### Learning Outcomes

**At the end of this section, you should be able to:**

- Define entropy and Gibb’s free energy
- Calculate entropy change for reactions
- Calculate Gibb’s free energy change for reactions using ΔG = ΔH – TΔS
- Predict the spontaneity of reactions

### Spontaneous Processes

**Everyday Examples**:**Ice Melting**: If you take an ice cube out of the freezer and leave it at room temperature, it melts into water. You don’t need to do anything; it happens on its own.**Ball Rolling Down a Hill**: If you place a ball at the top of a hill and let go, it will roll down by itself without you pushing it.

**Characteristics**:- These processes tend to increase disorder or randomness in the system. In scientific terms, we say they increase entropy.
- They usually move towards a more stable state (like water at room temperature is more stable than ice).

### Nonspontaneous Processes

**Everyday Examples**:**Freezing Water into Ice**: To freeze water into ice, you need to put it in the freezer, which uses energy to keep it cold.**Pushing a Ball Uphill**: If you want to move a ball up a hill, you have to push it all the way up. It won’t go up on its own.

**Characteristics**:- These processes tend to decrease disorder or randomness in the system, meaning they decrease entropy.
- They require energy input to go against natural tendencies (like making water into ice).

### Entropy

All spontaneous processes share a common trait: an increase in **entropy**.

**, is a measure of molecular randomness or disorder.**

*S*Imagine your room. If your room is messy, with clothes and books scattered everywhere, it has **high entropy** because it’s **very disordered**. If your room is neat and everything is in its place, it has **low entropy** because it’s **well-ordered**.

#### How Does Entropy Work in Chemistry?

**Particles and Movement**:- In chemistry, we often talk about molecules and atoms. The more ways these tiny particles can move around and arrange themselves, the higher the entropy.
- For example, in a solid, particles are tightly packed and can’t move much, so the entropy is low. In a gas, particles can move freely and spread out, so the entropy is high.

**Changes in Entropy**:- When ice melts into water, the water molecules can move around more freely than in the solid ice, so the entropy increases.
- When water boils and turns into steam, the molecules spread out even more, so the entropy increases even further.

### Calculating entropy changes

**Standard entropies, S°**

Entropy is given the symbol S, and standard entropy (measured at 298 K and a pressure of 1 bar) is given the symbol S°. The units of entropy are J K^{-1}mol^{-1}.

**Entropy changes, ΔS°**

ΔS° = ΣS°(products) – ΣS°(reactants)

Where Σ (sigma) simply means “the sum of”.

**Example**

Calculate the entropy change of the system for

the reaction:

2Ca(s) + O_{2}(g) → 2CaO(s)

**Solution:**

The standard entropy values are:

S^{ϴ} [Ca(s)] = 41.40 J K^{–1} mol^{–1}

S^{ϴ} [O_{2}(g)] = 205.0 J K^{–1} mol^{–1}

S^{ϴ} [CaO(s)] = 39.70 JK^{–1} mol^{–1}

ΔS^{ϴ}_{system} = S^{ϴ}_{products} – S^{ϴ}_{reactants}

= 2 × S^{ϴ} [CaO(s)] – {2 × S^{ϴ} [Ca(s)] + S^{ϴ} [O_{2}(g)]}

= 2 × 39.70 – ((2 × 41.40) + 205.0)

= 79.40 – 287.8

ΔS^{ϴ}_{system} = –208.4 JK^{–1} mol^{–1}

**Calculating the entropy change of the surroundings.**

An **exothermic reaction** releases heat to the **surroundings**, and that extra heat increases the entropy of the surroundings.

An **endothermic reaction** absorbs heat from the surroundings, and so the entropy of the surroundings decreases.

Total entropy change is the sum of the entropy changes of the system and the surroundings.

ΔS_{total} = ΔS_{surroundings} + ΔS_{system}

ΔS_{surroundings} =

ΔH is the enthalpy change for the reaction. T is the temperature in Kelvin.

**Example:**

In the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition of the ore. For example, iron is used to reduce antimony in sulfide ores:

Sb_{2}S_{3}(*s*) + 3Fe(*s*) → 2Sb(s) + 3FeS(s) ΔH = -125 kJ

Calculate ΔS_{surroundings} for the reaction.

**Solution:**

We use

ΔS_{surroundings} =

where

T = 25 + 273 = 298 K

ΔS_{surroundings} = = 0.419 kJ/K = +419 J/K

**Free Energy**

Another thermodynamic function that helps predict whether a process or reaction will proceed spontaneously is Gibb’s free energy (*G*).

The formula is *G = H−TS.*

For a process that occurs at constant temperature, the change in free energy (∆*G*) is given by the equation

∆*G* = ∆*H* – T∆*S*

where:

- ∆
*H*is the change in enthalpy (heat content). - T is the absolute temperature (in Kelvin).
- Δ
*S*is the change in entropy (disorder).

#### Free Energy and Spontaneity

- If ΔG<0, the process or the reaction is spontaneous (it can happen on its own).
- If ΔG>, the process is or the reaction nonspontaneous (it requires energy input to occur).
- If ΔG=0, the system is or the reaction at equilibrium (no net change occurs).

**Example**

Calculate the Gibbs free energy change for the decomposition of zinc carbonate at 298 K.

ZnCO_{3}(s)→ ZnO(s) + CO_{2}(g) ΔH_{r} = +71.0 kJ mol^{–1}

(Values for S in J K^{–1} mol^{–1}: CO_{2}(g) = +213.6,

ZnCO_{3}(s) = +82.4, ZnO(s) = +43.6)

**Solution**

**Step 1** Convert the value of ΔH_{r} to J mol^{–1}:

+71.0 × 1000 = 71 000 J mol^{–1}

**Step 2** Calculate ΔS_{system}:

ΔS_{system} = S_{products} – S_{reactants}

= S[ZnO(s)] + S[CO_{2}(g)] – S[ZnCO_{3}(s)] = 43.6 + 213.6 – 82.4

ΔS_{system} = +174.8 J K^{–1} mol^{–1}

**Step 3** Calculate ΔG:

ΔG = ΔH_{reaction} – TΔS_{system}

= 71 000 – 298 × (+174.8)

ΔG = +18 909.6 J mol^{–1}

= +18.9 kJ mol^{–1} (to 3 significant figures)

As the value of ΔG is positive, the reaction is not spontaneous at 298 K.