In the previous post, we discussed the correlation between the entropy change of the surroundings, Δ*S*_{surr,} and the enthalpy change of the system.

This relationship is useful for determining whether a given reaction is spontaneous or not because it allows calculating the Δ*S*_{univ }if the Δ*S*_{sys }is known or can be determined.

Remember, according to the second law of thermodynamics, the process is spontaneous only if the entropy of the universe is increasing, Δ*S*_{univ }> 0:

Now, it is not very practical **to deal with the entropy of the surroundings** and calculate the entropy change of the universe every time we need to determine whether a reaction is spontaneous or not. In chemistry, our focus is usually on the reaction, which is the system, so it would be more convenient to work with the parameters of the system without worrying about the surroundings.

And this is where the famous Gibbs free (G) energy comes into play as it combines the ** system’s** enthalpy and entropy:

*G *= *H *− *TS*

The Gibbs free energy change (ΔG) is what is used for determining if the process is spontaneous or not.

To derive the expression for the Gibbs free energy, we combine the following two equations that we have already discussed:

\[\Delta {S_{surr}}\, = \;\frac{{ – \Delta {H_{sys}}}}{T}\]

Δ*S*_{univ }= Δ*S*_{surr }+ Δ*S*_{sys }

Plugging this expression in the equation for Δ*S*_{surr}, and multiplying everything by –*T, *we get:

\[ – T\Delta {S_{univ}}\, = \;\frac{{ – \cancel{T}\; \times \;( – \Delta {H_{sys}})}}{{\cancel{T}}}\; – \;T\Delta {S_{sys}}\; = \, – T\Delta {S_{univ}}\; = \;\Delta {H_{sys}}\, – \;T\Delta {S_{sys}}\]

The term, (-*T*Δ*S*_{univ}) is the **Gibbs free energy change (**Δ** G^{o})**, and it is a comprehensive quantity to assess the spontaneity of a process because it takes into account the

**enthalpy, entropy,**and

**temperature**.

Notice that all the terms on the right are related to the system and therefore, to calculate the Δ*G ^{o} we only need to have data pertaining to the system.*

Just like for the enthalpy and entropy the superscript indicates ** standard-state** conditions.

* *

**Δ***G*** and Spontaneity of Reaction**

*G*

So, Δ*G*^{o} is calculated by the parameters of the system, but it silently includes the Δ*S*_{univ} just with the opposite sign. Therefore, because a spontaneous process is associated with a positive Δ*S*_{univ}, **negative **Δ** G **(exergonic) indicates a

**spontaneous**, while

**positive**Δ

**(endergonic) means a**

*G***nonspontaneous**process. When

*G*= 0*,*the system is at

**equilibrium.**

Another way to look at this is that G is energy and remember, any process tends to go downhill energetically because lower energy means more stability.

*To summarize, remember that: *

♦ Δ*G* is proportional to the negative of Δ*S*_{univ.}

♦ **Δ G < 0: **A negative Δ

*G*corresponds to a

**spontaneous**process.

♦

**Δ**A positive Δ

*G >*0:*G*corresponds to a

**nonspontaneous**process.

♦

**Δ**The system is at

*G*= 0:**equilibrium**. There is no net change.

**Example: **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

**Solution**

**a)** Make sure the units of Δ*H *and Δ*S *match in terms of Joules.

Δ*H *= +35 kJ = 35 x 10^{3} J

Δ*G* = 35 x 10^{3} J – 300. K Δ*S *x 15.0 J/K = 30,500 J

**Δ****G > ****0**, therefore, the process is **nonspontaneous**.

* *

**b) **Again convert the units to match the Joules.

Δ*H *= +45 kJ = 45 x 10^{3} J

Δ*G* = 45 x 10^{3} J – 312 K x 153 J/K = -2736 J

**Δ****G < ****0**, therefore, the process is **spontaneous**.

In the next article, we will discuss the correlation between the *Δ**G, ΔH, and ΔS *and the spontaneity of the reaction depending on the sign of these quantities.