ProblemSet8 Solutions

8/12/2019 ProblemSet8 Solutions

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MSE 3001: Applied Thermodynamics

Homework Set #8 Solutions

1) Gaskell #11.6(i) Define the problem

We consider the dissociation reaction of nitrogen at 3000 K and determine the

partial pressure of atomic nitrogen gas at a total pressure of 1 atm and the total

pressure of the gas at a partial pressure of N2of 90 % of the total pressure.

(ii) Plan a solutionThe key to the solution is the standard Gibbs free energy of the dissociation

reaction and the relation to the equilibrium constant:

N2= 2N

==2N

2

Np0p

p

lnRTKlnRTG

The total pressure is given as the sum of the partial pressures of the atomic and the

diatomic nitrogen.

(iii) Execute the solution

1. G0(3000K) = 945,000-114.9*3000 J = 600,300 JFor a total pressure of 1 atm the partial pressure of N2is 1 pNand hence:

=

N

2

Np1

pln*3000*R300,600 , therefore pN= 5.9*10-6atm

2. pN2= 0.9P, pN= 0.1P and hence

= P

9.0

1.0ln*3000*R300,600

2

, thus P = 3.2*10-9

atn

2) Gaskell #11.7(i) Define a solution

We calculate first the total pressure at which the equilibrium mol fraction of N2is

equal to 0.2 when ammonia gas is heated up to a temperature of 300C. Next we

determine the standard enthalpy and entropy changes for the dissociation reaction ofammonia gas.

(ii) Plan a solution

The reaction is:2NH3(g) = N2(g) + 3 H2(g)


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The stoichiometry of the reaction tells us that per unit of N2in equilibrium three times

that unit must exist as H2gas. If the mol fraction of N2in equilibrium is 0.2 the molfraction of H2is therefore 0.6 and the mol fraction of NH3is then 1 0.2 0.6 = 0.2.

From the textbook appendix A-1 we know that

T7.31TlnT8.25030,87G0 =for this reaction. With

==

23NH

2N3

2Hp

0

p

pplnRTKlnRTG and Daltons Law we can

solve the problem.

To obtain the standard entropy of the reaction we use the thermodynamic relation

ST

G=

and once we have the entropy and Gibbs free energy we

obtain the enthalpy from G = H TS.

(iii) Execute the solution

At 300C or 573K we have: J8.021,255737.31573ln5738.25030,87G0 ==

Therefore lnKp= 5.25 and Kp= 191.

From XN2= 0.2, XH2= 0.6, and XNH3= 0.2 it follows that

pN2= 0.2P, pH2= 0.6P, and pNH3= 0.2P and thus

22

2

3

P08.1P2.0

2.06.0191 =

= and thus P = 13.3 atm.

7.31)Tln1(8.25)T(ST

G 00

++==

and hence

T4.221021,25STG)T(H 000 +=+= . At 300C the standard entropy of thedissociation reaction is thus -221.4 J/K and the standard enthalpy is -101,814 J.


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3) Determine the maximum temperature to which molybdenum can be heatedwithout formation of the oxide MoO2in an atmosphere of 69 percent hydrogen

and 31 percent by volume of steam (G0= -585,760- 19.2*TlnT + 233.7*T)


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4) The equilibrium constant K for the hydrogenation of pyridine (used forpharmaceuticals, herbicides, pesticides, food flavorings, dyes, adhesives, paints,

etc) to piperidine (used for pharmaceuticals, for the production of vulcanization

accelerator, etc) in the temperature range between 140C and 260C is:

lnK = -46.699 + 24320/T.

Calculate Hm0, Sm

0, cp m

0at 200C.


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