As was previously demonstrated in this chapter’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:
$$ΔG=ΔH-TΔS$$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.
There are four possibilities regarding the signs of enthalpy and entropy changes.
Predicting the Temperature Dependence of Spontaneity
The incomplete combustion of carbon is described by the following equation:
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.
Check Your Learning
Popular chemical hand warmers generate heat by the air-oxidation of iron:
How does the spontaneity of this process depend upon temperature?
Answer:
ΔH and ΔS are negative; the reaction is spontaneous at low 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:
$$ΔG=ΔH-TΔS$$ $$y=b+mx$$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:
$$ΔG=0=ΔH-TΔS$$ $$T=\frac{ΔH}{ΔS}$$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.
These plots show the variation in ΔG with temperature for the four possible combinations of arithmetic sign for ΔH and ΔS.
Equilibrium Temperature for a Phase TransitionAs defined in the chapter 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 Appendix to estimate the boiling point of water.
Solution
The process of interest is the following phase change:
When this process is at equilibrium, ΔG = 0, so the following is true:
$$0=ΔH-TΔS\quad \text{or} \quad T=\frac{ΔH°}{ΔS°}$$Using the standard thermodynamic data from Appendix,
$$ΔH°=1\;mol \times ΔH_f^°(H_2O(g)) – 1\;mol \times ΔH_f^°(H_2O(l))$$ $$ΔH°=(1\;mol)(-241.82\;kJ/mol)-(1\;mol)(-285.83\;kJ/mol)=44.01\;kJ$$$$ΔS°=1\;mol \times ΔS°(H_2O(g)) – 1\;mol \times ΔS°(H_2O(l))$$ $$ΔS°=(1\;mol)(188.8\;J/K·mol)-(1\;mol)(70.0\;J/K·mol)=118.8\;J/K$$
$$T=\frac{ΔH}{ΔS}=\frac{44.01×10^3\;J}{118.8\;J/K}=378.8\;K=97.3\;°C$$
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 (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.
Check Your Learning
Use the information in Appendix to estimate the boiling point of CS2.
Answer:
313 K (accepted value 319 K)