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.
Expert Answer
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Answer 1

The dissolution of KOH in water is a favorable process as it releases heat (delta H < 0) and increases in disorder (delta S > 0), resulting in a spontaneous reaction (delta G < 0), as reflected in choice B.

Step-by-step explanation:

The dissolution of KOH in water is a thermodynamically favorable process because it is exothermic, indicated by the increase in temperature after KOH dissolves. The heat released (delta H) is negative, and since the ions become more widely dispersed upon dissolving, the entropy (delta S) increases, leading to a negative Gibbs free energy change (delta G), suggesting a spontaneous process.

The correct answer is B: 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 delta H < 0. Also, the ions become widely dispersed as KOH(s) dissolves, thus delta S > 0. Therefore, delta G < 0.