THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<
9th Edition
ISBN: 9781266657610
Author: CENGEL
Publisher: MCG CUSTOM
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 16.6, Problem 1P
Why is the criterion for chemical equilibrium expressed in terms of the Gibbs function instead of entropy?
Expert Solution & Answer
To determine
Reason behind the criterion for chemical equilibrium expressed in terms of the Gibbs function instead of entropy.
Answer to Problem 1P
The equilibrium criteria can be expressed in terms of properties alone when the Gibbs function is used.
Explanation of Solution
Conclusion:
The reacting system depends upon the heat transfer, change in entropy principle relation requires a detail knowledge about the heat transfer between the system and its surrounding, which is not practical.
Thus, the equilibrium criteria can be expressed in terms of properties alone when the Gibbs function is used.
Want to see more full solutions like this?
Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
6. (a) Write down an expression for the volume difference dV between
two equilibrium states in which the temperature and pressure differ by
dT and dp respectively, T and P being independent.
(b) Write down a similar expression for the pressure difference dP
between two equilibrium states in which I and V are independent
(c) Proceed to show that the partial derivatives are related by:
-1
(3), (FF), (7), -
What is the differential of the Gibbs function (G = H – TS) at constant temperature and pressure ?
(a) In reaching equilibrium, how much heat transfer occurs from 1.00 kg of water at 40.0º C when it is placed in contact with 1.00 kg of 20.0º C water in reaching equilibrium? (b) What is the change in entropy due to this heat transfer? (c) How much work is made unavailable, taking the lowest temperature to be 20.0º C ? Explicitly show how you follow the steps in the Problem-Solving Strategies for Entropy.
Chapter 16 Solutions
THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<
Ch. 16.6 - Why is the criterion for chemical equilibrium...Ch. 16.6 - Write three different KPrelations for reacting...Ch. 16.6 - Is a wooden table in chemical equilibrium with the...Ch. 16.6 - A reaction chamber contains a mixture of CO2, CO,...Ch. 16.6 - A reaction chamber contains a mixture of N2and N...Ch. 16.6 - A reaction chamber contains a mixture of CO2, CO,...Ch. 16.6 - Which element is more likely to dissociate into...Ch. 16.6 - Prob. 8PCh. 16.6 - Prob. 9PCh. 16.6 - Prob. 10P
Ch. 16.6 - Prob. 11PCh. 16.6 - Prob. 12PCh. 16.6 - Prob. 13PCh. 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 - Prob. 22PCh. 16.6 - Prob. 23PCh. 16.6 - Determine the equilibrium constant KP for the...Ch. 16.6 - Prob. 26PCh. 16.6 - Prob. 27PCh. 16.6 - Carbon monoxide is burned with 100 percent excess...Ch. 16.6 - Prob. 30PCh. 16.6 - Prob. 31PCh. 16.6 - Estimate KP for the following equilibrium reaction...Ch. 16.6 - Prob. 33PCh. 16.6 - A mixture of 3 mol of N2, 1 mol of O2, and 0.1 mol...Ch. 16.6 - Prob. 35PCh. 16.6 - Prob. 36PCh. 16.6 - Prob. 37PCh. 16.6 - Prob. 38PCh. 16.6 - Prob. 40PCh. 16.6 - What is the equilibrium criterion for systems that...Ch. 16.6 - Prob. 43PCh. 16.6 - Prob. 44PCh. 16.6 - Prob. 45PCh. 16.6 - Prob. 47PCh. 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. 58PCh. 16.6 - Prob. 59PCh. 16.6 - Prob. 60PCh. 16.6 - Prob. 61PCh. 16.6 - Using the Henrys constant data for a gas dissolved...Ch. 16.6 - Prob. 63PCh. 16.6 - Prob. 64PCh. 16.6 - Prob. 65PCh. 16.6 - Prob. 66PCh. 16.6 - A liquid-vapor mixture of refrigerant-134a is at...Ch. 16.6 - Prob. 68PCh. 16.6 - Prob. 69PCh. 16.6 - An oxygennitrogen mixture consists of 30 kg of...Ch. 16.6 - Prob. 71PCh. 16.6 - Prob. 72PCh. 16.6 - Prob. 73PCh. 16.6 - Prob. 74PCh. 16.6 - Prob. 75PCh. 16.6 - Prob. 76PCh. 16.6 - An ammoniawater absorption refrigeration unit...Ch. 16.6 - Prob. 78PCh. 16.6 - Prob. 79PCh. 16.6 - Prob. 80PCh. 16.6 - One lbmol of refrigerant-134a is mixed with 1...Ch. 16.6 - Prob. 82RPCh. 16.6 - Prob. 83RPCh. 16.6 - Prob. 84RPCh. 16.6 - Prob. 85RPCh. 16.6 - Prob. 88RPCh. 16.6 - Prob. 89RPCh. 16.6 - Prob. 90RPCh. 16.6 - Prob. 91RPCh. 16.6 - Prob. 92RPCh. 16.6 - A constant-volume tank contains a mixture of 1 mol...Ch. 16.6 - Prob. 94RPCh. 16.6 - Prob. 95RPCh. 16.6 - Prob. 96RPCh. 16.6 - Prob. 97RPCh. 16.6 - Prob. 99RPCh. 16.6 - Consider a glass of water in a room at 25C and 100...Ch. 16.6 - Prob. 101RPCh. 16.6 - Prob. 102RPCh. 16.6 - Prob. 105RPCh. 16.6 - Prob. 106RPCh. 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...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- Calculate the entropy of refrigerant 134a (R134a) at 7.0282 bar and 117 K. At 7.0282 bar, the saturation temperature is 300 K. At vapor/liquid equilibrium at 7.0282 bar the entropy of the saturated liquid is 1.1287 J/(g K) and the entropy of the saturated vapor is 1.7156 J/(g K). CP of the liquid is 1.4324 J/(g K) and CP of the vapor is 1.0438 J/(g K). Report your answer in units of J/(g K) using three decimal places.arrow_forwardOxygen gas initially at 355 K and 1.50 bar undergoes the following cyclic processes in a closed system:Step 1: Adiabatic compression to 7.50 barStep 2: Addition of 13 240.0 J/mol of heat to raise the temperature at constant pressureStep 3: Adiabatic expansion to 1.50 barStep 4: Isobaric cooling back to original state.Determine ΔU, ΔH, Q and W for each step, and for the overall reversible process. Express all answers in kJ/mol and up to decimal placesarrow_forwardOne mole of a gas is expanded isothermally at 300 Kelvins from 5.00 MegaPascal to 0.10 MegaPascal. If this gas obeys the equation of state given below and from this B is equal to 2.0 x10^4 L/(mol-K^2), answer the question that follow. Calculate the change in molar Gibbs free energy (in KJ/mole) for this process. Express answer in THREE SIGNIFICANT FIGURES.arrow_forward
- A gold block initially at 62.9°C is put into 120 grams of H20 at 25.8°C in a container. The mixture at thermal equilibrium reached a temperature of 28.3°C. What is the mass of gold used?arrow_forwardThe change in the molar volume accompanying fusion of solid benzene is 0.5 cm3 mol−1. Determine the change in Gibbs energy of fusion when the pressure is increased from 1 bar to 5000 bar.arrow_forwarddefine the standard-state Gibbs function change?arrow_forward
- Apply the Gibbs phase rule to determine the number of independent variables associated with a multicomponent, multiphase system.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_forwardHello Expert, Please assist with this Thermodynamics Question. Kindly answer correctly and show ALL the steps. Thanks in advancearrow_forward
- 3) An ideal gas consisting of one mole of molecules of type A is in contact with the surroundings at T=300 K, and under a constant pressure of 1 atm. The gas undergoes a spontaneous isomeric chemical reaction, wherein some fraction x of molecules of type A change shape, become isomers of type B. This results in an equilibrium mixture of 1-x moles of A and x moles of B. The enthalpy of B is lower than the enthalpy of A by 3 kJ/mole, such that AH = -xɛ; & = 3kJ/mole . The change in entropy is given by AS ==R(xln x+ (1–x)ln(1– x)). Use the computer to graph the change in Gibbs free energy as a function of x. Determine, by inspection of your graph, or otherwise, the concentration of A and the concentration of B when equilibrium is obtained. What is the maximum work that could be extracted from this process (aside from PV work)? P= latm (consternt) T: 300karrow_forward6. A 100kg of commercial tungsten (C=138J/kg-K) at a temperature of 180°C and 45kg of diamond (C=520J/kg-K) at a temperature of 135°C_is placed inside an insulated container containing ice at freezing temperature. If heat is restricted to escape, determine the mass of the ice if the temperature of the mixture when equilibrium is attained is about 73.53•C. Draw a figure and explain each step solutionarrow_forwardYou are testing some novel fuel mixtures for potential use in automobiles. Your testing apparatus is a 2.54 cm inner diameter cylinder fitted with a piston, designed to mimic a cylinder found in a car engine. A fuel-air mixture is ignited while a force of F = 273.0 N is applied to the piston. There are 0.200 mol of gas after combustion at 290.0 °C, and the atmospheric pressure is 1.01325 x 105 Pa. Initial Final conditions conditions What is the pressure of gas inside the cylinder after combustion while a force of Fj = 273.0 N is applied to the cylinder? Assume combustion products behave ideally. 6.4 x105 Ра What is the volume of gas in the cylinder after combustion with a force of 273.0 N applied to the cylinder? 1.463 x10-3 V = m3 The force applied to the piston is then reduced to Ff = 111.0 N, and the gas responds in a two-step process. The first, quick step is adiabatic. During the second, slower step, the gas returns the change in kinetic and potential energies in both steps is…arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University Press
Mechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSON
Thermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEY
Mechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage Learning
Engineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
First Law of Thermodynamics, Basic Introduction - Internal Energy, Heat and Work - Chemistry; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=NyOYW07-L5g;License: Standard youtube license