Combustion R.A

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Solution Manual for Combustion 4th Edition

Irvin Glassman

Department of Mechanical and Aerospace Engineering Princeton University

Princeton, New Jersey

Richard Yetter

Department of Mechanical and Nuclear Engineering The Pennsylvania State University

University Park, PA

This solution manual is a revision to the solution manual for "Combustion, 3rd. Ed." that was written in its final form by Prof. Queiroz and Dr. Black of Brigham Young University.

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2

A Solution Manual for: COMBUSTION. 4th Ed.

by I. Glassman, and R.A. Yetter

CHAPTER 1 Problem 1

Calculate the heat of reaction when methane and air in stoichiometric proportions are brought into a calorimeter at 500 K. The product composition is brought to the ambient temperature (298 K) by the cooling water. The pressure in the calorimeter is assumed to remain at 1 atm, but the water formed has condensed.

Solution: The heat of reaction can be calculated using Eqn. (10) in the text. However, as also suggested in the textbook, this equation can be simplified for the general case when

0

0 0

T 298

H =H .

The modified equation becomes:

(

2

)

( )

(

1

)

( )

0 0 0 0 0 0 i T 298 f ₂₉₈ j T 298 ₂₉₈ f ₂₉₈ i j i prod j react Q H n ⎡ H H H ⎤ n ⎡ H H H ⎤ − = Δ =

∑

_⎣ − + Δ _⎦ −

∑

_⎣ − + Δ _⎦ (1)

In this Equation, T2 and T1 are the temperatures of the products and reactants respectively. The general equation for the complete combustion of one mole of methane in air is:

(

)

4 th 2 2 2 2 2

CH +a O +3.76N →a CO +b H O c N+

where ath is a coefficient describing how much air is needed for stoichiometric combustion. Balancing the above reaction one has:

(

)

4 2 2 2 2 2

CH +2 O +3.76N →CO +2 H O 7.52 N+

In order to apply Eqn. (1) to the above reaction, the following properties from Tables in the textbook are needed (note that for this problem T2 = 298 K and T1 = 500 K):

gases 0 f H Δ (kJ/mol) 1 0 0 T 298 H −H (kJ/mol) 2 0 0 T 298 H −H (kJ/mol) CH4 −74.90 8.21 0.0 O2 0.0 6.09 0.0 N2 0.0 5.92 0.0 H2O(liq) −286.04 0.0 0.0 CO2 −393.77 0.0 0.0

Equation (1) then becomes:

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(

)

(

)

(

)

(

)

(

)

₂2

(

)

₂ 2 2 4 CO H O N CH O N 2 0.0 6.09 7.52 0.0 5.92 − + − + or Q=955.85 kJ

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4

Problem 2

Calculate the flame temperature of normal octane (liquid) burning in air at an equivalence ratio of 0.5. For this problem assume there is no dissociation of the stable products formed. All reactants are at 298 K and the system operates at 1-atm pressure. Compare your results with those given in the graphs in the text. Explain any differences.

Solution: We will assume that adiabatic conditions exist and that the water in the products is in the vapor phase. In this case, Eqn. (10) in the text can be used in an iterative approach to solve for the adiabatic flame temperature. However, this equation can be simplified for the general case when

0

0 0

T 298

H =H . For Q = 0, the modified equation becomes:

(

2

)

( )

(

1

)

( )

0 0 0 0 0 0 i T 298 f ₂₉₈ j T 298 f ₂₉₈ i j i prod j react n _⎣⎡ H −H + ΔH _⎦⎤ = n _⎣⎡ H −H + ΔH ⎤_⎦

∑

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants respectively. Since the conditions for the reactants are specified, the solution approach is to find T2 such that the above equality will be satisfied.

The general equation for the complete combustion of one mole of octane with excess air is:

( )

( )( )(

)

8 18 liq th 2 2 2 2 2 2

C H + m a O +3.76N →a CO +b H O c O+ +d N

where ath is the coefficient describing how much air is needed for stoichiometric combustion and m is the coefficient used to represent the amount of excess air, respectively. For the case when c = 0 and m = 1 (no excess air), a simple mass balance yields ath = 12.5. Therefore, for an excess air of 100% (m = 2) or an equivalence ratio of 0.5, the chemical reaction becomes:

( )

(

)

8 18 liq 2 2 2 2 2 2

C H +25 O +3.76N →8 CO +9 H O 94 N+ +12.5 O

Since the conditions of the reactants are specified (T1 = 298 K), Table 2 in the text and Appendix A provide the following thermodynamic properties:

gases 0 f H Δ (kJ/mol) 1 0 0 T 298 H −H (kJ/mol) C8H18(liq) −250.12 0.0 O2 0.0 0.0 N2 0.0 0.0

The enthalpy of the reactants (Hr, given by the right-hand side of Eqn (1)) can be calculated as:

(

)

(

)

(

)

8 18 2 2

r C H O N

H = −1 250.12 0.0+ +11 0.0 0.0+ +41.36 0.0 0.0+ = −250.12kJ For the products it follows from thermodynamic properties in the text that:

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gases 0 f H Δ (kJ/mol) O2 0.0 N2 0.0 H2O(vap) −242.00 CO2 −393.77

Therefore, the expression for the enthalpy of the products (Hp, given by the left-hand side of Eqn. (1) in terms of the enthalpy of the individual gases in the products becomes

(

2

)

0 0 T 298 let Δ =H H −H :

(

2

) (

2

)

(

2

) (

2

)

p CO H O O N H = −8 393.77+ ΔH + −9 242.00+ ΔH +12.5 0.0+ ΔH +94 0.0+ ΔH Simplifying the above equation yields:

2 2 2 2

CO H O O N

8 HΔ + Δ9 H +12.5 HΔ + Δ94 H =5, 078.04

A first guess for the products temperature can be obtained by assuming that all product gases are N2. This approach gives the following as an initial guess for T2 (Note that 123.5 = 8 + 9 + 12.5 + 94): 123.5 2 N H Δ = 5,078.04 so 2 N H Δ = 41.12 kJ/mol therefore T2 = 1577

We will then assume that T2 is between 1500 and 1600 K. If that is so, we can develop the following table (all ΔH’s are in kJ/mol and (Hp - Hr)'s are in kJ):

T2 2 CO H Δ (kJ/mol) 2 H O H Δ 2 O H Δ 2 N H Δ Hp – Hr (kJ) 1500 61.76 48.13 40.64 38.43 -30.37 1550 64.69 50.51 42.48 40.18 202.00 1506 62.11 48.41 40.86 38.64 -2.55 Interpolate 1500 -30.37 T2 00 1550 202.00 = so

The enthalpy difference for this temperature was also calculated in the above table to show that

2 T 1500 50 − 30.37 232.37 T2 = 1506 K

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6

once you have a good idea of the temperature interval where the adiabatic temperature should be in, linear interpolation does a good job of predicting the temperature.

Comparing the results for the adiabatic flame temperature found above to that found using Fig. 3 on page 22 of the textbook is done as follows:

Enthalpy per gram C8H18 =

250.12 96 18 − + = 2.194 kJ/gm H/C = 18 8 = 2.25

With these values Fig. 3 yields T2 ≅ 1505 K, which is very close to the calculated adiabatic temperature.

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Carbon monoxide is oxidized to carbon dioxide in an excess of air (1 atm) in an afterburner so that the final temperature is 1300 K. Under the assumption of no dissociation determine the air-fuel ratio required. Report the results on both a molar and mass basis. For the purposes of this problem assume that air has the composition of 1 mole of oxygen to 4 moles of nitrogen. The carbon monoxide and air enter the system at 298 K.

Solution: We will assume that adiabatic conditions exist. In this case, Eqn. (1.10) in the textbook can be used to calculate the initial composition of the mixture which will give an adiabatic flame temperature of T2 = 1300 K. This equation can be simplified for the general case when

0

0 0

T 298

H =H For Q = 0, the modified equation becomes:

(

2

)

( )

(

1

)

( )

∑

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants, respectively. Since the temperature for the reactants and products have been given (T1 = 298 K and T2 = 1300 K), the problem simplifies to finding x in the reaction shown below, which will satisfy Eqn. (1) above.

(

2 2

)

2

(

)

2 2

CO+x O +4 N =CO + x−0.5 O +4 Nx

The following table can be developed from information in the textbook.

Gases 0 f H Δ (kJ/mol) 1300 0 0 T 298 H −H (kJ/mol) CO2 −393.77 50.19 CO −110.62 Not needed O2 0.0 33.37 N2 0.0 31.52

Substituting these values in Eqn. (1) one has:

(

)

(

)

(

)

(

)

(

)

(

)

2 2 2 2 2 CO O N CO O N 2 1 1 393.77 50.19 0.0 33.37 4 0.0 31.52 2 1 110.62 0.0 0.0 0.0 4 0.0 0.0 0.0 343.58 33.37 16.69 126.08 110.62 0.0 159.45 249.65 1.566 − − + + + + + − − + − + + + = − + − + + = = = x x x x x x x x

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8

4) = 7.83 moles (which is the A/F ratio on a molar basis). The air/fuel ratio on a mass basis is:

air fuel

m 7.83 moles of air 29 kg / mol AF m 1 mole of CO 28 kg / mol × = = × so AF 8.11kg of air kg of CO =

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The exhaust of a carbureted engine, which is operated slightly fuel rich, has an efflux of unburned hydrocarbons entering the exhaust manifold. One can assume that all the hydrocarbons are equivalent to ethylene (C2H4) and all the remaining gases are equivalent to inert nitrogen (N2). On a molar basis there are 40 moles of nitrogen for every mole of ethylene. The hydrocarbons are to be burned over an oxidative catalyst and converted to carbon dioxide and water vapor only. In order to accomplish this objective, ambient (298 K) air must be injected into the manifold before the catalyst. If the catalyst is to be maintained at 1000 K, how many moles of air per mole of ethylene must be added? Take the temperature of the manifold gases before air injection as 400 K. Assume the composition of air to be 1 mole of oxygen to 4 moles of nitrogen.

Solution: We will assume that adiabatic conditions exist and that no dissociation of the stable products will take place. In this case, Eqn. (1.10) in the textbook can be used to calculate the initial composition of the mixture which will give an adiabatic flame temperature of T2 = 1000 K. This equation can be simplified for the general case when

0

0 0

T 298

(

2

)

( )

(

1

)

( )

∑

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants, respectively. Since the temperature for the reactants and products have been given (Tl = 400 K and T2 = 1000 K), the problem simplifies to finding x in the reaction shown below, which will satisfy equation (1) above.

C2H4 + 40N2 + x(O2 + 4N2) → 2 CO2 + 2 H2O + (x - 3) O2 + (40 + 4x) N2 (2) The following table can be developed from information in the text.

Gases 0 f H Δ (kJ/mol 400 0 0 T 298 H −H (kJ/mol) 1000 0 0 T 298 H −H (kJ/mol) C2H4 52.34 4.89 Not needed N2 0.0 2.97 21.47 O2 0.0 3.03 22.72 CO2 −393.77 Not needed 33.43

H2O(vap) −242.00 Not needed 26.00

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10

(

)

(

)

(

)

(

)

(

)

(

)

(

)(

)

(

)(

)

2 4 2 2 2 2 2 2 2 C H N O N CO H O O N 1 52.34 4.89 40 0.0 2.97 0.0 0.0 4 0.0 0.0 2 393.77 33.43 2 242.00 26.00 3 0.0 22.72 40 4 0.0 21.47 0.0 57.22 118.91 720.68 432.00 22.72 68.17 858.96 85.90 0.0 108.62 538.01 and 4 + + + + + + + − − + − − + − − + − + + = + + + − + − − = = = x x x x x x x x .953

Therefore, the total number of moles of air needed to react with one mole of C2H4 is

4.953 x (1 + 4) = 24.75 moles.

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A combustion test was performed at 20-atm pressure in a hydrogen-oxygen system. Analysis of the combustion products, which were considered to be in equilibrium, was as follows:

Compound Mole fraction

H2O 0.4855 H2 0.4858 O2 0.0 O 0.0 H 0.0216 OH 0.0069

What must the combustion temperature have been in the test?

Solution: Since all the products are in equilibrium, any equilibrium reaction between the products can be used to solve for the temperature. The simplest reaction to choose (and the only one that is an equilibrium reaction of formation) is:

2

1

H H

2 ⇔

The expression for the equilibrium constant of formation is then given by

( ) 2 1/ 2 H P,f H 1/ 2 H n P K n n ⎛ ⎞ = ⎜_⎜ ⎟_⎟ ⎝

∑

⎠

Since values are given in mole fraction

∑

n=1. Therefore,

( )

( ) 1/ 2 P,f H 1/ 2 10 P,f H 0.0216 K 20 0.138593 0.4858 log K 0.85826 = = = −

Looking up the temperature in Table 2 in the appendix for H which corresponds to this value yields

T ~ 2959 K

If one desires to use the other possible equilibrium reaction given by

2 2

1

H OH H O

2 + ⇔

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12

The equilibrium constant is given by

(

) (

)

2 2 1/ 2 H O P 1/ 2 1/ 2 H OH n _P _0.4855 ₁ K 22.573349 n n n 0.0069 0.4858 20 ⎛ ⎞ = ⎜_⎜ ⎟_⎟ = = ⎝

∑

⎠ or log10 KP = 1.353596

Then, in order to solve for the final temperature of the mixture one would have to calculate the value for the equilibrium constant for the above reaction and find the corresponding temperature as shown below. ( ) ( ) 2 P,f H O P P,f OH K K K = Temp. K_{P,f OH}₍ ₎ ₍ ₎ 2 P,f H O K KP Log10 KP 2900 K 1.1066 31.2608 28.2494 1.4510 3000 K 1.1614 22.0293 18.9679 1.2780

Interpolating for the temperature between 2900 and 3000 K that would give log10 KP = 1.353596 yields:

T ~ 2956 K

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Whenever carbon monoxide is present in a reacting system it is possible for it to disproportionate into carbon dioxide according to the equilibrium 2CO ⇔ Cs + CO2. Assume that such an equilibrium can exist in some crevice in an automotive cylinder or manifold. Determine whether raising the temperature decreases or increases the amount of carbon present. Determine the KP for the reaction equilibrium and the effect of raising the pressure on the amount of carbon formed.

Solution: One simply considers how the equilibrium is shifted by the temperature and pressure to solve the problem.

(i) Temperature effect

2 CO ⇔ Cs + CO2 (1) 2 CO P 2 CO p K p = (2)

in terms of equilibrium constants of formation for C + O2 ⇔ CO2 one has

( 2) 2 2 CO P,f CO O p K p = (3) and for C 1O₂ CO 2 + ⇔ ( ) 2 CO P,f CO 1 2 O p K p = (4)

For Eqn. (1) it follows that

( ) ( ) 2 P,f CO P 2 P,f CO K K K =

or log10 KP =log10 KP,f CO₍ ₂₎−2 log10 KP,f CO_{( )}

Evaluating this log equation in two different temperatures one has

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14

( 2)

10 P,f CO

log K =20.68 and 2 log K10 P,f CO_{( )} =2 10.46

(

)

=20.92 so

log Kp = −0.24 or KP = 0.58 (b) At 2000 K

( 2)

10 P,f CO

log K =10.35 and 2 log K10 P,f CO_{( )}=2 7.47

(

)

=14.94 so

log KP = −4.59 or KP = 2.57 × 10−5

Therefore, KP drops as temperature increases.

Equation (1) and (2) show that as KP decreases, the reaction given by Eqn. (1) is shifting to the left, thus less carbon forms as the temperature increases.

(ii) Pressure effect

2 2 1 CO CO P 2 2 CO CO p n P K p n n − ⎛ ⎞ = = ⎜_⎜ ⎟_⎟ ⎝

∑

⎠

Thus, as the pressure increases more CO2 forms, shifting the reaction from left to right and thus more carbon forms as the pressure is increased.

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Determine the equilibrium constant KP at 1000 K for the following reaction 2 CH4 ⇔ 2 H2 + C2H4

Solution: As explained on page 14 of the text, the equilibrium constant (given in terms of the equilibrium constant of formation) is:

( ) ( ) ( ) 2 2 4 4 2 P,f H P,f C H P 2 P,f CH K K K K =

At 1000 K it follows from Table 2 in the appendix that (note Table 2 provides the logl0 of K):

( ) ( ) ( ) 2 2 4 4 0 P,f H 6.213 7 P,f C H 1.011 P,f CH K 10 1 K 10 6.1235 10 K 10 0.0975 − − − = = = = × = =

so evaluating Eqn. (1) for KP one has

(

)

(

)

7 5 P 2 1 6.1235 10 K 6.442 10 0.0975 − − × × = = × so K P = 6.442 × 10-5

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16

Problem 8

The atmosphere of Venus is said to contain 5% carbon dioxide and 95% nitrogen by volume. It is possible to simulate this atmosphere for Venus reentry studies by burning gaseous cyanogen (C2N2) and oxygen and diluting with nitrogen in the stagnation chamber of a continuously operating wind tunnel. If the stagnation pressure is 20 atm, what is the maximum stagnation temperature that could be reached and still have Venus atmosphere conditions? If the stagnation pressure were 1 atm, what would be the maximum temperature? Assume all gases enter the chamber at 298 K. Take the heat of formation of cyanogen as

( )

0_f

298

H 381.84

Δ = k/mol.

Solution: In order to maintain Venus atmosphere, the composition of 100 moles of products would have to be 5 CO2 + 95 N2. Eqn. (10) in the textbook can be used to calculate the stagnation temperature of this system. This equation can be simplified for the general case when

0

0 0

T 298

(

2

)

( )

(

1

)

( )

∑

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants respectively. Since the conditions for the reactants are specified the solution approach is to find T2 such that the above equality will be satisfied.

(a) for Venus atmosphere the general combustion reaction would be

a C2N2 + b O2 + c N2 → 5 CO2 + 95 N2

which can be balanced for one mole of C2N2 as follows: C2N2 + 2 O2 + 36 N2 → 2 CO2 + 38 N2,

In preparation to apply Eqn. (1) the following tables can be generated.

gases 0

(

)

f H kJ mol Δ

(

)

1 0 0 T 298 H −H kJ mol C2N2 381.84 0.0 O2 0.0 0.0 N2 0.0 0.0

An expression for the enthalpy of the reactants (Hr, given by the right-hand side of Eqn. (1)) in terms of the enthalpy of the individual gases in the reactants becomes:

(

)

(

)

(

)

2 2 2 2

r C N O N

H =1 381.84 0.0+ +2 0.0 0.0+ +36 0.0 0.0+ = −381.84 kJ

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gases 0

(

)

f H kJ mol Δ N2 0.0 CO2 −393.77

Thus, an expression for the enthalpy of the products (Hp, given by the left-hand side of Eqn. (1)) in terms of the enthalpy of the individual gases in the products becomes

(

)

2 0 0 T 298 let Δ =H H −H :

(

2

) (

2

)

p CO N H =2 −393.77+ ΔH +38 0.0+ ΔH Equating Hr and Hp yields:

2 2

CO N

2 HΔ +38 HΔ =1,169.37

A first guess for the products temperature can be obtained by assuming that all product gases are N2. This approach gives the following as an initial guess for T2 (Note that 40 = 2 + 38):

2 N 40 HΔ =1,169.37 so 2 N H 29.23kJ mol Δ = , therefore, T2 ≈ 1232 K

We will then assume that T2 is between 1200 and 1300 K. If that is so, we can develop the following table (all ΔH’s are in kJ/mol and (Hp - Hr)'s are in kJ):

T2

(

)

2 CO H kJ mol Δ ΔHN₂ Hp − Hr (kJ) 1200 44.51 28.13 −11.51 1300 50.19 31.52 128.87 Interpolate 1200 −11.51 T2 00 1300 128.31 2 T 1200 11.51 100 139.88 − = so T2 = 1208 K Since ₍ ₎ 2 17 P,f CO

K =1.75 10× at 1200 K, no significant dissociation of the products occurs. Therefore, pressure has no effect and T2 would remain the same at all pressures.

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18

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A mixture of 1 mole of N2 and 0.5 mole O2 is heated to 4000 K at 0.5 atm pressure and results in an equilibrium mixture of N2, O2, and NO only. If the O2 and N2 were initially at 298 K and the process is one of steady heating, how much heat was required to bring the final mixture to 4000 K on the basis of one initial mole of N2?

Solution: The first step is to find the equilibrium concentration of products. Once this is done, one can use Eqn. (1.10) to calculate the amount of heat necessary to bring the final mixture to 4000 K. Consider the following reaction at equilibrium.

2 2 2 2

1

1 N O a N b O c NO 2

+ → + + (1)

balancing this reaction yields the following relationship between a, b and c.

N: 2 = 2 a + c 0: 1 = 2 b + c

Solving these two equations for b and c in terms of a it follows that

2a 1 b

2 −

= and c = 2 − 2a Therefore, Eqn. ( 1) becomes:

(

)

2 2 2 2 2a 1 1N 0.5 O a N O 2 2a NO 2 − + → + + −

For the equilibrium reaction given by 1N₂ 1O₂ NO

2 +2 ⇔ , the expression for the equilibrium constant of formation is given by:

( ) 2 2 1 1 2 1 2 NO P,f NO 1/ 2 1/ 2 N O n P K n n n − − ⎛ ⎞ = ⎜_⎜ ⎟_⎟ ⎝

∑

⎠

The total number of moles is given by n a 2a 1 2 2a 3 2

−

= + + − =

∑

. At 4000 K, 1og10 KP,f(NO) = - 0.524 or KP,f(NO) = 0.29923. Squaring Eqn. (3) would give the following equation for the variable a:

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20

(

)

( ) 2 2 P,f NO 2 2a K 0, 2a 1 a 2 − − = −

Solving for a in the above equation and then using the mass balance equations to solve for b and c yields: a = 0.909, b = 0.409 and c = 0.182. Equation (2) now becomes

1 N2 + 0.5 O2 → 0.909 N2 + 0.409 O2 + 0.182NO

The amount of heat required to bring the above mixture to 4000 K is calculated using Eqn. (1.10) in the text by taking T2 = 298 K and T1 = 4000 K:

(

2

)

( )

(

1

)

( )

0 0 0 0 0 0 i T 298 f ₂₉₈ j T 298 f ₂₉₈ i j i prod j react Q n ⎡ H H H ⎤ n ⎡ H H H ⎤ − =

∑

_⎣ − + Δ _⎦ −

∑

_⎣ − + Δ _⎦ (4)

from tables in the text one has:

gases 0

(

)

f H kJ mol Δ

(

)

1 0 0 T 298 H −H kJ mol

(

)

2 0 0 T 298 H −H kJ mol O2 0.0 139.01 0.0 N2 0.0 130.16 0.0 NO 90.43 132.76 0.0

so the equation for Q becomes

(

)

(

)

(

)

(

)

(

)

2 2 2 2 NO O N N O Q 0.182 90.43 132.76 0.409 0.0 139.01 0.909 0.0 130.16 1 0.0 0.0 1/ 2 0.0 0.0 − = + + + + + − + − + so Q = - 215.79 kJ

if no dissociation occurred the value of Q would be

(

)

(

)

(

)

2

(

)

2 2 2 O N N O Q 1/ 2 0.0 139.01 1 0.0 130.16 1 0.0 0.0 1/ 2 0.0 0.0 − = + + + − + − + so Q = - 199.67 kJ

Therefore, more energy is required to raise the temperature of the mixture to 4000 K due to dissociation.

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Calculate the adiabatic decomposition temperature of benzene under the constant pressure condition of 20 atm. Assume that benzene enters the decomposition chamber in the liquid state at 298 K and decomposes into the following products: carbon (graphite), hydrogen, and methane.

Solution: The solution for this problem is an iterative one. In order to calculate the final temperature of the mixture, one needs the composition of the mixture at equilibrium. But the composition of the mixture can only be found if the final temperature is known. Therefore, one will guess the final temperature and calculate the mixture composition using the equilibrium concept. Once the composition is know at the guessed temperature, the adiabatic flame temperature is calculated and checked against the guessed value. This process is repeated until the guessed and calculated temperatures are obtained. Consider the following reaction at equilibrium:

C6H6 (1) → a C (s) + b H2 + c CH4 (1)

balancing this reaction yields the following relationship between a, b and c.

C: 6 = a + c H: 6 = 2 b + 4 c

solving these two equations for a and b in terms of c it follows that

6 4c

b 3 2c

2 −

= = − and a = 6 − c Therefore, Eqn. (1) becomes:

( )

(

)

( ) 6 6 1 s 2 4 6 4c C H 6 c C H c CH 2 − → − + + (2)

Now consider the equilibrium reaction given by

C(s) + 2 H2 ⇔ CH4

For this equilibrium reaction, the expression for the equilibrium constant of formation of CH4 is

( 4) 4 2 1 2 CH P P,f CH 2 H n _P K K n n − ⎛ ⎞ = = ⎜_⎜ ⎟_⎟ ⎝

∑

⎠ (3)

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22

The total number of moles is given by n 6 4c c 3 c 2

−

= + = −

∑

(note that the number of moles of graphite are not included in the calculation for the total number of moles at equilibrium). Equation (3) then becomes:

P 2 c b c K 0 b 20 + ₋ ₌

expressing b in terms of c gives:

(

)

2 P c 3 c K 0 20 3 2c + ₋ ₌ − (4)

In order to calculate the adiabatic flame temperature one has to apply Eqn. (1.11) in the text taking T1 = 298 K.

(

2

)

( )

(

1

)

( )

0 0 0 0 0 0 i T 298 f ₂₉₈ j T 298 f ₂₉₈ i j i prod j react Q 0 n ⎡ H H H ⎤ n ⎡ H H H ⎤ − = =

∑

_⎣ − + Δ _⎦ −

∑

_⎣ − + Δ _⎦ (5)

With Eqns. (4) and (5) the iterative process may now start. Since the conditions for the reactants are specified (T1 = 298 K), Table 1 in the text provides the following thermodynamic properties:

gases 0

(

)

f H kJ mol Δ

(

)

1 0 0 T 298 H −H kJ mol C6H6 (1) 49.03 0.0

And the enthalpy of the reactants (Hr, given by the right-hand side of Eqn. (5)) can be calculated as:

Hr = 1 (49.03)C6H6 (1) = 49.03 kJ

Let us first guess that T2= 1000 K; therefore, logl0 KP = - 1.011 or KP = 0.097499. Solving Eqn. (4) for c using this value of KP and then solving the mass balance equations for a and b results in:

c = 0.9943, b = 1.0114, and a = 5.0057

For a product temperature of 1000 K, application of Eqn. (5) using the values of a, b, and c found above yields: T2 ( )s

(

)

C H kJ mol Δ ΔHH₂ ΔHCH₄ Hp Hp − Hr (kJ) 1000 8.47 20.70 38.20 26.87 −22.16

(23)

and then solving the mass balance equations for a and b gives:

c = 0.7409, b = 1.5182, and a = 5.2591

Following the same procedure as previously outlined yields:

T2 ( )s

(

)

C H kJ mol Δ ΔHH₂ ΔHCH4 Hp Hp − Hr (kJ) 1100 14.01 23.74 45.58 88.01 38.98

Interpolation for the value of temperature where Hp − Hr = 0 yields T2 ~ 1036 K.

At T2 = 1036K, logl0 Kp = -1.1654 or Kp = 0.06832, c = 0.9101, b = 1.1798, a = 5.0899 and T2 ( )s

(

)

C H kJ mol Δ ΔHH₂ ΔHCH₄ Hp Hp − Hr (kJ) 1036 10.47 21.79 40.86 48.02 −1.01 Since Hp − Hr ~0 T 2 ~ 1036 K

(24)

24

Problem 11

Calculate the flame temperature and the product composition of liquid ethylene oxide decomposing at 20-atm pressure by the irreversible reaction:

C2H4O(1) → a CO + b CH4 + c H2 + d C2H4

The four products are as specified. The equilibrium known to exist is

2 CH4 ⇔ 2 H2 + C2H4

The heat of formation ΔH0_f of liquid ethylene oxide is = - 76.62 kJ/mol. It enters the decomposition chamber at 298 K.

Solution: The solution for this problem is an iterative one. In order to calculate the final temperature of the mixture, one needs the composition of the mixture at equilibrium. But the composition of the mixture can only be found if the final temperature is known. Therefore, one will guess the final temperature and calculate the mixture composition using the equilibrium concept. Once the composition is know at the guessed temperature, the adiabatic flame temperature is calculated and checked against the guessed value. This process is repeated until the guessed and calculated temperatures are the same. Consider the following reaction at equilibrium:

C2H4O(1) → a CO + b CH4 + c H2 + d C2H4

balancing this reaction yields the following equations for a, b, c and d.

C: 2 = a + b +2 d O: 1 = a

H: 4 = 4 b + 2 c + 4 d

simplifying it follows that:

b + 2 d = 1 (1)

4 b + 2 c + 4 d = 4 (2)

The other equation for the solution of b, c, and d will come from the Kp expression for the given equilibrium reaction.

2 CH4 ⇔ 2 H2 + C2H4

For this equilibrium reaction, the expression for the equilibrium constant in terms of the equilibrium constant of formation of each individual species is,

(25)

( ) ( ) ( ) 2 2 4 2 2 4 4 4 P,f H P,f C H H C H P 2 2 CH P,f CH K K n n _P K K n n ⎛ ⎞ = = ⎜_⎜ ⎟_⎟ ⎝

∑

⎠ (3)

or in terms of b, c, and d ( with P = 20 atm.) one has

2 P 2 c d 20 K b 1 b c d = + + + (4)

By combining the equation above with the mass balance equations the following expression can be written for KP:

(

)

2 P 2 1 b 20 K b 5 b − = − (5)

The approach now is to guess T2, look up KP and solve for the three unknowns in Eqns. (1), (2), and (5). The correct guess will be the one that satisfies the adiabatic flame temperature equation.

In order to calculate the adiabatic flame temperature, Eqn. (1.11) in the text can be used by taking T1 = 298 K.

(

2

)

( )

(

1

)

( )

∑

_⎣ − + Δ _⎦ −

∑

_⎣ − + Δ _⎦ (6)

gases 0

(

)

f H kJ mol Δ C2H4O(1) −76.62 CO −110.62 CH4 −74.90 C2H4 52.34

Applying Eqn. (6) yields

(

)

2 0 0 T 298 let Δ =H H −H :

(

CO

)

(

CH₄

) (

H₂

) (

C H₂ ₄

)

76.62 a 110.62 H b 74.90 H c 0.0 H d 52.34 H − = − + Δ + − + Δ + + Δ + + Δ (7)

(26)

26 ( ) ( ) ( ) 2 2 4 4 10 P,f H 10 P,f C H 10 P,f CH log K 1 log K 5.664 log K 2.372 = = − = −

so from Eqn. (3), log10 KP = 0.12023. Solving Eqn. (5) for b and then Eqns. (1) and (2) for c and d gives:

a = 1.0, b = 0.7557, c = 0.2443, and d = 0.1221

Application of Eqn. (7) using T2 = 1400 K and the values of a, b, c, and d just calculated yields: T2 ΔHCO (kJ/mol) ΔHCH₄ ΔHH₂ ΔHC H₂ ₄ Hp (kJ)

1400 35.36 69.66 33.08 91.26 -53.60 If we now take T2 = 1300 K, from Tables in the text one has:

( ) ( ) ( ) 2 2 4 4 10 P,f H 10 P,f C H 10 P,f CH log K 1 log K 5.766 log K 2.107 = = − = −

so from Eqn. (3), log10 KP = - 1.552 or KP = 0.028054. Solving Eqn. (5) for b and then Eqns. (1) and (2) for c and d gives:

a = 1.0, b = 0.8398, c = 0.1603, and d = 0.0801

Application of Eqn. (7) using T2 = 1300 K and the values of a, b, c, and d just calculated yields: T2 ΔHCO (kJ/mole) ΔHCH₄ ΔHH₂ ΔHC H₂ ₄ Hp (kJ)

1300 31.89 61.34 29.93 80.63 -74.65 Extrapolation to the value of T2 where Hp ~ Hr (−76.62) yields:

T 2 ~ 1291 K

(27)

Liquid hydrazine (N2H4) decomposes exothermically in a monopropellant rocket operating at 100-atm chamber pressure. The products formed in the chamber are N2, H2, and ammonia (NH3) according to the irreversible reaction

N2H4(1) → a N2 + b H2 + c NH3

Determine the adiabatic decomposition temperature and the product composition a, b, and c. Take the standard heat of formation of liquid hydrazine as 50.03 kJ/mole. The hydrazine enters the system at 298 K.

Solution: The solution for this problem is an iterative one. In order to calculate the final temperature of the mixture, one needs the composition of the mixture at equilibrium. But the composition of the mixture can only be found if the final temperature is known. Therefore, one will guess the final temperature and calculate the mixture composition using the equilibrium concept. Once the composition is known at the guessed temperature, the adiabatic flame temperature is calculated and checked against the guessed value. This process is repeated until the guessed and calculated temperatures are the same. Consider the following reaction at equilibrium:

N2H4(1) → a N2 + b H2 + c NH3

balancing this reaction yields the following equations for a, b, and c.

N: 2 = 2 a + c (1)

H: 4 = 2 b + 3 c (2)

The other equation necessary for the solution of a, b, and c will come from the Kp for the following equilibrium reaction.

2 2 3

1 3

N H NH

2 +2 ⇔

For this equilibrium reaction, the equilibrium constant of formation of NH3 can be expressed in terms of the mixture composition at equilibrium as shown:

( 3) 3 2 2 1 1 2 3 2 NH p P,f NH 1 2 3 2 H N n _P K K n n n − − ⎛ ⎞ = = ⎜_⎜ ⎟_⎟ ⎝

∑

⎠ (3)

(28)

28 p 1/ 2 3/ 2 c a b c K a b 100 + + = (4)

The approach now is to guess T2, look up Kp and solve for the three unknowns in Eqns. (1), (2), and (4). The correct guess will be the one that satisfies the adiabatic flame temperature equation.

(

2

)

( )

(

1

)

( )

∑

_⎣ − + Δ _⎦ −

∑

_⎣ − + Δ _⎦ (5)

With Eqns. (4) and (5) the iterative process may now start. Since the conditions for the reactants are specified (T1 = 298 K), Table 1.1 in the text provides the following thermodynamic

properties: gases 0

(

)

f H kJ mol Δ N2H4(1) 50.03 N2 0.0 H2 0.0 NH3 −46.22

(

)

2 0 0 T 298 let Δ =H H −H :

(

N2

) (

H2

)

(

NH3

)

50.03=a 0.0+ ΔH +b 0.0+ ΔH + −c 46.22+ ΔH (6)

Let us first guess that T2 = 900 K, from Tables in the text it follows that:

( 3)

10 P,f NH

log K = −2.915

Solving Eqns. (l), (2) and (4) results in:

a = 0.9487, b = 1.8462, and c = 0.1026

For a product temperature of 900 K, application of Eqn. (6) using the values of a, b, and c, found above, yields (all ΔH's are in kJ/mol and Hp is in kJ):

T2

(

)

2 N H kJ mol Δ ΔHH₂ ΔHNH₃ Hp (kJ) 900 K 18.23 17.69 27.09 47.98

First guess that T2 = 1000 K, from Tables in the text it follows that:

(29)

( 3)

10 P,f NH

log K = −3.233

Solving Eqns. (1), (2) and (4) one has:

a = 0.9739, b = 1.9218, and c = 0.05216

For a product temperature of 1000 K, application of Eqn. (6) using values of a, b, and c, found above yields: T2

(

)

2 N H kJ mole Δ ΔHH₂ ΔHNH₃ Hp (kJ) 1000 K 21.47 20.70 32.60 60.00

Interpolation to the value of T2 where Hp ~ Hr (50.03) yields:

(30)

30

Problem 13

Gaseous hydrogen and oxygen are burned at l-atm pressure under the rich conditions designated by the following combustion reaction:

O2 + 5 H2 → a H2O + b H2 + c H

The gases enter at 298 K. Calculate the adiabatic flame temperature and the product composition a, b, and c.

Solution: The solution for this problem is an iterative one. In order to calculate the final temperature of the mixture, one needs the composition of the mixture at equilibrium. But the composition of the mixture can only be found if the final temperature is known.

Therefore, one will guess the final temperature and calculate the mixture composition using the equilibrium concept. Once the composition is known at the guessed temperature, the adiabatic flame temperature is calculated and checked against the guessed value. This process is repeated until the guessed and calculated temperatures are the same. Consider the following reaction at equilibrium:

O2 + 5 H2 → a H2O + b H2 + c H

balancing this reaction yields the following equations for a, b, and c.

O: 2 = a (1)

H: 10 = 2 a + 2 b + c (2)

The other equation necessary for the solution of a, b, and c will come from the KP expression for the following equilibrium reaction.

2

1

H H

2 ⇔

For this equilibrium reaction, the equilibrium constant of formation of H can be expressed in terms of the mixture composition at equilibrium as shown:

( ) 2 1 1/ 2 H P P,f H 1/ 2 H n P K K n n − ⎛ ⎞ = = ⎜_⎜ ⎟_⎟ ⎝

∑

⎠ (3)

or in terms of a, b, and c (with P = 1 atm) one has

P 1/ 2 c 1 K b a b c = + + (4)

(31)

The approach now is to guess T2, look up Kp and solve for the three unknowns in Eqns. (1), (2), and (4). The correct guess will be the one that also satisfies the adiabatic flame temperature equation.

(

2

)

( )

(

1

)

( )

∑

_⎣ − + Δ _⎦ −

∑

_⎣ − + Δ _⎦ (5)

With Eqns. (4) and (5) the iterative process may now start. Since the conditions for the reactants are specified (T1 = 298 K), Table 1.1 in the text provides the following thermodynamic properties: gases 0

(

)

f H kJ mol Δ O2 0.0 H 218.09 H2 0.0 H2O −242.00

(

)

2 0 0 T 298 let Δ =H H −H :

(

H O₂

) (

H₂

)

(

H

)

0.0= −a 242.00+ ΔH +b 0.0+ ΔH +c 218.09+ ΔH (6)

Let us first guess that T2 = 2500 K, from Table 2 it follows that, log10 Kp,f(H) = −1.601

a = 2.0, b = 2.9516, and c = 0.0967

For a product temperature of 2500 K application of Eqn. (6) using the values of a, b, and c, yields (all ΔH's are in kJ/mole and Hp is in kJ):

T2

(

)

2 H O H kJ mol Δ ΔHH₂ ΔHH Hp (kJ) 2500 K 99.03 70.54 45.80 −52.21

(32)

32

log10 Kp,f(H) = −1.247

a = 2.0, b = 2.8912, and c = 0.2176

For a product temperature of 2700 K application of Eqn. (6) using the values of a, b, and c, found above yields (all ΔH's are in kJ/mol and Hp is in kJ):

T2

(

)

2 H O H kJ mol Δ ΔHH₂ ΔHH Hp (kJ) 2700 K 109.89 77.77 49.96 18.96

Interpolation to the value of T2 where Hp ~ Hr (Hr = 0.0) yields:

T2 ~ 2647 K

(33)

The liquid propellant rocket combination nitrogen tetroxide (N2O4) and UDMH (unsymmetrical dimethyl hydrazine) has optimum performance at an oxidizer to fuel weight ratio of 2 at a chamber pressure of 67 atm. Assume that the products of combustion of this mixture are N2, CO2, H2O, CO, H2, O H, OH, and NO. Set down the equations necessary to calculate the adiabatic combustion temperature and the actual product composition under these conditions. These equations should contain all the numerical data in the description of the problem and in the tables in the appendices. The heats of formation of the reactants are:

( ) ( ) 0 2 4 1 f 298 0 f 1 N O H 2.90 kJ mol UDMH ₂₉₈ 53.17 kJ mol Δ = − ΔΗ = +

The propellants enter the combustion chamber at 298 K.

Solution: The solution for this problem consists of finding the correct number of equations necessary to solve for the unknown quantities in this problem. There are 9 unknown species coefficients and the temperature is unknown. Therefore, 10 equations must be found to solve the problem. These equations consist of the energy, 4 mass balance, and 5 equilibrium equations.

To specify the general reaction equation, the mole proportion of reactants must be found first. This can be done as follows.

The molecular weights of the reactants are:

for N2O4, MW = 92 for C2N2H8, MW = 60

From the problem statement one has:

2 4

mass UDMH 2 mass N O =

By assuming 1 mole N2O4 this equation becomes mass UDMH

2 60 = or

(34)

34

So

nUDMH = 1.304

The general equation for the reaction then becomes:

C2N2H8 + 1.304 N2O4 → a N2 + b CO2 + c H2O + d CO + e H2 + f O + g H + h OH + i NO The mass balance equations can be written as:

C: 2 = b + d (1)

N: 4.609 = 2a + i (2)

H: 8 = 2c + 2e + g + h (3)

O: 5.216 = 2b + c + d + f + h + i (4)

The energy equation would be the equation for adiabatic flame temperature (Eq. (11)) found in the text.

(

2

)

( )

(

1

)

( )

∑

The following information can be used to simplify the equation

2 4 fN O H 2903 kJ mol Δ = − f UDMH H 53.172 kJ mol Δ =

The right side of the equation is:

1(53.172) + 1.304 (-2.093) = 50.44 kJ

Therefore, the energy equation becomes:

(

2

)

( )

0 0 0 i T 298 f ₂₉₈ i i prod 50.44=

∑

n ⎡_⎣ H −H + ΔH ⎤_⎦ (5)

Some of the possible equilbrium reactions include:

H2 ⇔ 2 H CO + O ⇔ CO2 N2 + 2 O ⇔ 2 NO H2 + 2 OH ⇔ 2 H2O

(35)

These reactions give the following equilibrium equations: 1 2 2 H p H n P K n n ⎛ ⎞ = ⎜_⎜ ⎟_⎟ ⎝

∑

⎠ 2 i g 67 e n ⎛ ⎞ = ⎜_⎜ ⎟_⎟ ⎝

∑

⎠ (6) 2 2 1 2 NO p 2 N O n P K n n n − ⎛ ⎞ = ⎜_⎜ ⎟_⎟ ⎝

∑

⎠ 2 i 2 n i a f 67 ⎛ ⎞ = ⎜_⎜ ⎟_⎟ ⎝ ⎠

∑

(7) 2 3 1 CO p CO O n _P K n n n − ⎛ ⎞ = ⎜_⎜ ⎟_⎟ ⎝

∑

⎠ i n b d f 67 ⎛ ⎞ = _⎜_⎜ _⎟_⎟ ⎝ ⎠

∑

(8) 2 4 2 1 2 H O p 2 H OH n _P K n n n − ⎛ ⎞ = ⎜_⎜ ⎟_⎟ ⎝

∑

⎠ 2 i 2 n c e h 67 ⎛ ⎞ = ⎜_⎜ ⎟_⎟ ⎝ ⎠

∑

(9) 2 2 5 1 H O CO p 2 CO OH n n _P K n n n − ⎛ ⎞ = ⎜_⎜ ⎟_⎟ ⎝

∑

⎠ i 2 n c b d h 67 ⎛ ⎞ = _⎜_⎜ _⎟_⎟ ⎝ ⎠

∑

(10)

where

∑

n_i = a + b + c + d + e + f + g + h + i, and the Kp's are evaluated at the adiabatic flame temperature.

(36)

36

Problem 15

Consider a fuel burning in inert airs and oxygen where the combustion requirement is only 0.21 moles of oxygen. Order the following mixtures as to their adiabatic flame temperatures with the given fuel.

a) pure O2

b) 0.21 O2 + 0.79 N2 (air) c) 0.21 O2+0.79 Ar

d) 0.21 O2 + 0.79 CO2

Solution: The temperatures order with respect to the Cp of the inert and excess oxygen. The lower the Cp of the inert gas the higher the temperature. Thus, Ar being monoatomic has the lowest Cp, O2 and N2 are about the same, but higher than Ar since they are diatomic and have vibrational and rotational degrees of freedom, and CO2 being triatomic has the highest Cp. Thus the order from highest to lowest is

c, a and/or b, d

(37)

Propellant chemists have proposed a new high energy liquid oxidizer, penta-oxygen O5, which is also a monopropellant. Calculate the monopropellant decomposition temperature at a chamber pressure of 10 atm if it assumed the only products are O atoms and O2 molecules. The heat of formation of the new oxidizer is estimated to be very high, + 1025 kJ / mol. Obviously the amounts of O2 and O must be calculated for one mole of O5 decomposing. The O5 enters the system at 298 K. Hint: The answer will lie somewhere between 4000 and 5000K.

Solution: The solution for this problem is an iterative one. In order to calculate the final temperature of the mixture, one needs the composition of the mixture at equilibrium. But the composition of the mixture can only be found if the final temperature is known. Therefore, one will guess the final temperature and calculate the mixture composition using the equilibrium concept. Once the composition is known at the guessed temperature, the adiabatic flame temperature is calculated and checked against the guessed value. This process is repeated until the guessed and calculated temperatures are the same. Consider the following reaction at equilibrium:

O5→ a O2 + b O

balancing this reaction yields the following equation for a.

O: 5 = 2a + b or b = 5 - 2a (1)

The other equation necessary for the solution of a and b will come from the Kp expression for the following equilibrium reaction.

2

1

O O

2 ⇔

For this equilibrium reaction, the expression for the equilibrium constant in terms of the equilibrium constant of formation of each individual species is,

( ) ( )2 2 2 1 P,f O _O P 1/ 2 1/ 2 O P,f O K _n _P K K n n − ⎛ ⎞ = = ⎜_⎜ ⎟_⎟ ⎝

∑

⎠ (2)

or in terms of a and b ( with P = 10 atm.) one has

1/ 2 p 1/ 2 b 10 K a a b ⎡ ⎤ = _⎢ _⎥ + ⎣ ⎦ (3)

By combining the equation above with the mass balance equation the following expression can be written for Kp:

(38)

38 1/ 2 p 1/ 2 5 2a 10 K a 5 1 − ⎡ ⎤ = _⎢ _⎥ − ⎣ ⎦ (4)

The approach now is to guess T2 look up Kp and solve for the two unknowns in Eqns. (1) and (4). The correct guess will be the one that satisfies the adiabatic flame temperature equation.

(

2

)

( )

(

1

)

( )

0 0 0 0 0 0 i T 298 f ₂₉₈ j T 298 f ₂₉₈ i j i prod j react n _⎣⎡ H −H + ΔH _⎦⎤ − n _⎣⎡ H −H + ΔH ⎤_⎦

∑

(5)

gases 0

(

)

f H kJ mol Δ O 249.36 O2 0 O5 1025

(

)

2 0 0 T 298 let Δ =H H −H :

(

O₂

)

(

O

)

1025=a 0.0+ ΔH +b 249.36+ ΔH (6)

First guess that T2 = 4500 K, from Tables in the text one has:

( ) ( ) 2 10 P,f O 10 P,f O log K 1 log K 0.543 = =

so from Eqn. (2), log10 Kp = 0.543 or Kp = 3.491. Solving Eqn. (4) for a and then Eqn. (1) for b gives:

a = 1.29, b = 2.42

Application of Eqn. (6) using T2 = 4500 K and the values of a and b just calculated yields:

T2

(

)

2 O H kJ mol Δ ΔHO Hp (kJ/mol) 4500 159.9 88.45 1023.77

(39)

Since the guessed value of T2 give Hp ~ H (1025 kJ/mol) the decomposition temperature is T2 ~ 4500 K