (formation) = 97540 joules/mole at 298 K

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

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

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

Answer 1

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