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

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, Po =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, Po =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

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