EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 8220102809444
Author: CENGEL
Publisher: YUZU
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Chapter 16.6, Problem 20P

( a)

To determine

The equilibrium constant for the reported reaction at 537 R.

( a)

Expert Solution
Check Mark

Answer to Problem 20P

The equilibrium constant for the reported reaction at 537 R is 1.12×1040.

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.

Conclusion:

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

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

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

Substitute 1 for vH2, 0.5 for vO2, 1 for vH2O, 0 for g¯H2(T), 0 for g¯O2(T), and 228,590kJ/molg¯H2O(T) in Equation (II).

ΔG(T)=(1)(228,590kJ/mol)(1)(0)(0.5)(0)=228,590kJ/mol1Btu/lbmol2.236kJ/mol=98,350Btu/lbmol

Substitute 98,350Btu/lbmol for ΔG(T), 1.986Btu/lbmol for Ru, and 537 R for T in Equation (I).

lnkp=(98,350Btu/lbmol)(1.986Btu/lbmol)(537R)=92.22

kp=e92.22=1.12×1040

Thus, the equilibrium constant for the reported reaction at 537 R is 1.12×1040.

Refer table A-28, “Natural logarithm of equilibrium constants”, obtain the value of lnkp for water as 92.22 at 537R.

(b)

To determine

The equilibrium constant for the reported reaction at 3240 R.

(b)

Expert Solution
Check Mark

Answer to Problem 20P

The equilibrium constant for the reported reaction at 3240 R is 1.9×104.

Explanation of Solution

Write the Gibbs function of water (g¯H2O) at 3240 R.

g¯H2O=h¯H2O+[h¯4320h¯537Ts]H2O (IV)

Here, enthalpy of water at 537 R is h¯H2O, enthalpy of water vapor at 4,320 R is h¯4320, enthalpy of water vapor at 537 R is h¯537, and entropy is s.

Write the Gibbs function of H2(g¯H2) at 3,240 R.

g¯H2=h¯H2+[h¯4320h¯298Ts]H2 (V)

Here, enthalpy of H2 at 537 R is h¯H2, enthalpy of H2 gas at 4,320 R is h¯4320, and enthalpy of H2 gas at 298 R is h¯298.

Write the Gibbs function of O2(g¯O2) at 3,240 R.

g¯O2=h¯O2+[h¯4320h¯298Ts]O2 (VI)

Here, enthalpy of O2 at 537 R is h¯O2, enthalpy of O2 gas at 4,320 R is h¯4320, enthalpy of O2 gas at 298 R is h¯298.

Conclusion:

Refer table A-26E, “Enthalpy formation table”,obtain the enthalpy formation of water vapor (h¯H2O) as 104,040Btu/lbmol.

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

Enthalpy of water vapor at 4,320 R as 31,205Btu/lbmol.

Enthalpyof water vapor at 537 R as 4,258Btu/lbmol.

Entropy of water vapor at 3,240 R as 61.9Btu/lbmolK.

Substitute 104,040Btu/lbmol for h¯H2O, 31,205Btu/lbmol for h¯4320, 4,258Btu/lbmol for h¯537, 3,240 K for T, and 61.9Btu/lbmolK for s in Equation (IV).

g¯H2O={104,040Btu/lbmol+[31,205Btu/lbmol4,258Btu/lbmol(3,240K)(61.9Btu/lbmolK)]}=277,805Btu/lbmol

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

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

Enthalpy of hydrogen gas at 4320 R as 23,485Btu/lbmol.

Enthalpy of hydrogen gas at 298 R as 3,640Btu/lbmol.

Entropy of hydrogen gas at 3,240 R as 44.125Btu/lbmolK.

Substitute 0 for h¯H2, 23,485Btu/lbmol for h¯4320, 3,640Btu/lbmol for h¯298, 3,240 K for T, and 44.125Btu/lbmolK for s in Equation (V).

g¯H2={0+[23,485Btu/lbmol3,640Btu/lbmol(3,240K)(44.125Btu/lbmolK)]}=123,120Btu/lbmol

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

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

Enthalpy of Oxygengas at 4,320 R as 25,972Btu/lbmol.

Enthalpy of Oxygen gas at 298 R as 3,725Btu/lbmol.

Entropy of Oxygen gas at 3240 R as 63.22Btu/lbmolK.

Substitute 0 for h¯O2, 25,972Btu/lbmol for h¯4320, 3,725Btu/lbmol for h¯298, 3,240 K for T, and 63.22Btu/lbmolK for s in Equation (VI).

g¯O2={0+[25,972Btu/lbmol3,725Btu/lbmol(3,240K)(63.22Btu/lbmolK)]}=182,585Btu/lbmol

Substitute 1 for vH2, 0.5 for vO2, 1 for vH2O, 277,805Btu/lbmol for g¯H2O, 123,120Btu/lbmol for g¯H2, and 182,585Btu/lbmol for g¯O2 in Equation (II).

ΔG(T)={(1)(277,805Btu/lbmolK)(1)(123,120Btu/lbmolK)(0.5)(182,585Btu/lbmolK)}=63,392Btu/lbmol

Substitute 63,392Btu/lbmol for ΔG(T), 1.986Btu/lbmol for Ru, and 3,240 R for T in Equation (I).

lnkp=(63,392Btu/lbmol)(1.986Btu/lbmol)(3,240R)=9.85

kp=e9.85=1.9×104

Thus, the equilibrium constant for the reported reaction at 3,240 R is 1.9×104.

Refer table A-28, “Natural logarithm of equilibrium constants”, obtain the value of lnkp for water as 9.826 at 3,240R.

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

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Chapter 16 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

Ch. 16.6 - Prob. 11PCh. 16.6 - 16–12 Determine the temperature at which 5 percent...Ch. 16.6 - 16–12 Determine the temperature at which 5 percent...Ch. 16.6 - Prob. 14PCh. 16.6 - Prob. 15PCh. 16.6 - Prob. 16PCh. 16.6 - Prob. 17PCh. 16.6 - Prob. 18PCh. 16.6 - Prob. 19PCh. 16.6 - Prob. 20PCh. 16.6 - Prob. 21PCh. 16.6 - Determine the equilibrium constant KP for the...Ch. 16.6 - Prob. 24PCh. 16.6 - Carbon monoxide is burned with 100 percent excess...Ch. 16.6 - Prob. 27PCh. 16.6 - Prob. 28PCh. 16.6 - Prob. 29PCh. 16.6 - Prob. 30PCh. 16.6 - Prob. 31PCh. 16.6 - A mixture of 3 mol of N2, 1 mol of O2, and 0.1 mol...Ch. 16.6 - Prob. 33PCh. 16.6 - Prob. 34PCh. 16.6 - Prob. 35PCh. 16.6 - Prob. 37PCh. 16.6 - Estimate KP for the following equilibrium reaction...Ch. 16.6 - Prob. 40PCh. 16.6 - What is the equilibrium criterion for systems that...Ch. 16.6 - Prob. 42PCh. 16.6 - Prob. 43PCh. 16.6 - Prob. 44PCh. 16.6 - Prob. 48PCh. 16.6 - Prob. 51PCh. 16.6 - Prob. 52PCh. 16.6 - Prob. 53PCh. 16.6 - Prob. 54PCh. 16.6 - Prob. 55PCh. 16.6 - Prob. 56PCh. 16.6 - Prob. 57PCh. 16.6 - Prob. 59PCh. 16.6 - Prob. 60PCh. 16.6 - Prob. 61PCh. 16.6 - Prob. 62PCh. 16.6 - Using the Henrys constant data for a gas dissolved...Ch. 16.6 - Prob. 65PCh. 16.6 - Prob. 66PCh. 16.6 - Prob. 67PCh. 16.6 - Prob. 68PCh. 16.6 - Prob. 69PCh. 16.6 - Prob. 70PCh. 16.6 - Prob. 71PCh. 16.6 - Prob. 72PCh. 16.6 - An oxygennitrogen mixture consists of 30 kg of...Ch. 16.6 - Prob. 74PCh. 16.6 - Prob. 75PCh. 16.6 - Prob. 76PCh. 16.6 - Prob. 77PCh. 16.6 - An ammoniawater absorption refrigeration unit...Ch. 16.6 - Prob. 79PCh. 16.6 - Prob. 81PCh. 16.6 - Prob. 82PCh. 16.6 - Prob. 83RPCh. 16.6 - Prob. 84RPCh. 16.6 - Prob. 85RPCh. 16.6 - Consider a glass of water in a room at 25C and 100...Ch. 16.6 - Prob. 87RPCh. 16.6 - 16–90 Propane gas is burned steadily at 1 atm...Ch. 16.6 - Prob. 91RPCh. 16.6 - Prob. 92RPCh. 16.6 - Prob. 93RPCh. 16.6 - Prob. 94RPCh. 16.6 - Prob. 95RPCh. 16.6 - A constant-volume tank contains a mixture of 1 mol...Ch. 16.6 - Prob. 101RPCh. 16.6 - Prob. 103RPCh. 16.6 - Prob. 104RPCh. 16.6 - Prob. 107RPCh. 16.6 - Prob. 108RPCh. 16.6 - Prob. 109FEPCh. 16.6 - Prob. 110FEPCh. 16.6 - Prob. 111FEPCh. 16.6 - Prob. 112FEPCh. 16.6 - Prob. 113FEPCh. 16.6 - Prob. 114FEPCh. 16.6 - Propane C3H8 is burned with air, and the...Ch. 16.6 - Prob. 116FEPCh. 16.6 - Prob. 117FEPCh. 16.6 - The solubility of nitrogen gas in rubber at 25C is...
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