1

Question 1:

Part 1. O2 ���� 2O

a. The effect of pressure:

K� � χ��χ���P�� ���P� ���� � exp��∆G��R�T � � only a function of temperature

At a fixed temperature:

χ��χ��P�� ��� � Constant

Thus, increasing pressure results in an increase in the mole fraction of O2 molecules. A decrease

in pressure results in an increase in the mole fraction of O atoms.

Consistent with the principal of Le Châtelier, an increase in pressure shifts the equilibrium

towards the smaller number of moles (the reactant side in this example).

b. The effect of temperature:

The enthalpy of the reactant /� ∆h12,��4 is smaller than the enthalpy of the product /�2h16,�� 7 2∆h12,�4 for 200K<T<5000K (Tables A.11 and A.12). Thus the dissociation of 1 mole of O2

molecules into 2 moles of O atoms is endothermic. Based on the Principal of Le Châtelier, an

increase in temperature increases the dissociation and the concentration of O atoms.

Part 2. N2 + O2 ���� 2NO

a. the effect of pressure

K� � χ8��χ8� . χ���P�� ���P� ���� � exp��∆G��R�T � � only a function of temperature

At a fixed temperature:

χ8��χ8� . χ��� Constant

The mole fractions of the species are independent of pressure.

2

b. The effect of temperature:

The enthalpy of the reactants /� ∆h12,�� 7 ∆h12,8�4 is smaller than the enthalpy of the product

/� 2∆h16,8�� 7 2∆h12,8�4 for 200K<T<5000K (see Tables A.7, A.9 and A.11). Thus the formation of

2 moles of NO molecules from 1 mole of O2 and 1 mole of N2 is endothermic. Based on the

Principal of Le Châtelier, the concentration of NO increases with an increase in temperature.

Part 3. CO + ½ O2 ���� CO2

a. the effect of pressure

K� � χ:��χ:�. χ��

�/� �P�� ���P� ���</� � exp��∆G��R�T � � only a function of temperature

At a fixed temperature:

χ:��χ:�. χ��

�/� P�� �����/� � Constant

With an increase in pressure, the mole fractions of the CO2 increases and the mole fraction of

the reactants decrease.

b. The effect of temperature:

The difference between the enthalpy of products and reactants /� h1� � h1=4 is negative for

200K<T<5000K (Tables A.1, A.2 and A.11). Thus the formation of CO2 from O2 and CO is

exothermic. Based on the Principal of Le Châtelier, the concentration of CO2 decreases with an

increase in temperature.

3

Problem 2:

Given: CH3OH and air, φ<1, A/F=8 kgf/kgair

Find: φ, χCO2

Assumptions: no dissociation

Solution:

a) For φ=1: CH3OH+a(O2+3.76N2)�CO2+2 H2O+3.76aN2

O-Balance: 1+2a=2+2; a=1.5

�AF�@A� �4.76aMWG��MW6��H � 4.76 I 1.5 I 28.85

32.04 � 6.429

φ � /A FQ 4@A�A FQ� 6.429

8 � 0.801

b) CH3OH+(a/φ)(O2+3.76N2)�CO2+2 H2O+bO2+3.76(a/ φ)N2

O-Balance: 1+2a/φ=2+2+2b; b=0.3725

χ:�� � N:��N S � 11 7 2 7 0.3725 7 3.76T1.5/.801U �

110.416 � 0.096

4

Problem 3:

Given: stoichiometric butanol-air mixture at 298K

Find: Adiabatic flame temperature, Tad

Assumptions: No dissociation

Solution:

Stoichiometric reaction: C4H10O+aO2+3.76aN2�4CO2+5H2O+3.76aN2

For φ=1, a=6; C4H10O+6O2+22.56N2�4CO2+5H2O+22.56N2

First law for adiabatic reactions: HR@298K=HP@Tad

VNh1W:XYZ[� 7 VNh1W�� 7 VNh1W8� � VNh1W:�� 7 VNh1WY�� 7 VNh1W8�

Where

h1 � h16� 7 ∆h12 � h16� 7 C�TT � 298U \1]� for diatomic oxygen and nitrogen molecules is zero. Sensible enthalpy change of the

reactants is also zero. Tad is guessed to be 2100K and Cp is evaluated at (2100+298)/2.

From Tables A.2, A.6, A.7: Cp,CO2 @ 1200K=56.205, CpH2O @ 1200K=43.874, CpN2 @ 1200K=33.707

V�277000 7 0W:XYZ[� 7 6V0W�� 7 22.56V0W8�� 4V�393546 7 56.205TTG^ � 298UW:��7 5V�241847 7 43.874TTG^ � 298UWY�� 7 22.56V0 7 33.707TTG^ � 298UW8�

Solve for Tad: Tad=2378.172K

By doing a few more iterations, a better result for Tad is obtained.

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Problem 4:

Given: Butanol combustion

Find: LHV

Assumptions: the oxidizer is air, stoichiometric mixture, TR = TP = 298K

Solution:

C4H10O+6O2+22.56N2�4CO2+5H2O+22.56N2

\1]� for diatomic oxygen and nitrogen molecules is zero. Sensible enthalpy change of both

products and reactants is zero.

LHV � H= � Hb � 1h6,:XYZ[�� � /4h6,:��� 7 5h6,Y��TcU� 4

� �277000 � T4 I �393546 7 5 I �241847U � 2,506,419 kJ/kmol 2506419 kJkmol74.1216 kgkmol

� 33814.96 kJkg � 33.8MJ

kg

The heating value of ethanol is 26.9MJ/kg (Practice Problems, Q. 2.7)

Butanol has a higher energy density; thus it is a better alternative for gasoline.

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Problem 5:

Given: H2+1/2 O2 � H2O, H/O=1

Find: Equilibrium composition at T=2000K and P=1atm

Assumption: This reaction is the only one involving O and H atoms

Solution:

∆G�� � 1Tg16,�� UY�� � 1Tg16,�� UY� � 1/2Tg16,�� U��

∆G�� � �135643 Tusing Appendix AU K� � exp h�∆G��R�T i � exp j 135643

8.315 I 2000k � 3.486 I 10<

K� � χY��χY� . χ��

�/� � PP����/�

#H#O � 1 � 2χY�� 7 2χY�χY�� 7 2χ��

nχ � χY�� 7 χY� 7 χ�� � 1

Solve 2 and 3 for χY� and χY�� in terms of χ��, and substitute into 1. Then solve 1 for χ��by

trial and error.

χY�� � 0.6662

χY� � 0.0003

χ�� � 0.3334

Eq. 1

Eq. 2

Eq. 3

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Problem 6:

Here is a picture of the STANJAN output page:

Pressure and temperature are constants, 1 atm and 2000K, respectively. In the initial mixture,

the ratio of H atoms to O atoms must be 1. Hence, you can use 1 mol of H and 1 mol of O, or 1

mol of H2 and 1 mol of O2, or 1 mol of H2, 1.5 mol of O2, and 1 mol of H2O, without affecting the

equilibrium state.

By comparing these results with those of Problem 2.47, you can see the effect of dissociation

and formation of additional species.

Here is a picture of the STANJAN output page for ɸ=1.

Problem 7:

Pressure and enthalpy are constants, initial conditions are 1 atm and 298 K. In the initial mixture the molar fractions are chosen based on ɸ. As the Adiabatic flame temperature is greater than 3000K, dissociation of H2, O2 and H2O to H, O and OH should also be considered.

Phi N (C2H2) N (O2) T(adiabatic) K X(CO2) X(CO) X(O2) X(H2) X(H2O) X(OH) X(O) X(H) 0.5 1 5 3.11E+03 1.74E-01 1.33E-01 3.91E-01 9.68E-03 9.49E-02 7.65E-02 9.99E-02 2.12E-02 0.6 1 4.166667 3.18E+03 1.65E-01 1.79E-01 3.13E-01 1.42E-02 9.94E-02 8.55E-02 1.12E-01 3.14E-02 0.7 1 3.571429 3.23E+03 1.54E-01 2.22E-01 2.51E-01 1.92E-02 1.02E-01 9.10E-02 1.18E-01 4.23E-02 0.8 1 3.125 3.28E+03 1.42E-01 2.62E-01 2.00E-01 2.47E-02 1.04E-01 9.37E-02 1.20E-01 5.38E-02 0.9 1 2.777778 3.31E+03 1.30E-01 2.99E-01 1.58E-01 3.07E-02 1.04E-01 9.43E-02 1.18E-01 6.57E-02

1 1 2.5 3.34E+03 1.18E-01 3.33E-01 1.25E-01 3.71E-02 1.03E-01 9.30E-02 1.14E-01 7.78E-02 1.1 1 2.272727 3.37E+03 1.06E-01 3.64E-01 9.71E-02 4.41E-02 1.01E-01 9.03E-02 1.07E-01 8.99E-02 1.2 1 2.083333 3.39E+03 9.49E-02 3.93E-01 7.48E-02 5.15E-02 9.83E-02 8.63E-02 9.93E-02 1.02E-01 1.3 1 1.923077 3.40E+03 8.44E-02 4.19E-01 5.69E-02 5.94E-02 9.48E-02 8.13E-02 9.04E-02 1.14E-01 1.4 1 1.785714 3.41E+03 7.45E-02 4.43E-01 4.25E-02 6.78E-02 9.05E-02 7.54E-02 8.08E-02 1.25E-01 1.5 1 1.666667 3.42E+03 6.52E-02 4.65E-01 3.12E-02 7.68E-02 8.57E-02 6.89E-02 7.09E-02 1.36E-01 1.6 1 1.5625 3.43E+03 5.66E-02 4.85E-01 2.24E-02 8.63E-02 8.02E-02 6.20E-02 6.10E-02 1.47E-01 1.7 1 1.470588 3.43E+03 4.86E-02 5.03E-01 1.55E-02 9.64E-02 7.42E-02 5.47E-02 5.12E-02 1.56E-01 1.8 1 1.388889 3.43E+03 4.10E-02 5.20E-01 1.03E-02 1.07E-01 6.75E-02 4.71E-02 4.17E-02 1.65E-01 1.9 1 1.315789 3.43E+03 3.40E-02 5.36E-01 6.54E-03 1.19E-01 6.02E-02 3.94E-02 3.29E-02 1.72E-01

2 1 1.25 3.42E+03 2.75E-02 5.50E-01 3.90E-03 1.32E-01 5.23E-02 3.19E-02 2.49E-02 1.77E-01

3.05E+03

3.10E+03

3.15E+03

3.20E+03

3.25E+03

3.30E+03

3.35E+03

3.40E+03

3.45E+03

0 0.5 1 1.5 2 2.5

T ad

iaba

tic (K

)

ɸ

T(adiabatic) K

0.00E+00

1.00E-01

2.00E-01

3.00E-01

4.00E-01

5.00E-01

6.00E-01

0 0.5 1 1.5 2 2.5 ɸ

Species Molar Fraction X O2

X CO