Question

At 700 K if $\Delta G = - 11.5KJ$, what is the value of $\Delta {S_{(universe)}}$?
A.$ - 11.5kJ{K^{ - 1}}$
B.$ - 16.4J{K^{ - 1}}$
C.$16.4J{K^{ - 1}}$
D.None of the above

Answer 1

Hint: The Gibbs free energy of a system is defined as the system's enthalpy minus the product of the temperature times the entropy of the system at any given time. Because it is described in terms of thermodynamic parameters that are state functions, the system's Gibbs free energy is a state function.

Complete answer: Enthalpy- When heat and work must be calculated, enthalpy is a concept utilized in science and engineering. Enthalpy reveals how much heat and work was added or withdrawn from a substance while it changed at constant pressure. Enthalpy is comparable to but not the same as energy.
Entropy: Entropy is the amount of thermal energy in a system per unit temperature that can't be used to conduct beneficial work. Because work is generated by ordered molecular motion, entropy is also a measure of a system's molecular disorder, or unpredictability.
Change in entropy: The ratio of heat transfer to temperature is the change in entropy. RT. In terms of entropy, the second law of thermodynamics states that a system's total entropy either grows or remains constant; it never decreases.
Gibbs free energy equation,
$\Delta G = \Delta H - T\Delta S$
Here,$\Delta G$is the change in Gibbs free energy, T is the temperature,$\Delta S$is the change in entropy, $\Delta H$is the change in enthalpy.
As given in the question $\Delta G = - 11.5KJ$, $T = 700K$,$\Delta H = 0$
Now putting these values in Gibbs free energy equation,
$ - 11.5KJ = 0 - 700\Delta S$
Now by solving we get
$\Delta S = - 16.4J{K^{ - 1}}$
So, the final answer is option (B) i.e., $ - 16.4J{K^{ - 1}}$

Note:
Free energy change criteria are preferable to entropy change criteria for predicting spontaneity because the former just requires free energy change in the system, whereas the latter requires entropy change in the system and surroundings.