AOT - Entropy and Spontaneity (Lesson)

Entropy and Spontaneity

Up to this point, changes in heat energy accompanying certain physical and chemical processes have been addressed. As shown in energy diagrams from previous lessons as well as other lessons, a release of energy from the system to the surroundings is often viewed as favorable. In other words, exothermic processes seem to be favorable processes while endothermic seem to be unfavorable. However, enthalpy is not the ultimate deciding factor as to whether a process occurs because there are many examples of endothermic reactions that happen spontaneously.

One example of this would be instant cold packs. When instant cold packs are crushed, a sample of ammonium nitrate mixes with water to form a solution. This is an extremely endothermic process that results in the bag becoming cold enough to the touch to be used as a treatment for sports injuries.

NH₄NO₃ (s) $LaTeX: \longrightarrow$ $\longrightarrow$ NH₄⁺ (aq) + NO₃^- (aq)

ColdPack

Since this and other endothermic reactions occur spontaneously, there must be an additional parameter that is a contributing factor to the spontaneity of chemical and physical processes. Chemists refer to this second parameter as entropy or S which describes the amount of disorder in a substance. While entropy can be described and thought of in a wide variety of ways, in chemical terms, it is a description of the orderliness of a system or how this order increases or decreases during various processes. While this still gives a vague idea of what this concept is, certain guidelines can be used to determine if entropy is increasing or decreasing for chemical processes.

General Concepts Related to Changes in Entropy

A change in entropy, S, for a system can be thought of as follows:

$LaTeX: \Delta$ $\Delta$ S = S_final - S_initial

These changes in entropy can be thought of as changes in the degree or order of changes in randomness of the system. A few helpful guidelines that will be used are as follows:

The entropy of a substance increases as it undergoes phase transitions from solid to liquid to gas. The solid state is characterized by tightly packed, slow moving particles with very little randomness. As a substance transitions to the liquid phase, the molecules are still held together via intermolecular forces; however, they now have the ability to flow which introduces a higher level of chaos. Finally, as a liquid transitions into the gas phase, it experiences the highest amount of entropy. The gas phase is characterized as a highly chaotic state with randomly moving particles and a high level of randomness.
Entropy increases when solids, liquids, or gases dissolve in a solvent. Similar to the phase transitions discussed above, dissolving one substance into another increases the entropy of the system. In the case of solids, not only do they lose all of the order of their crystalline states, but they are also now mixed with a second substance thereby increasing the entropy even further.
Entropy increases as the temperature increases. According to kinetic molecular theory, an increase in the temperature of a substance results in an increase in the kinetic energy of the sample. This increase in kinetic energy results in an increase in the average speed of the particles. In terms of entropy, faster moving particles are more chaotically moving particles; therefore, an increase in temperature leads to an increase in the entropy of the system.

You Try It!

In the following self-assessment activity, state whether the $LaTeX: \Delta S_{rxn}^{^{\circ}}$ $\Delta S_{rxn}^{^{\circ}}$ is positive or negative. Click on the plus sign to check your answer!

Calculating Entropy Changes

When discussing Hess's law as it related to enthalpy, it was shown that this law could be applied to formation reactions that allowed for the calculation of the $LaTeX: \Delta H^{^{\circ}}_{rxn}$ $\Delta H^{^{\circ}}_{rxn}$ for an unknown chemical reaction. The reason why this law can be applied in this manner relies upon the fact that enthalpy is a state function. Entropy, S, much like enthalpy, is also a state function, and for this reason, the change in entropy, $LaTeX: \Delta$ $\Delta$ S, can be calculated using a very similar equation shown below.

$LaTeX: \Delta S^{^{\circ}}_{rxn}$ $\Delta S^{^{\circ}}_{rxn}$ = $LaTeX: \Sigma S_f^{^{\circ}}\left(products\right)-\Sigma S^{^{\circ}}_f\left(reactants\right)$ $\Sigma S_f^{^{\circ}}\left(products\right)-\Sigma S^{^{\circ}}_f\left(reactants\right)$

The equation below and the tabulated entropy of formation values can be used as an example of how to calculate the entropy change associated with this process. Notice in the table that the entropy of formation for molecular oxygen (O₂) is not zero as it would be for enthalpy. This is due to the fact that the reference state for entropy is different than that of enthalpy.

C₆H₁₂O₆ (s) + 6 O₂ (g) $LaTeX: \longrightarrow$ $\longrightarrow$ 6 CO₂ (g) + 6 H₂O (l)

Table
Substance	$LaTeX: S^{^{\circ}}_f$ $S^{^{\circ}}_f$ (J/mol $LaTeX: \cdot$ $\cdot$ K)
C₆H₁₂O₆ (s)	212
O₂ (g)	205.03
CO₂ (g)	213.63
H₂O (l)	69.91

$LaTeX: \Delta S^{^{\circ}}_{rxn}$ $\Delta S^{^{\circ}}_{rxn}$ = {6 S^{$^\circ$}[CO₂ (g)] + 6 S^{$^\circ$}[H₂O (l)]} LaTeX: - $-$ {S^{$^\circ$}[C₆H₁₂O₆ (s)] + 6 S^{$^\circ$}[O₂ (g)]}

$LaTeX: \Delta S^{^{\circ}}_{rxn}$ $\Delta S^{^{\circ}}_{rxn}$ = [6(213.63) + 6(69.91)] LaTeX: - $-$ [1(212) + 6(205.03)]

$LaTeX: \Delta S^{^{\circ}}_{rxn}$ $\Delta S^{^{\circ}}_{rxn}$ = 259.06 J/mol $LaTeX: \cdot$ $\cdot$ K

You Try It!

In the following self-assessment activity, calculate $LaTeX: \Delta S_{rxn}^{^{\circ}}$ $\Delta S_{rxn}^{^{\circ}}$ . Click the plus sign to check your answer!

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