A 5.00 g-sample of KOH at 25.0 C was added to 100.0 g of H2O(l) at room temperature inside an insulated cup calorimeter, and the contents were stirred. After all the KOH(s) dissolved, the temperature of the solution had increased. Based on the information given, which of the following best justifies the claim that the dissolution of KOH is a thermodynamically favorable process? A. The forces between the ions and the water molecules are stronger than the forces between water molecules, thus, delta H < 0. Also, the ions become less dispersed as KOH(s) dissolves, thus delta S > 0. Therefore, delta G < 0. B. The energy require to break the bonds between the ions in the solid is less than that released as the ion-dipole attractions form during solvation, thus delta H < 0. Also, the ions become widely dispersed as KOH(s) dissolved, thus delta S > 0. Therefore, delta G < 0. C. The average kinetic energy of the particles increases, resulting in delta H > 0. Also, the ions become more widely dispersed as KOH(s) dissolves, thus delta S > 0. Therefore, delta G > 0. D. The average kinetic energy of the particles increases, resulting in delta H > 0. Also, the ions become more widely dispersed as KOH(s) dissolves, thus delta S < 0. Therefore, delta G > 0.
A 5.00 g-sample of KOH at 25.0 C was added to 100.0 g of H2O(l) at room temperature inside an insulated cup calorimeter, and the contents were stirred. After all the KOH(s) dissolved, the temperature of the solution had increased. Based on the information given, which of the following best justifies the claim that the dissolution of KOH is a thermodynamically favorable process? A. The forces between the ions and the water molecules are stronger than the forces between water molecules, thus, delta H < 0. Also, the ions become less dispersed as KOH(s) dissolves, thus delta S > 0. Therefore, delta G < 0. B. The energy require to break the bonds between the ions in the solid is less than that released as the ion-dipole attractions form during solvation, thus delta H < 0. Also, the ions become widely dispersed as KOH(s) dissolved, thus delta S > 0. Therefore, delta G < 0. C. The average kinetic energy of the particles increases, resulting in delta H > 0. Also, the ions become more widely dispersed as KOH(s) dissolves, thus delta S > 0. Therefore, delta G > 0. D. The average kinetic energy of the particles increases, resulting in delta H > 0. Also, the ions become more widely dispersed as KOH(s) dissolves, thus delta S < 0. Therefore, delta G > 0.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
Question
A 5.00 g-sample of KOH at 25.0 C was added to 100.0 g of H2O(l) at room temperature inside an insulated cup calorimeter, and the contents were stirred. After all the KOH(s) dissolved, the temperature of the solution had increased. Based on the information given, which of the following best justifies the claim that the dissolution of KOH is a thermodynamically favorable process?
A. The forces between the ions and the water molecules are stronger than the forces between water molecules, thus, delta H < 0. Also, the ions become less dispersed as KOH(s) dissolves, thus delta S > 0. Therefore, delta G < 0.
B. The energy require to break the bonds between the ions in the solid is less than that released as the ion-dipole attractions form during solvation, thus delta H < 0. Also, the ions become widely dispersed as KOH(s) dissolved, thus delta S > 0. Therefore, delta G < 0.
C. The average kinetic energy of the particles increases, resulting in delta H > 0. Also, the ions become more widely dispersed as KOH(s) dissolves, thus delta S > 0. Therefore, delta G > 0.
D. The average kinetic energy of the particles increases, resulting in delta H > 0. Also, the ions become more widely dispersed as KOH(s) dissolves, thus delta S < 0. Therefore, delta G > 0.
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