## CHEMISTRY THE CENTRAL SCIENCE

**19 CHEMICAL THERMODYNAMICS**

**19.6 FREE ENERGY AND TEMPERATURE**

Tabulations of , such as those in Appendix C, make it possible to calculate Δ*G*° for reactions at the standard temperature of 25 °C, but we are often interested in examining reactions at other temperatures. To see how Δ*G* is affected by temperature, let's look again at Equation 19.11:

Notice that we have written the expression for Δ*G* as a sum of two contributions, an enthalpy term, Δ*H*, and an entropy term, –*T*Δ*S*. Because the value of –*T*Δ*S* depends directly on the absolute temperature *T*, Δ*G* varies with temperature. We know that the enthalpy term, Δ*H*, can be either positive or negative and that *T* is positive at all temperatures other than absolute zero. The entropy term, –*T*Δ*S*, can also be positive or negative. When Δ*S* is positive, which means the final state has greater randomness (a greater number of microstates) than the initial state, the term –*T*Δ*S* is negative. When Δ*S* is negative, –*T*Δ*S* is positive.

The sign of Δ*G*, which tells us whether a process is spontaneous, depends on the signs and magnitudes of Δ*H* and –*T*Δ*S*. The various combinations of Δ*H* and –*T*Δ*S* signs are given in **TABLE 19.3**.

Note in Table 19.3 that when Δ*H* and –*T*Δ*S* have opposite signs, the sign of Δ*G* depends on the magnitudes of these two terms. In these instances temperature is an important consideration. Generally, Δ*H* and Δ*S* change very little with temperature. However, the value of *T* directly affects the magnitude of – *T*Δ*S*. As the temperature increases, the magnitude of –*T*Δ*S* increases, and this term becomes relatively more important in determining the sign and magnitude of Δ*G*.

As an example, let's consider once more the melting of ice to liquid water at 1 atm:

This process is endothermic, which means that Δ*H* is positive. Because the entropy increases during the process, Δ*S* is positive, which makes –*T*Δ*S* negative. At temperatures below 0 °C (273 K), the magnitude of Δ*H* is greater than that of –*T*Δ*S*. Hence, the positive enthalpy term dominates, and Δ*G* is positive. This positive value of Δ*G* means that ice melting is not spontaneous at *T* < 0 °C, just as our everyday experience tells us; rather, the reverse process, the freezing of liquid water into ice, is spontaneous at these temperatures.

What happens at temperatures greater than 0 °C? As *T* increases, so does the magnitude of –*T*Δ*S*. When *T* > 0 °C, the magnitude of –*T*Δ*S* is greater than the magnitude of Δ*H*, which means that the –*T*Δ*S* term dominates and Δ*G* is negative. The negative value of Δ*G* tells us that ice melting is spontaneous at *T* > 0 °C.

At the normal melting point of water, *T* = 0 °C, the two phases are in equilibrium. Recall that Δ*G* = 0 at equilibrium; at *T* = 0 °C, Δ*H* and –*T*Δ*S* are equal in magnitude and opposite in sign, so they cancel and give Δ*G* = 0.

**TABLE 19.3 • How Signs of** Δ** H and** Δ

*S*Affect Reaction Spontaneity**GIVE IT SOME THOUGHT**

The normal boiling point of benzene is 80 °C. At 100 °C and 1 atm, which term is greater in magnitude for the vaporization of benzene, Δ*H* or *T*Δ*S*?

Our discussion of the temperature dependence of **Δ***G* is also relevant to standard free-energy changes. We can calculate the values of Δ*H*° and Δ*S*° at 298 K from the data in Appendix C. If we assume that these values do not change with temperature, we can then use Equation 19.12 to estimate Δ*G*° at temperatures other than 298 K.

**SAMPLE EXERCISE 19.9 Determining the Effect of Temperature on Spontaneity**

The Haber process for the production of ammonia involves the equilibrium

Assume that Δ*H*° and Δ*S*° for this reaction do not change with temperature. **(a)** Predict the direction in which Δ*G*° for the reaction changes with increasing temperature. **(b)** Calculate Δ*G*° at 25 °C and 500 °C.

**SOLUTION**

**Analyze** In part (a) we are asked to predict the direction in which Δ*G*° changes as temperature increases. In part (b) we need to determine Δ*G*° for the reaction at two temperatures.

**Plan** We can answer part (a) by determining the sign of **Δ***S* for the reaction and then using that information to analyze Equation 19.12. In part (b) we first calculate Δ*H*° and Δ*S*° for the reaction using data in Appendix C and then use Equation 19.12 to calculate Δ*G*°.

**Solve**

**(a)** The temperature dependence of Δ*G*° comes from the entropy term in Equation 19.12, Δ*G*° = Δ*H*° **–***T*Δ*S*°. We expect Δ*S*° for this reaction to be negative because the number of molecules of gas is smaller in the products. Because Δ*S*° is negative, **–***T*Δ*S*° is positive and increases with increasing temperature. As a result, Δ*G*° becomes less negative (or more positive) with increasing temperature. Thus, the driving force for the production of NH_{3} becomes smaller with increasing temperature.

**(b)** We calculated Δ*H*° for this reaction in Sample Exercise 15.14 and Δ*S*° in Sample Exercise 19.5: Δ*H*° = –92.38 kJ and Δ*S*° = –198.3 J/K. If we assume that these values do not change with temperature, we can calculate Δ*G*° at any temperature by using Equation 19.12. At *T* = 25 °C = 298 K, we have

At *T* = 500°C = 773 K, we have

Notice that we had to convert the units of –*T*Δ*S*° to kJ in both calculations so that this term can be added to the Δ*H*° term, which has units of kJ.

**Comment** Increasing the temperature from 298 K to 773 K changes Δ*G*° from –33.3 kJ to + 61 kJ. Of course, the result at 773 K assumes that Δ*H*° and Δ*S*° do not change withtemperature. Although these values do change slightly with temperature, the result at 773 K shouldbe a reasonable approximation.

The positive increase in Δ*G*° with increasing *T* agrees with our prediction in part (a). Our result indicates that in a mixture of N_{2}(*g*), H_{2}(*g*), and NH_{3}(*g*), each present at a partial pressure of 1 atm, the N_{2}(*g*) and H_{2}(*g*) react spontaneously at 298 K to form more NH_{3}(*g*). At 773 K, the positive value of Δ*G*° tells us that the reverse reaction is spontaneous. Thus, when the mixture of these gases, each at a partial pressure of 1 atm, is heated to 773 K, some of the NH_{3}(*g*) spontaneously decomposes into N_{2}(*g*) and H_{2}(*g*).

**PRACTICE EXERCISE**

**(a)** Using standard enthalpies of formation and standard entropies in Appendix C, calculate Δ*H*° and Δ*S*° at 298 K for the reaction 2 SO_{2}(*g*) + O_{2}(*g*) *→**2* SO_{3}(*g*). **(b)** Use your values from part (a) to estimate Δ*G*° at 400 K.

*Answers:*