11 « < 11 of 30 >Question 11A 5.00 g-sample of KOH(s) at 25.0 C was added to 100.0 g of H,O(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 favorableprocess?The forces between the ions and the water molecules are stronger than the forces between water molecules, thus AH < 0. Also, the ions become less dispersed as KOH(s)Adissolves, thus AS > 0. Therefore, AG < 0.SubrThe 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 ionsBbecome 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.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,DAG > 0.

Given that, KOH is dissolved in water. After all the KOH is dissolved, the temperature of the…

11 Q nple of KOH(s) at 25.0 C was added to 100.0 g of H,O(1) at room temperature inside an insulated cup calorimeter, and the contents were stirred. After all the KOH(s) dissolvedi ure 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 The forces between the ions and the water molecules are stronger 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. 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 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.

11 Q nple of KOH(s) at 25.0 C was added to 100.0 g of H,O(1) at room temperature inside an insulated cup calorimeter, and the contents were stirred. After all the KOH(s) dissolvedi ure 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 The forces between the ions and the water molecules are stronger 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. 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 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.

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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