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 15P

(a)

To determine

The equilibrium constant for the reported reaction at 1000 K.

(a)

Expert Solution
Check Mark

Answer to Problem 15P

The equilibrium constant for the reported reaction at 1000 K is 8.66×1011.

Explanation of Solution

Write the stoichiometric reaction for the reported process.

H2OH2+1/2O2

From the stoichiometric reaction, infer that the stoichiometric coefficients for hydrogen (vH2) is 1, for oxygen (vO2) is 0.5, and for water (vH2O) is 1 respectively.

Write the formula for equilibrium constant (kp).

kp=eΔG(T)/RuTlnkp=ΔG(T)/RuT (I)

Here, temperature is T, Gibbs function is ΔG(T) and universal gas constant is Ru.

Write the formula for Gibbs energy (ΔG(T)).

ΔG(T)=vH2g¯H2(T)+vO2g¯O2(T)vH2Og¯H2O(T) (II)

Here, Gibbs function of hydrogen is g¯H2, Gibbs function of oxygen is g¯O2 and Gibbs function of H2O is g¯H2O.

Write the Gibbs function of H2(g¯H2) at 1000 K.

g¯H2=h¯H2+[h¯1000h¯298Ts]H2 (III)

Here, enthalpy of formation of H2 is h¯H2, enthalpy of H2 gas at 1000 K is h¯1000, enthalpy of H2 gas at 298 K is h¯298.

Write the Gibbs function of O2(g¯O2) at 1000 K.

g¯O2=h¯O2+[h¯1000h¯298Ts]O2 . (IV)

Here, enthalpy of formation of O2 is h¯O2, enthalpy of O2 gas at 1000 K is h¯1000, enthalpy of O2 gas at 298 K is h¯298.

Write the Gibbs function of water (g¯H2O) at 1000 K.

g¯H2O=h¯H2O+[h¯1000h¯298Ts]H2O (V)

Here, enthalpy of formation of water vapor is h¯H2O, enthalpy of water vapor at 1000 K is h¯1000, enthalpy of water vapor at 298 K is h¯298, and entropy is s.

Conclusion:

Refer table A-26, “Enthalpy formation table”, obtain the enthalpy of H2(h¯H2) as 0.

Refer table A-22, “Ideal gas properties of Hydrogen”, obtain the following properties of hydrogen gas.

Enthalpy of hydrogen gas at 1000 K, (h¯)H2=29,154kJ/kmol.

Enthalpy of hydrogen gas at 298 K, (h¯)H2=8,468kJ/kmol.

Entropy of hydrogen gas at 1000 K, s=166.114kJ/kmolK.

Substitute 0 for h¯H2, 29,154kJ/kmol for h¯1000, 8,468kJ/kmol for h¯298, 1000 K for T, and 166.114kJ/kmolK for s in Equation (III).

g¯H2={0+[29,154kJ/kmol8,468kJ/kmol(1,000K)(166.114kJ/kmolK)]}=145,428kJ/kmol

Refer table A-26, “Enthalpy formation table”, obtain the enthalpy of O2(h¯O2) as 0.

Refer table A-19, “Ideal gas properties of Oxygen”, obtain the following properties of Oxygen gas.

Enthalpy of oxygen gas at 1000 K, (h¯)O2=31,389kJ/kmol.

Enthalpy of oxygen gas at 298 K, (h¯)O2=8,682kJ/kmol.

Entropy of oxygen gas at 1000 K, s=243.471kJ/kmolK.

Substitute 0 for h¯O2, 31,389kJ/kmol for h¯1000, 8,682kJ/kmol for h¯298, 1000 K for T, and 243.471kJ/kmolK for s in Equation (IV).

g¯O2={0+[31,389kJ/kmol8,682kJ/kmol(1,000K)(243.471kJ/kmolK)]}=220,764kJ/kmol

Refer table A-26, “Enthalpy formation table”, obtain the enthalpy formation of water vapor (h¯H2O) as 241,820kJ/kmol.

Refer table A-23, “Ideal gas properties of water vapor”, obtain the following properties of water vapor.

Enthalpy of water vapor at 1000 K, (h¯)H2O=35,882kJ/kmol.

Enthalpy of water vapor at 298 K, (h¯)H2O=9,904kJ/kmol.

Entropy of water vapor at 1000 K, s=232.597kJ/kmolK.

Substitute 241,820kJ/kmol for h¯H2O, 35,882kJ/kmol for h¯1000, 9,904kJ/kmol for h¯298, 1000 K for T, and 232.597kJ/kmolK for s in Equation (V).

g¯H2O={241,820kJ/kmol+[35,882kJ/kmol9,904kJ/kmol(1,000K)(232.597kJ/kmolK)]}=448,439kJ/kmol

Substitute 1 for vH2, 0.5 for vO2, 1 for vH2O, 448,439kJ/kmol for g¯H2O, 145,428kJ/kmol for g¯H2, and 220,764kJ/kmol for g¯O2 in Equation (II).

ΔG(T)=vH2g¯H2(T)+vO2g¯O2(T)vH2Og¯H2O(T)={(1)(145,428kJ/kmol)+(0.5)(220,764kJ/kmol)(1)(448,439kJ/kmol)}=192,629kJ/kmol

Substitute 192,629kJ/kmol for ΔG(T), 8.314kJ/kmol for Ru, and 1000 K for T in Equation (I).

lnkp=ΔG(T)/RuT=(192,629kJ/kmol)(8.314kJ/kmol)(1,000K)=23.1692

kp=e23.1692=8.66×1011

Thus, the equilibrium constant for the reported reaction at 1000 K is 8.66×1011.

Refer table A-28, “Natural logarithm of equilibrium constants”, obtain the value of lnkp for water as 23.163 at 1000 K.

(b)

To determine

The equilibrium constant for the reported reaction at 2000 K.

(b)

Expert Solution
Check Mark

Answer to Problem 15P

The equilibrium constant for the reported reaction at 2000 K is 2.88×104.

Explanation of Solution

Write the Gibbs function of H2(g¯H2) at 2000 K.

g¯H2=h¯H2+[h¯2000h¯298Ts]H2 (VI)

Here, enthalpy of H2 gas at 2000 K is h¯2000.

Write the Gibbs function of O2(g¯O2) at 2000 K.

g¯O2=h¯O2+[h¯2000h¯298Ts]O2 (VII)

Here, enthalpy of O2 gas at 2000 K is h¯2000.

Write the Gibbs function of water (g¯H2O) at 2000 K.

g¯H2O=h¯H2O+[h¯2000h¯298Ts]H2O (VIII)

Here, enthalpy of water vapor at 2000 K is h¯2000.

Conclusion:

Refer table A-26, “Enthalpy formation table”, obtain the enthalpy of H2(h¯H2) as 0.

Refer table A-22, “Ideal gas properties of Hydrogen”, obtain the following properties of hydrogen gas.

Enthalpy of hydrogen gas at 2000 K, (h¯)H2=61,400kJ/kmol.

Entropy of hydrogen gas at 2000 K, s=188.297kJ/kmolK.

Substitute 0 for h¯H2, 61,400kJ/kmol for h¯2000, 8,468kJ/kmol for h¯298, 2000 K for T, and 188.297kJ/kmolK for s in Equation (VI).

g¯H2={0+[61,400kJ/kmol8,468kJ/kmol(2000K)(188.297kJ/kmolK)]}=323,662kJ/kmol

Refer table A-26,“Enthalpy formation table”, obtain the enthalpy of O2(h¯O2) as 0.

Refer table A-19, “Ideal gas properties of Oxygen”, obtain the following properties of Oxygen gas.

Enthalpy of oxygen gas at 2000 K, (h¯)O2=67,881kJ/kmol.

Entropy of oxygen gas at 2000 K, s=268.655kJ/kmolK.

Substitute 0 for h¯O2, 67,881kJ/kmol for h¯2000, 8,682kJ/kmol for h¯298, 2000 K for T, and 268.655kJ/kmolK for s in Equation (VII).

g¯O2={0+[67,881kJ/kmol8,682kJ/kmol(2000K)(268.655kJ/kmolK)]}=478,111kJ/kmol

Refer table A-26,“Enthalpy formation table”, obtain the enthalpy formation of water vapor (h¯H2O) as 241,820kJ/kmol.

Refer table A-23, “Ideal gas properties of water vapor”, obtain the following properties of water vapor.

Enthalpy of water vapor at 2000 K, (h¯)H2O=82,593kJ/kmol.

Entropy of water vapor at 2000 K, s=264.571kJ/kmolK.

Substitute 241,820kJ/kmol for h¯H2O, 82,593kJ/kmol for h¯2000, 9,904kJ/kmol for h¯298, 2000 K for T, and 264.571kJ/kmolK for s in Equation (VIII).

g¯H2O={241,820kJ/kmol+[82,593kJ/kmol9,904kJ/kmol(2000K)(264.571kJ/kmolK)]}=698,273kJ/kmol

Substitute 1 for vH2, 0.5 for vO2, 1 for vH2O, 698,273kJ/kmol for g¯H2O, 323,662kJ/kmol for g¯H2, and 478,111kJ/kmol for g¯O2 in Equation (II).

ΔG(T)={(1)(323,662kJ/kmol)+(0.5)(478,111kJ/kmol)(1)(698,273kJ/kmol)}=135,555.5kJ/kmol

Substitute 135,555.5kJ/kmol for ΔG(T), 8.314kJ/kmol for Ru, and 2000 K for T in Equation (I).

lnkp=ΔG(T)/RuT=(135,555.5kJ/kmol)(8.314kJ/kmol)(2000K)=8.1522

kp=e8.1522=2.88×104

Thus, the equilibrium constant for the reported reaction at 2000 K is 2.88×104.

Refer table A-28, “Natural logarithm of equilibrium constants”, obtain the value of lnkp for water as 8.5 at 2000 K.

Thus, both Gibbs function data and equilibrium constants table provide the same data.

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Chapter 16 Solutions

Thermodynamics: An Engineering Approach

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