### Learning Objectives

- To gain an understanding of the relationship between spontaneity, free energy, and temperature.
- To be able to calculate the temperature at which a process is at equilibrium under standard conditions.

In the Gibbs free energy change equation, the only part we as scientists can control is the temperature. We have seen how we can calculate the standard change in Gibbs free energy, Δ*G*⁰, but not all reactions we are interested in occur at exactly 298 K. The temperature plays an important role in determining the Gibbs free energy and spontaneity of a reaction.

Δ*G* = Δ*H* – *T*Δ*S*

If we examine the Gibbs free energy change equation, we can cluster the components to create two general terms, an enthalpy term, Δ*H*, and an entropy term, –*T*Δ*S*. Depending on the sign and magnitude of each, the sum of these terms determines the sign of Δ*G* and therefore the spontaneity (Table 18.2 “Spontaneity and the Signs of Enthalpy and Entropy Terms”).

Table 18.2. Spontaneity and the Signs of Enthalpy and Entropy Terms

Since all temperature values are positive in the Kelvin scale, the temperature affects the magnitude of the entropy term. As shown in Table 18.2 “Spontaneity and the Signs of Enthalpy and Entropy Terms,” the temperature can be the deciding factor in spontaneity when the enthalpy and entropy terms have opposite signs. If Δ*H* is negative, and –*T*Δ*S* positive, the reaction will be spontaneous at low temperatures (decreasing the magnitude of the entropy term). If Δ*H* is positive, and –*T*Δ*S* negative, the reaction will be spontaneous at high temperatures (increasing the magnitude of the entropy term).

Sometimes it can be helpful to determine the temperature when Δ*G*⁰ = 0 and the process is at equilibrium. Knowing this value, we can adjust the temperature to drive the process to spontaneity or alternatively to prevent the process from occurring spontaneously. Remember that, at equilibrium:

Δ*G*⁰ = 0 = Δ*H*⁰- *T*Δ*S*⁰

We can rearrange and solve for the temperature *T*:

*T*Δ*S*⁰ =Δ*H*⁰

*T* =[latex]\frac{{{\rm \ }\Delta {\rm H}}^{o}}{{\rm \ }\Delta {\rm S}^{o}}[/latex]

**Example 6**

Using the appendix table of standard thermodynamic quantities, determine the temperature at which the following process is at equilibrium:

CHCl_{3}(ℓ) ⇌ CHCl_{3}(g)

How does the value you calculated compare to the boiling point of chloroform given in the literature?

Solution

At equilibrium: Δ*G*⁰ = 0 = Δ*H*⁰- *T*Δ*S*⁰

We must estimate Δ*H*⁰ and *S*⁰ from their enthalpies of formation and standard molar entropies, respectively.

Δ*H*⁰ = ∑*nΔH*⁰* _{f}*(products) – ∑

*mΔH*⁰

*(reactants)*

_{f}Δ*H*⁰=* –*102.7 kJ/mol – (–134.1 kJ/mol)

Δ*H*⁰ = + 31.4 kJ/mol

Δ*S*⁰ = ∑*nΔS*⁰ (products) – ∑*mΔS*⁰ (reactants)

Δ*S*⁰ = 295.7 J/mol K – (201.7 J/mol K)

Δ*S*⁰ = 94.0 J/mol K (or 94.0 x 10^{-3} kJ/mol K)

Now we can use these values to solve for the temperature:

*T* =[latex]\frac{{{\rm \ }\Delta {\rm H}}^{o}}{{\rm \ }\Delta {\rm S}^{o}}[/latex]

*T* = * [latex]\textit{ } \frac{31.4\ kJ/mol}{94.0\ x\ {10}^{-3}\ kJ/mol\ K} \textit{ }[/latex]*

*T *= 334 K = 60.9⁰C

The literature boiling point of chloroform is 61.2⁰C. The value we have calculated is very close but slightly lower due to the assumption that Δ*H*⁰ and *S*⁰ do not change with temperature when we estimate the Δ*H*⁰ and *S*⁰ from their enthalpies of formation and standard molar entropies.

### Key Takeaways

- The temperature can be the deciding factor in spontaneity when the enthalpy and entropy terms have opposite signs:

– If Δ*H* is negative, and *–T*Δ*S *positive, the reaction will be spontaneous at low temperatures (decreasing the magnitude of the entropy term).

– If Δ*H* is positive, and *–T*Δ*S *negative, the reaction will be spontaneous at high temperatures (increasing the magnitude of the entropy term).