An ice cube at 0 Celcius melts when placed inside a room at 22 Celcius. 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, delta H < 0, and delta S < 0 because the motion of the H2O molecules decreases as it transitions from solid to liquid. At a temperature higher than 0 Celcius, the term T delta S is smaller than delta H, resulting in a thermodynamically favorable process with delta G < 0. B. Melting ice releases energy, delta H < 0, and delta S > 0 because the motion of the H2O molecules increases as it transitions from solid to liquid. At a temperature higher than 0 Celcius, the term T delta S is greater than delta H, resulting in a thermodynamically favorable process with delta G > 0. C. Melting ice requires energy, delta H > 0, and delta S < 0 because the motion of the H2O molecules decreases as it transitions from solid to liquid. At a temperature higher than 0 Celcius, the term T delta S is smaller than delta H, resulting in a thermodynamically favorable process with delta G > 0. D. Melting ice requires energy, delta H > 0, and delta S > 0 because the motion of the H2O molecules increases as it transitions from solid to liquid. At a temperature higher than 0 Celcius, the term T delta S is greater than delta H, resulting in a thermodynamically favorable process with delta G < 0.

Chemistry for Engineering Students

4th Edition

ISBN:9781337398909

Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter10: Entropy And The Second Law Of Thermodynamics

Section: Chapter Questions

Problem 10.80PAE

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An ice cube at 0 Celcius melts when placed inside a room at 22 Celcius. 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, delta H < 0, and delta S < 0 because the motion of the H2O molecules decreases as it transitions from solid to liquid. At a temperature higher than 0 Celcius, the term T delta S is smaller than delta H, resulting in a thermodynamically favorable process with delta G < 0. B. Melting ice releases energy, delta H < 0, and delta S > 0 because the motion of the H2O molecules increases as it transitions from solid to liquid. At a temperature higher than 0 Celcius, the term T delta S is greater than delta H, resulting in a thermodynamically favorable process with delta G > 0. C. Melting ice requires energy, delta H > 0, and delta S < 0 because the motion of the H2O molecules decreases as it transitions from solid to liquid. At a temperature higher than 0 Celcius, the term T delta S is smaller than delta H, resulting in a thermodynamically favorable process with delta G > 0. D. Melting ice requires energy, delta H > 0, and delta S > 0 because the motion of the H2O molecules increases as it transitions from solid to liquid. At a temperature higher than 0 Celcius, the term T delta S is greater than delta H, resulting in a thermodynamically favorable process with delta G < 0.

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