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Question 12W A 5.00 g-sample of KOH(s) at 25.0 C was added to 100.0 g of H2O(1) temperature of the solution had increased. Based on the information given, which of the following best justifies the claim that the dissolution of KOH(s) is a thermodynamically favorable process? room temperature inside an insulated cup calorimeter, and the contents were stirred. After all the KOH(s) dissolved, the The forces between the ions and the water molecules are strọnger than the forces between water molecules, thus AH < 0. Also, the ions become less dispersed as KOH(s) dissolves, thus A AS >0. Therefore, AG < 0. The energy required to break the bonds between the ions in the solid is less than that released as the ion-dipole attractions form during solvation, thus AH < 0. Also, the ions become more B widely dispersed as KOH(s) dissolves, thus AS > 0. Therefore, AG < 0. The average kinetic energy of the particles increases, resulting in AH >0. Also, the ions become more widely dispersed as KOH(s) dissolves, thus AS > 0. Therefore, AG > 0. The average kinetic energy of the particles increases, resulting in AH >0. Also, the ions become more widely dispersed as KOH(s) dissolves, thus AS < 0. Therefore, AG > 0. D

Question 12W A 5.00 g-sample of KOH(s) at 25.0 C was added to 100.0 g of H2O(1) temperature of the solution had increased. Based on the information given, which of the following best justifies the claim that the dissolution of KOH(s) is a thermodynamically favorable process? room temperature inside an insulated cup calorimeter, and the contents were stirred. After all the KOH(s) dissolved, the The forces between the ions and the water molecules are strọnger than the forces between water molecules, thus AH < 0. Also, the ions become less dispersed as KOH(s) dissolves, thus A AS >0. Therefore, AG < 0. The energy required to break the bonds between the ions in the solid is less than that released as the ion-dipole attractions form during solvation, thus AH < 0. Also, the ions become more B widely dispersed as KOH(s) dissolves, thus AS > 0. Therefore, AG < 0. The average kinetic energy of the particles increases, resulting in AH >0. Also, the ions become more widely dispersed as KOH(s) dissolves, thus AS > 0. Therefore, AG > 0. The average kinetic energy of the particles increases, resulting in AH >0. Also, the ions become more widely dispersed as KOH(s) dissolves, thus AS < 0. Therefore, AG > 0. D

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter6: Thermochemisty
Section: Chapter Questions
Problem 6.109QP: A 21.3-mL sample of 0.977 M NaOH is mixed with 29.5 mL of 0.918 M HCl in a coffee-cup calorimeter...
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Question 12 W A 5.00 g-sample of KOH(s) at 25.0°C was added to 100.0 g of H2O(1) 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(s) is a thermodynamically favorable process? The forces between the ions and the water molecules are strọnger than the forces between water molecules, thus AH < 0. Also, the ions become less dispersed as KOH(s) dissolves, thus AS >0. Therefore, AG < 0. less than that released as the ion-dipole attractions form during solvation, thus AH < 0. Also, the ions become more The energy required to break the bonds between the ions in the solid widely dispersed as KOH(s) dissolves, thus AS > 0. Therefore, AG < 0. The average kinetic energy of the particles increases, resulting in AH > 0. Also, the ions become more widely dispersed as KOH(s) dissolves, thus AS > 0. Therefore. AG > 0. D The average kinetic energy of the particles increases, resulting in AH > 0. Also, the ions become more widely dispersed as KOH(s) dissolves, thus AS < 0. Therefore. AG > 0. esc Q Search or enter website name @ # $ % &
Transcribed Image Text:Question 12 W A 5.00 g-sample of KOH(s) at 25.0°C was added to 100.0 g of H2O(1) 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(s) is a thermodynamically favorable process? The forces between the ions and the water molecules are strọnger than the forces between water molecules, thus AH < 0. Also, the ions become less dispersed as KOH(s) dissolves, thus AS >0. Therefore, AG < 0. less than that released as the ion-dipole attractions form during solvation, thus AH < 0. Also, the ions become more The energy required to break the bonds between the ions in the solid widely dispersed as KOH(s) dissolves, thus AS > 0. Therefore, AG < 0. The average kinetic energy of the particles increases, resulting in AH > 0. Also, the ions become more widely dispersed as KOH(s) dissolves, thus AS > 0. Therefore. AG > 0. D The average kinetic energy of the particles increases, resulting in AH > 0. Also, the ions become more widely dispersed as KOH(s) dissolves, thus AS < 0. Therefore. AG > 0. esc Q Search or enter website name @ # $ % &
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