Temperature Dependence of Spontaneity

As was previously demonstrated in this tutorial’s section on entropy, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions can also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:

\(\text{Δ}G=\text{Δ}H-T\text{Δ}S\)

The spontaneity of a process, as reflected in the arithmetic sign of its free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:

Both ΔH and ΔS are positive. This condition describes an endothermic process that involves an increase in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is greater than ΔH. If the TΔS term is less than ΔH, the free energy change will be positive. Such a process is spontaneous at high temperatures and nonspontaneous at low temperatures.
Both ΔH and ΔS are negative. This condition describes an exothermic process that involves a decrease in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is less than ΔH. If the TΔS term’s magnitude is greater than ΔH, the free energy change will be positive. Such a process is spontaneous at low temperatures and nonspontaneous at high temperatures.
ΔH is positive and ΔS is negative. This condition describes an endothermic process that involves a decrease in system entropy. In this case, ΔG will be positive regardless of the temperature. Such a process is nonspontaneous at all temperatures.
ΔH is negative and ΔS is positive. This condition describes an exothermic process that involves an increase in system entropy. In this case, ΔG will be negative regardless of the temperature. Such a process is spontaneous at all temperatures.

These four scenarios are summarized in the figure below.

A table with three columns and four rows is shown. The first column has the phrase, “Delta S greater than zero ( increase in entropy ),” in the third row and the phrase, “Delta S less than zero ( decrease in entropy),” in the fourth row. The second and third columns have the phrase, “Summary of the Four Scenarios for Enthalpy and Entropy Changes,” written above them. The second column has, “delta H greater than zero ( endothermic ),” in the second row, “delta G less than zero at high temperature, delta G greater than zero at low temperature, Process is spontaneous at high temperature,” in the third row, and “delta G greater than zero at any temperature, Process is nonspontaneous at any temperature,” in the fourth row. The third column has, “delta H less than zero ( exothermic ),” in the second row, “delta G less than zero at any temperature, Process is spontaneous at any temperature,” in the third row, and “delta G less than zero at low temperature, delta G greater than zero at high temperature, Process is spontaneous at low temperature.”

There are four possibilities regarding the signs of enthalpy and entropy changes.

Example

Predicting the Temperature Dependence of Spontaneity

The incomplete combustion of carbon is described by the following equation:

\(\text{2C}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2CO}\left(g\right)\)

How does the spontaneity of this process depend upon temperature?

Solution

Combustion processes are exothermic (ΔH < 0). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, ΔS > 0). The reaction is therefore spontaneous (ΔG < 0) at all temperatures.

When considering the conclusions drawn regarding the temperature dependence of spontaneity, it is important to keep in mind what the terms “high” and “low” mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature.

A process that is nonspontaneous at one temperature but spontaneous at another will necessarily undergo a change in “spontaneity” (as reflected by its ΔG) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which ΔG is plotted on the y axis versus T on the x axis:

\(\text{Δ}G=\text{Δ}H-T\text{Δ}S\)

\(y=b+mx\)

Such a plot is shown in the figure below. A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependent spontaneity as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative ΔG) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the x-intercept of the line, that is, the value of T for which ΔG is zero:

\(\text{Δ}G=0=\text{Δ}H-T\text{Δ}S\)

\(T=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{Δ}H}{\text{Δ}S}\)

And so, saying a process is spontaneous at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which ΔG for the process is zero. As noted earlier, the condition of ΔG = 0 describes a system at equilibrium.

A graph is shown where the y-axis is labeled, “Free energy,” and the x-axis is labeled, “Increasing temperature ( K ).” The value of zero is written midway up the y-axis with the label, “delta G greater than 0,” written above this line and, “delta G less than 0,” written below it. The bottom half of the graph is labeled on the right as, “Spontaneous,” and the top half is labeled on the right as, “Nonspontaneous.” A green line labeled, “delta H less than 0, delta S greater than 0,” extends from a quarter of the way up the y-axis to the bottom right of the graph. A yellow line labeled, “delta H less than 0, delta S less than 0,” extends from a quarter of the way up the y-axis to the middle right of the graph. A second yellow line labeled, “delta H greater than 0, delta S greater than 0,” extends from three quarters of the way up the y-axis to the middle right of the graph. A red line labeled, “delta H greater than 0, delta S less than 0,” extends from three quarters of the way up the y-axis to the top right of the graph.

These plots show the variation in ΔG with temperature for the four possible combinations of arithmetic sign for ΔH and ΔS.

Example

Equilibrium Temperature for a Phase Transition

As defined in the tutorial on liquids and solids, the boiling point of a liquid is the temperature at which its liquid and gaseous phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in this appendix to estimate the boiling point of water.

Solution

The process of interest is the following phase change:

\({\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(g\right)\)

When this process is at equilibrium, ΔG = 0, so the following is true:

\(0=\text{Δ}H\text{°}-T\text{Δ}S\text{°}\phantom{\rule{5em}{0ex}}\text{or}\phantom{\rule{5em}{0ex}}T=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{Δ}H\text{°}}{\text{Δ}S\text{°}}\)

Using the standard thermodynamic data from this appendix,

\(\begin{array}{ccc}\hfill \text{Δ}H\text{°}& =& \text{Δ}{H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\left(g\right)\right)\phantom{\rule{0.2em}{0ex}}-\text{Δ}{H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\left(l\right)\right)\hfill \\ & =& -\text{241.82 kJ/mol}-\left(-\text{286.83 kJ/mol}\right)=\text{44.01 kJ/mol}\end{array}\)

\(\begin{array}{ccc}\hfill \text{Δ}S°& =& \text{Δ}{S}_{298}^{°}\left({\text{H}}_{2}\text{O}\left(g\right)\right)\phantom{\rule{0.2em}{0ex}}-\text{Δ}{S}_{298}^{°}\left({\text{H}}_{2}\text{O}\left(l\right)\right)\hfill \\ & =& \text{188.8 J/K·mol}-\text{70.0 J/K·mol}=\text{118.8 J/K·mol}\hfill \end{array}\)

\(T=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{Δ}H\text{°}}{\text{Δ}S\text{°}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{44.01\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J/mol}}{118.8\phantom{\rule{0.2em}{0ex}}\text{J/K·mol}}\phantom{\rule{0.2em}{0ex}}=370.5\phantom{\rule{0.2em}{0ex}}\text{K}=97.3\phantom{\rule{0.2em}{0ex}}\text{°C}\)

The accepted value for water’s normal boiling point is 373.2 K (100.0 °C), and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (this appendix). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point.

Temperature Dependence of Spontaneity