(a)
The logarithm equilibrium constant for the reaction at
Compare the results for the values of
(a)
Answer to Problem 19P
The logarithm equilibrium constant for the reaction at
The equilibrium constant obtained from the equilibrium constants of Table A-28 at 1440Ris
Explanation of Solution
Express the standard-state Gibbs function change.
Here, the Gibbs function of components
Write the equation to calculate the natural logarithms of equilibrium constant for the chemical equilibrium of ideal-gas mixtures.
Here, universal gas constant is
Write the equation to calculate the equilibrium constant for the chemical equilibrium of ideal-gas mixtures.
Conclusion:
From the equilibrium reaction, the values of
Refer Table A-26, obtain the values of
Substitute 1 for
Substitute
Substitute
Thus, the equilibrium constant obtained from the equilibrium reaction at 298 K is
Refer Table A-28, “Natural logarithms of the equilibrium constant” obtain the equilibrium constant for the reaction at the temperature of 298 K as
Substitute
Thus, the equilibrium constant obtained from the table A-28 at 1440 R is
The value obtained for equilibrium constant at 298 K from the definition of the equilibrium constant is
(b)
The logarithm equilibrium constant for the reaction at 2000 K.
Compare the results for the values of
(b)
Answer to Problem 19P
The logarithm equilibrium constant for the reaction at 2000 K is
The equilibrium constant obtained from the equilibrium constants of Table A-28 at 2000K is.
Explanation of Solution
Express the standard-state Gibbs function change.
Here, the Gibbs function of components
Write the equation to calculate the natural logarithms of equilibrium constant for the chemical equilibrium of ideal-gas mixtures.
Here, universal gas constant is
Write the equation to calculate the equilibrium constant for the chemical equilibrium of ideal-gas mixtures.
Conclusion:
From the equilibrium reaction, the values of
Refer Table A-26, obtain the values of
Refer to Table A-22, obtain the value of
Refer to Table A-22, obtain the value of
Refer to Table A-19E, obtain the value of
Refer to Table A-19E, obtain the value of
Refer to Table A-23, obtain the value of
Refer to Table A-23, obtain the value of
Substitute 1 for
Substitute
Substitute
Thus, the equilibrium constant obtained from the equilibrium reaction at 2000K is
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
Substitute
Thus, the equilibrium constant obtained from the table A-28 at2000K is
The value obtained for equilibrium constant at 2000K from the definition of the equilibrium constant is
Want to see more full solutions like this?
Chapter 16 Solutions
Thermodynamics: An Engineering Approach
- Relate the chemical equilibrium constant to the enthalpy of reaction.arrow_forwardShow that a mixture of saturated liquid water and saturated water vapor at 300 kPa satisfies the criterion for phase equilibrium.arrow_forwardThe standard enthalpy of formation of solid barium oxide, BaO, is –553.5 kJ/mol, andthe standard enthalpy of formation of barium peroxide, BaO2, is –634.3 kJ/mol.(a) Calculate the standard enthalpy change for the following reaction. Is the reactionexothermic or endothermic?2 BaO2(s) → 2 BaO(s) + O2(g)(b) Draw an energy level diagram that shows the relationship between the enthalpychange of the decomposition of BaO2, to BaO and O2, and the enthalpies of formationof BaO2(s) and BaO2(s)arrow_forward
- Derive the expression for equilibrium constant for ideal - gas mixtures?arrow_forwardAt 25°C a solution consists of 0.450 mole of pentane, C5H12, and 0.250 mole of cyclopentane, C5H10. What is the mole fraction of cyclopentane in the vapor that is in equilibrium with this solution? The vapor pressure of the pure liquids at 25°C are 451 torr for pentane and 321 torr for cyclopentane. Assume that the solution is an ideal solution. (a) 0.284 (b) 0.551 (c) 0.716 (d) 0.643 (e) 0.357arrow_forwardThe heat of combustion of H2(g) to form H2O(l) under constant-pressure conditions is -285.83 kJ/mol at 25°C. If the water is formed at 1 bar and has a density of 1000 kg/m³, calculate the change in internal energy for this reactionA. -187 kJ/molB. -210 kJ/molC. -282 kJ/molD. -310 kJ/molarrow_forward
- Consider an unknown substance where the solid and liquid phases are at equilibriumwhen T = −50◦C, P = 20 atm and when T = −15◦C, P = 10 atm. (a) Calculate the slope of thesolid-liquid coexistence curve. (b) Determine whether or not the unknown substance expands orcontracts upon freezing. Give explicit equations to support your answer.arrow_forwardThirty pounds of ice at 32OF is placed in 100 lbs of water at 100 O The latent heat of ice may be taken as 144 BTU per lb. If no heat is lost or added to the mixture, the temperature (in Farhenheit) when equilibrium is reached is: a. None of these b. 51 c. 48 d. 49 e. 50arrow_forwardA newly purchased container that has a capacity of 1 m³ contains a mixture of liquid and steam in equilibrium at a temperature of 601 K. The mass of the liquid is found to be 15 kg. Determine the quality of the mixture in %. Use steam tables of Keena et al.arrow_forward
- For the reaction H2(g) + Br2(g) 2HBr(g), Kc = 81.4 at 385ºC. If [H2] = [Br2] = [HBr] = 1.6 × 10–2 M at 385ºC, which one of the following is correct? Group of answer choices The reaction is at equilibrium already. The reaction will proceed in the forward direction because K > 1. The reaction will proceed in the reverse direction to reach equilibrium because Q < K. The reaction will proceed in the forward direction to reach equilibrium.arrow_forwardA well-insulated rigid vessel, divided by a partition, has 0.2 kmol of carbon dioxide at 30 deg. C and 2 bar in one section and 3 kmol of hydrogen gas at 40 deg. C and 4 bar in the other section. The partition is removed and the gases mix. The specific heat capacity, pc of hydrogen and carbon dioxide are14.35 Jkg-1K-1 and 0.84 Jkg-1K-1 respectively. Determine the final equilibrium pressure of the gas mixture.arrow_forwardAn insulated rigid tank is divided into two compartments by a partition. One compartment contains 8 kg of oxygen gas at 42 ∘C and 100 kPa, and the other compartment contains 4 kg of nitrogen gas at 20 ∘C and 180 kPa. The partition is then removed and the two gases are allowed to mix. Part A Determine the mixture temperature after equilibrium has been reached. Express your answer to four significant figures. Part B Determine the mixture pressure after equilibrium has been reached. Express your answer to five significant figures and include the appropriate units.arrow_forward
- Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University PressMechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSONThermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
- Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEYMechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage LearningEngineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY