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.

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Question 11
A 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 favorable
process?
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)
A
dissolves, thus AS > 0. Therefore, AG < 0.
Subr
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
B
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,
D
AG > 0.
Transcribed Image Text:« < 11 of 30 > Question 11 A 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 favorable process? 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) A dissolves, thus AS > 0. Therefore, AG < 0. Subr 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 B 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, D AG > 0.
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