Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
8th Edition
ISBN: 9780073398174
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 16.6, Problem 21P

(a)

To determine

The logarithm equilibrium constant for the reaction at 25°C.

Compare the results for the values of KP obtained from the equation of equilibrium constants and the equilibrium constant Table A-28.

(a)

Expert Solution
Check Mark

Answer to Problem 21P

The logarithm equilibrium constant for the reaction at 25°C is 1.1695×1040_.

The equilibrium constant obtained from the equilibrium constants of Table A-28 at 1440Ris 1.1125×1040_.

Explanation of Solution

Express the standard-state Gibbs function change.

ΔG*(T)=vH2Og¯*H2O(T)vH2g¯*H2(T)vO2g¯*O2(T) (I)

Here, the Gibbs function of components H2,O2,andH2O at 1 atm pressure and temperature T are g¯*H2(T),g¯*O2(T),andg¯*H2O(T) respectively, temperature of H2,O2,andH2O are TH2O,TCO,TH2,andTCO2, and stoichiometric coefficients of components H2,O2,andH2O are vH2,vO2,andvH2O respectively.

Write the equation to calculate the natural logarithms of equilibrium constant for the chemical equilibrium of ideal-gas mixtures.

lnKP=ΔG*(T)RuT (II)

Here, universal gas constant is Ru and temperature of Gibbs function of formation is T.

Write the equation to calculate the equilibrium constant for the chemical equilibrium of ideal-gas mixtures.

KP=e(ΔG*(T)RuT) (III)

Conclusion:

From the equilibrium reaction, the values of vH2O,vH2,andvO2 are 1, 1,  and 1 respectively.

Refer Table A-26, obtain the values of g¯*H2O,g¯*H2,andg¯*O2 as below:

g¯*H2O=228,590 kJ/kmolg¯*H2=0kJ/kmol g¯*O2=0kJ/kmol

Substitute 1 for vH2O, 1 for vH2, 0.5 for vO2, 228,590 kJ/kmol for g¯*H2O, 0 for 0kJ/kmol , and 0 for g¯*O2 in Equation (I).

ΔG*(T)=1(228,590kJ/kmol)1(0)0.5(0)=228,590 kJ/kmol

Substitute 228,590kJ/kmol for ΔG*(T), 25°C for T, and 8.314kJ/kmolK for Ru in Equation (II).

lnKP=228,590kJ/kmol(8.314kJ/kmolK)×25°C=228,590kJ/kmol(8.314kJ/kmolK)×(25+273)K=92.26

Substitute 92.26 for KP in eEquation (III).

KP=e(92.26)=1.1695×1040

Thus, the equilibrium constant obtained from the equilibrium reaction at 298 K is 1.1695×1040_.

Refer Table A-28, “Natural logarithms of the equilibrium constant” obtain the equilibrium constant for the reaction at the temperature of 298 K as 92.21_.

Substitute 92.21 for lnKP in Equation (III).

KP=e(92.21)=1.1125×1040

Thus, the equilibrium constant obtained from the table A-28  at 1440 R is 1.1125×1040_.

The value obtained for equilibrium constant at 298 K from the definition of the equilibrium constant is 92.26 which is equal to the value obtained for equilibrium constant as 92.21 from the Table A-28.

(b)

To determine

The logarithm equilibrium constant for the reaction at 2000 K.

Compare the results for the values of KP obtained from the equilibrium constants of Table A-28.

(b)

Expert Solution
Check Mark

Answer to Problem 21P

The logarithm equilibrium constant for the reaction at 2000 K is 3471_.

The equilibrium constant obtained from the equilibrium constants of Table A-28 at 2000K is.

3446.1_

Explanation of Solution

Express the standard-state Gibbs function change.

ΔG*(T)=vH2Og¯*H2O(T)vH2g¯*H2(T)vO2g¯*O2(T)=vH2O(h¯Ts¯)H2OvH2(h¯Ts¯)H2vO2(h¯Ts¯)O2=[vH2O(h¯fo,H2O+(h¯H2Oh¯oH2O)Ts¯H2O)vH2(h¯f,H2o+(h¯H2h¯oH2)Ts¯H2)vO2(h¯fo,O2+(h¯O2h¯oO2)Ts¯O2)] (IV)

Here, the Gibbs function of components H2,O2,andH2O at 1 atm pressure and temperature T are g¯*H2(T),g¯*O2(T),andg¯*H2O(T) respectively, enthalpy on the unit mole basis of H2,O2,andH2O are h¯H2,h¯O2,andh¯H2O respectively, absolute entropy of H2,O2,andH2O are s¯H2,s¯O2,ands¯H2O, temperature of H2,O2,andH2O are TH2O,TCO,TH2,andTCO2, sensible enthalpy at the specified state of H2,O2,andH2O are h¯H2,h¯O2,andh¯H2O, the sensible enthalpy at the standard reference state of 25°C and 1 atm of H2,O2,andH2O are h¯H2o,h¯O2o,andh¯H2Oo, the enthalpy of formation at a specified states of 25°C and 1 atm for the components H2,O2,andH2O are h¯f,H2o,h¯f,O2o,andh¯f,H2Oo, stoichiometric coefficients of components H2,O2,andH2O are vH2,vO2,andvH2O respectively.

Write the equation to calculate the natural logarithms of equilibrium constant for the chemical equilibrium of ideal-gas mixtures.

lnKP=ΔG*(T)RuT (V)

Here, universal gas constant is Ru and temperature of Gibbs function of formation is T.

Write the equation to calculate the equilibrium constant for the chemical equilibrium of ideal-gas mixtures.

KP=e(ΔG*(T)RuT) (VI)

Conclusion:

From the equilibrium reaction, the values of vH2O,vCO,vH2,andvCO2 are 1, 1, 1, and 1 respectively.

Refer Table A-26, obtain the values of h¯f,H2o,h¯f,O2o,andh¯f,H2Oo as below:

h¯f,H2o=0h¯f,O2o=0 h¯f,H2Oo=241,820kJ/kmol

Refer to Table A-22, obtain the value of h¯H2o at temperature of 298K.

h¯H2o=9904kJ/kmol

Refer to Table A-22, obtain the value of h¯H2 and s¯H2 at temperature of 2000K.

h¯H2=61,400 kJ/kmols¯H2=188.297kJ/kmolK

Refer to Table A-19E, obtain the value of h¯O2o at temperature of 298K.

h¯O2o=8682kJ/kmol

Refer to Table A-19E, obtain the value of h¯O2 and s¯O2 at temperature of 2000K.

h¯O2=67,881kJ/kmols¯O2=268.655kJ/kmolK

Refer to Table A-23, obtain the value of h¯H2Oo at temperature of 298K.

h¯H2Oo=9904 kJ/kmol

Refer to Table A-23, obtain the value of h¯H2O and s¯H2O at temperature of 2000 K.

h¯H2Oo=82,593 kJ/kmols¯H2O=264.571 kJ/kmolK

Substitute 1 for vH2O, 241,820kJ/kmol for h¯fo,H2O, 82593 kJ/kmol for h¯H2O, 9904kJ/kmol for h¯oH2O, 2000 K for T, 264.571 kJ/kmolK for s¯H2O, 1 for vH2, 0 for h¯f,H2o, 61,400 kJ/kmol for h¯H2, 8468kJ/kmol for h¯oH2, 188.297 kJ/kmolK for Ts¯H2, 0.5 for vO2, 0 for h¯fo,O2, 67881 kJ/kmol for h¯O2, 8682 kJ/kmol for h¯oO2, 268.655 kJ/kmolK for s¯O2 in Equation (IV).

ΔG*(T)=[1(241,820kJ/kmol+82593 kJ/kmol9904kJ/kmol2000 K×264.571 kJ/kmolK)H2O1(0+61,400 kJ/kmol8468kJ/kmol2000K×188.297 kJ/kmolK)H20.5(0+(67881 kJ/kmol8682 kJ/kmol)2000K×268.655 kJ/kmolK)O2]=135.556 kJ/kmol

Substitute 53,436Btu/lbmol for ΔG*(T), 3960 R for T, and 1.986Btu/lbmolR for Ru in Equation (V).

lnKP=135,556kJ/kmol(1.986kJ/kmolK)×2000K=8.152

Substitute 8.152 for KP in Equation (III).

KP=e(8.152)=3471

Thus, the equilibrium constant obtained from the equilibrium reaction at 2000K is 3471_.

Refer Table A-28, “Natural logarithms of the equilibrium constant” obtain the equilibrium constant for the reaction by interpolating for the temperature of 2000 K as 8.145.

Substitute 8.145 for KP in Equation (VI).

KP=e(8.145)=3446.10

Thus, the equilibrium constant obtained from the table A-28 at2000K is 3446.10_.

The value obtained for equilibrium constant at 2000K from the definition of the equilibrium constant is 8.152 which is almost equal to the value obtained for equilibrium constant as 8.145 from the Table A-28.

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Chapter 16 Solutions

Thermodynamics: An Engineering Approach

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