Question

An ice cube at 0 degrees Celsius melts when placed inside a room at 22 degrees Celsius. Based on the concepts of enthalpy, entropy, and Gibbs free energy, which of the following best explains why the process is thermodynamically favorable?

A. Melting ice releases energy (ΔH<0) and entropy decreases (ΔS<0) because the motion of the H2O molecules decreases as it transitions from solid to liquid. At a temperature higher than 0 degrees Celsius, the term TΔS is smaller than ΔH, resulting in a thermodynamically favorable process with ΔG<0.

B. Melting ice releases energy (ΔH<0) and entropy increases (ΔS>0) because the motion of the H2O molecules increases as it transitions from solid to liquid. At a temperature higher than 0 degrees Celsius, the term TΔS is greater than ΔH, resulting in a thermodynamically favorable process with ΔG>0.

C. Melting ice requires energy (ΔH>0) and entropy decreases (ΔS<0) because the motion of the H2O molecules decreases as it transitions from solid to liquid. At a temperature higher than 0 degrees Celsius, the term TΔS is smaller than ΔH, resulting in a thermodynamically favorable process with ΔG>0.

D. Melting ice requires energy (ΔH>0) and entropy increases (ΔS>0) because the motion of the H2O molecules increases as it transitions from solid to liquid. At a temperature higher than 0 degrees Celsius, the term TΔS is greater than ΔH, resulting in a thermodynamically favorable process with ΔG<0.

Answer 1

The melting of ice is thermodynamically favorable because it results in an increase in entropy and requires energy; the transition from solid to liquid increases the disorder of water molecules. With a positive enthalpy change (ΔH>0) and a positive entropy change (ΔS>0), the negative Gibbs free energy change (ΔG<0) confirms the process is favorable. So the correct answer is Option D.

Step-by-step explanation:

The melting of an ice cube at 0 degrees Celsius in a room at 22 degrees Celsius is a thermodynamically favorable process because it involves an increase in entropy and requires energy input. When ice melts, it goes from a solid, orderly phase where water molecules are in a fixed position, to a less orderly, liquid phase with increased molecular motion. This transition from solid to liquid results in higher disorder, meaning the entropy (ΔS) increases. According to the entropy of fusion of ice, the process absorbs 6.01 kJ/mol of heat, indicating that the enthalpy change (ΔH) is positive. Since ΔH is greater than 0 and ΔS is greater than 0, Gibbs free energy change (ΔG) can be calculated using the equation ΔG = ΔH - TΔS.

At temperatures above 0 degrees Celsius, TΔS will be greater than ΔH because the temperature is acting on a positive entropy change. Thus, when ΔH and TΔS are combined in the ΔG equation, the result is a negative Gibbs free energy change (ΔG<0), making the process of melting ice thermodynamically favorable. Therefore, the correct explanation is D: Melting ice requires energy (ΔH>0) and entropy increases (ΔS>0) because the motion of the H2O molecules increases as it transitions from solid to liquid. At a temperature higher than 0 degrees Celsius, the term TΔS is greater than ΔH, resulting in a thermodynamically favorable process with ΔG<0.