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, Δ H < 0. Also, the ions become less dispersed as KOH(s) dissolves, thus delta S > 0. Therefore, Δ 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 Δ H < 0. Also, the ions become widely dispersed as KOH(s) dissolved, thus Δ S > 0. Therefore, Δ G < 0.

C. The average kinetic energy of the particles increases, resulting in Δ H > 0. Also, the ions become more widely dispersed as KOH(s) dissolves, thus Δ S > 0. Therefore, Δ G > 0.

D. The average kinetic energy of the particles increases, resulting in Δ H > 0. Also, the ions become more widely dispersed as KOH(s) dissolves, thus Δ S < 0. Therefore, Δ G > 0.

Answer 1

The dissolution of KOH in water releases heat, resulting in a negative enthalpy change (ΔH < 0) and an increase in entropy (ΔS > 0) as the ions become more dispersed, leading to a negative Gibbs free energy (ΔG < 0) and signifying a thermodynamically favorable process.

option c is the correct

Step-by-step explanation:

The dissolution of KOH in water is a thermodynamically favorable process as it involves the release of heat and an increase in entropy.

When 5.03 g of KOH is dissolved in water, the temperature rises from 23.0°C to 34.7°C, indicating an exothermic process where heat is released into the surroundings, making ΔH < 0. Additionally, the process involves breaking the ionic bonds in the solid KOH and forming new ion-dipole interactions with water, a step that requires less energy than the exothermic heat released upon hydration.

As the solid KOH dissolves, the individual ions become more widely dispersed, leading to an increase in entropy (ΔS > 0).

The correct claim justifying why the dissolution of KOH is thermodynamically favorable is that the energy required to break the bonds between the ions in the solid is less than that released during the formation of ion-dipole attractions, and because the ions become widely dispersed upon dissolution, which increases the entropy, thus ΔG < 0. This aligns with the principle that for a process to be spontaneous, it must result in a decrease in Gibbs free energy (ΔG), which is the case when ΔH is negative and ΔS is positive.