We have seen in the previous post, that the Gibbs free energy of a reaction is a comprehensive measure of the spontaneity of the reaction that can be calculated based on the parameters of the system.

Today, we are going to see how the enthalpy, temperature, and entropy of the reaction affect the Δ*G*^{o}*. *The first thing is you need to remember that a negative Δ*G** ^{o }*indicates a spontaneous reaction, and a positive Δ

*G*

*indicates a nonspontaneous reaction:*

^{o}* *

♦ **Δ****G ****< 0: **A negative Δ*G* corresponds to a **spontaneous** process.

♦ **Δ****G >**** 0: **A positive Δ*G* corresponds to a **nonspontaneous** process.

♦ **Δ****G ****= 0: **The system is at **equilibrium**. There is no net change.

* *

Let’s first look at the big picture by isolating the two terms in the Gibbs free energy equation: the Δ*H** ^{o }*and

*T*Δ

*S*

*terms:*

^{o}

* *

To make the (more) process spontaneous, you want to make the Δ*G*^{o} a negative large number. Therefore, a negative Δ*H*^{o}, meaning an exothermic reaction, is a good contributor to making the reaction spontaneous. On the other hand, increasing the entropy makes the Δ*G*^{o }smaller because of the negative sign and therefore, it also indicates a spontaneous reaction:

* *

* *

* ** *

In general, there are four possibilities for the signs of the Δ*H*^{o }and Δ*S*^{o} which are summarized below:

* *

* *

**Scenario 1. **When Δ*H *is negative and Δ*S *is positive, Δ*G *is negative at all temperatures because regardless of the values, the net product is a negative number.

Therefore, we can state that exothermic reactions with a positive entropy change are always spontaneous.

* *

**Scenario 2. **When Δ*H *is positive and Δ*S *is negative, Δ*G *is positive at all temperatures, and therefore, the reaction is nonspontaneous. This is expected because an endothermic reaction goes upwards energetically and decreasing the entropy makes the reaction less spontaneous too. To summarize, we can say that endothermic reactions with a negative entropy change are always nonspontaneous, regardless of the temperature.

* *

**Scenario 3. **If both Δ*H *and Δ*S *are positive, then Δ*G *will be negative only when the T Δ*S *> *H*. Increasing the temperature will favor spontaneity as it makes the T Δ*S *a larger negative number because of the negative sign*.*

* *

**Scenario 4. **If both Δ*H *and Δ*S *are negative, then Δ*G *will be negative only when the T Δ*S *< *H*. Decreasing the temperature will favor spontaneity as it makes the T Δ*S *a smaller positive number.

**For example**, given the values, Δ*H*°_{rxn} = -18,000 J, and Δ*S*°_{rxn} = -70 J/K, calculate the Δ*G*°_{rxn} at 25 ^{o}C and determine if the reaction is spontaneous or not.

**Solution**:

Δ*G*°_{rxn}* = *Δ*H*°_{rxn}* – T*Δ*S*°_{rxn} = -18,000 J – (25 + 273) (-70 J/K) = 2,860 J

* *

Δ*G*°_{rxn}* ***> 0**, therefore, the process is **nonspontaneous **at 25 ^{o}C.

**The temperature at Which the Reaction Becomes Spontaneous**

Now, you need to know that if the reaction is not spontaneous at a given temperature, it does not mean it won’t be spontaneous at a different temperature. Sometimes, you may get a question on a test asking to determine the temperature at which a given reaction becomes spontaneous.

To do that, you need to first, **set **Δ*G ***= 0 **which corresponds to the **equilibrium, **and determine the corresponding temperature. This is the crossover temperature at which a reaction switches between being spontaneous and nonspontaneous and we derive it from the Gibbs free energy equation.

**For example**, at what temperature the reaction in the previous example will become spontaneous?

**Solution**: To find the temperature at which the reaction becomes spontaneous, first, **set **Δ*G ***= 0 **which corresponds to the **equilibrium, **and determine the corresponding temperature.

Δ*G* = Δ*H* – *T*Δ*S*

0 = Δ*H* – *T*Δ*S*

\[T\, = \,\frac{{\Delta H}}{{\Delta S}}\]

\[T{\mkern 1mu} = {\mkern 1mu} \frac{{\Delta H}}{{\Delta S}}\; = \;\frac{{18,000\,{\rm{J}}}}{{70\,{\rm{J/K}}}}\, = \;257\,{\rm{K}}\,\]

Next, **identify the favorable term **that makes Δ*G *negative. In this case, Δ*H ***is the favorable term **because it is negative and therefore, it decreases the Δ*G. *Because the entropy is negative, *T*Δ*S ^{o}*

_{rxn}is positive, we need to make the

*T*Δ

*S*

^{o}_{rxn}a less significant contributor by

**decreasing the temperature**. It means the process will become

**spontaneous below 257 K**.

*Note:* If the Δ*H ^{o}*

_{rxn}and Δ

*S*

^{o}_{rxn}are not given in the problem, you need to calculate them first, using the data in the appendix, and then plug the values in the Gibbs free energy equation.

**Check Also**

- Standard Entropy Change (𝚫
*S*^{o}_{rxn}) of a Reaction - The Gibbs Free Energy
- Entropy and State Change
- Entropy Changes in the Surroundings
- 𝚫
*G*^{o}_{rxn}from the Free Energies of Formation - Gibbs Free Energy and Hess’s Law
- Gibbs Free Energy Under Nonstandard Conditions
- Gibbs Free Energy and Equilibrium Constant
**Entropy, Enthalpy, and Gibbs Free Energy Practice Problems**

#### Practice

Given the values of Δ*H *and Δ*S*, which of the following changes will be spontaneous at constant *T *and *P*?

**a)** Δ*H *= +35 kJ, Δ*S *= +10 J/K, *T *= 300. K

**b)** Δ*H *= +45 kJ, Δ*S *= +153 J/K, *T *= 312 K

**c)** Δ*H *= – kJ, Δ*S *= +8.0 J/K, *T *= 298 K

**d) **Δ*H *= – kJ, Δ*S *= -60. J/K, *T *= 250 K

At what temperatures will the following processes be spontaneous?

**a) **Δ*H *= +25.0 kJ, Δ*S *= +40.0 J/K

**b) **Δ*H *= -5.00 kJ, Δ*S *= -30.0 J/K

**c) **Δ*H *= -20.0 kJ, Δ*S *= +60.0 J/K

**c) **Δ*H *= +45.0 kJ, Δ*S *= -70.0 J/K