(a)
Interpretation:
The explanations for the given set of statements have to be given.
Concept introduction:
Free energy:
Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.
Relationship between
(a)
Explanation of Solution
To explain: The change in standard free energy for the given reaction
Given reaction
The relationship between
Above equation shows that, if the temperature remains constant then the free energy change will remains constant.
(b)
Interpretation:
The explanations for the given set of statements have to be given.
Concept introduction:
Free energy:
Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.
Relationship between
(b)
Explanation of Solution
To explain: The change in free energy for the given reaction
Given reaction
The relationship between
Above equation shows that, if the value of reaction quotient changes then the value of free energy also change.
(c)
Interpretation:
The explanations for the given set of statements have to be given.
Concept introduction:
Free energy:
Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.
Relationship between
(c)
Explanation of Solution
To explain: The spontaneity of the given reaction
Given reaction and information
Since, the value of equilibrium constant is very large, the given reaction is spontaneous.
(d)
Interpretation:
The explanations for the given set of statements have to be given.
Concept introduction:
Free energy:
Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.
Relationship between
(d)
Explanation of Solution
To check: The given statement
Given statement
The given statement is not correct. At equilibrium the value of free energy is zero, but the value of standard free energy change is constant. Hence, the given statement is not true.
(e)
Interpretation:
The explanations for the given set of statements have to be given.
Concept introduction:
Free energy:
Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.
Relationship between
(e)
Explanation of Solution
To check: The composition of the mixture at equilibrium
Given reaction
The value of equilibrium constant is very large, so at equilibrium the composition will mostly the product side
(f)
Interpretation:
The explanations for the given set of statements have to be given.
Concept introduction:
Free energy:
Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.
Relationship between
(f)
Explanation of Solution
To give: The values of
Given reaction
For the reverse reaction,
At equilibrium, the value of free energy change is zero. The reaction mixture will mostly reactant side
The given reverse reaction is non-spontaneous.
Want to see more full solutions like this?
Chapter 18 Solutions
General Chemistry - Standalone book (MindTap Course List)
- Explain how the entropy of the universe increases when an aluminum metal can is made from aluminum ore. Thefirst step is to extract the ore, which is primarily a formof A12O3, from the ground. After it is purified by freeingit from oxides of silicon and iron, aluminum oxide ischanged to the metal by an input of electrical energy. 2Al2O3(s)electricalenergy4Al(s)+3O2(g)arrow_forwardWhich contains greater entropy, a quantity of frozen benzene or the same quantity of liquid benzene at the same temperature? Explain in terms of the dispersal of energy in the substance.arrow_forwardThe decomposition of diamond to graphite [C(diamond) C(graphite)] is thermodynamically favored, but occurs slowly at room temperature. a. Use fG values from Appendix L to calculate rG and Keq for the reaction under standard conditions and 298.15 K. b. Use fH and S values from Appendix L to estimate rG and Keq for the reaction at 1000 K. Assume that enthalpy and entropy values are valid at these temperatures. Does heating shift the equilibrium toward the formation of diamond or graphite? c. Why is the formation of diamond favored at high pressures? d. The phase diagram shows that diamond is thermodynamically favored over graphite at 20,000 atmospheres pressure (about 2 GPa) at room temperature. Why is this conversion actually done at much higher temperatures and pressures?arrow_forward
- For the ammonia synthesis reaction ⇌ Does the entropy effect favor products? Explain your answer. Does the energy effect favor products? Explain your answer. Is the equilibrium concentration of NH3(g) greater at high or low temperature? Explain.arrow_forwardTetrachloromethane (carbon tetrachloride), CCl4, has a normal boiling point of 76.7C and an enthalpy of vaporization, Hvap, of 29.82 kJ/mol. Estimate the entropy of vaporization, Svap. Estimate the free energy of vaporization, Gvap, at 25C.arrow_forwardFor each of the following processes, identify the systemand the surroundings. Identify those processes that arespontaneous. For each spontaneous process, identify theconstraint that has been removed to enable the process to occur: Ammonium nitrate dissolves in water. Hydrogen and oxygen explode in a closed bomb. A rubber band is rapidly extended by a hangingweight. The gas in a chamber is slowly compressed by aweighted piston. A glass shatters on the floor.arrow_forward
- Consider the reaction of 2 mol H2(g) at 25C and 1 atm with 1 mol O2(g) at the same temperature and pressure to produce liquid water at these conditions. If this reaction is run in a controlled way to generate work, what is the maximum useful work that can be obtained? How much entropy is produced in this case?arrow_forwardConsider the reaction of 1 mol H2(g) at 25C and 1 atm with 1 mol Br2(l) at the same temperature and pressure to produce gaseous HBr at these conditions. If this reaction is run in a controlled way to generate work, what is the maximum useful work that can be obtained? How much entropy is produced in this case?arrow_forwardThe standard free energies of formation and the standard enthalpies of formation at 298 K for difluoroacetylene (C2F2) and hexafluorobenzene (C6F6) are Gfo(KJ/mol) Hfo(KJ/mol) C2F2(g) 191.2 241.3 Hexane 78.2 132.8 For the following reaction: C6F6(g)3C2F2(g) a. calculate S at 298 K. b. calculate K at 298 K. c. estimate K at 3000. K, assuming H and S do not depend on temperature.arrow_forward
- Cells use the hydrolysis of adenosine triphosphate, abbreviated as ATP, as a source of energy. Symbolically, this reaction can be written as ATP(aq)+H2O(l)ADP(aq)+H2PO4(aq) where ADP represents adenosine diphosphate. For this reaction, G =30.5 kJ/mol. a. Calculate K at 25C. b. If all the free energy from the metabolism of glucose C6H12O6(s)+6O2(g)6CO2(g)+6H2O(l) goes into forming ATP from ADP, how many ATP molecules can be produced for every molecule of glucose?arrow_forwardAccording to a source, lithium peroxide (Li2O2) decomposes to lithium oxide (Li2O) and oxygen gas at about 195C. If the standard enthalpy change for this decomposition is 33.9 kJ/mol, what would you give as an estimate for the standard entropy change for this reaction? Explain.arrow_forward
- Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningGeneral Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage Learning
- Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning