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

Solution:

N2O4(g) <==> 2NO2(g)

Let species ‘a’ is N2O4(g) and species ’b‘ is 2NO2(g)

From the reaction we have

    and nio=no=1

For this reaction

  ( these valves are available in literature)

Given temperature is 298 K

K=exp(= exp(5080/(8.314*298))=7.771

Basis: 1 mole of ‘a’ initially present

Then

At equilibrium the compositions are expressed in terms of reaction coordinate

yi=ni/n=( nio+ / (no+) where is stochometric number for the species i and is the stoichiometric number for the reaction. nio initial moles of the species present , no is the total initial moles.   is the recation coordinate.

mole fraction of a =ya= (1- )/(1+ )

mole fraction of b =yb= (0+2 )/(1+ )

The relation between equilibrium constant , composition and pressure is given by

----(1)

Here P=1 atm, P^o =std state pressure=1 atm

(2*)² /((1-(1+ )) =1*7.771=7.771

Solving for we have 0.8125

Equilibrium compositions are

Ya= (1-0.8125)/ (1+0.8125) =0.1034 and yb=1-0.1034=0.8965

Basis : 1 mole inert Argon in addition to 1 mole of N2O4.

mole fraction of a =ya= (1- )/(2+ )

mole fraction of b =yb= (0+2 )/(2+ )

The relation between equilibrium constant , composition and pressure is given by

----(1)

Here P=1 atm, P^o =std state pressure=1 atm

Substituting the above mole fractions in equation 1, we have

4² /((1-(2+ )) =1*7.771=7.771

Solving for we have 0.8654

Equilibrium compositions are

ya= (1-0.8654)/ (2+0.8654) =0.0469

and yb= (2*0.8654)/(2+.8654)=0.604

mole fraction of argon= 1-(0.0469+.604)=0.349