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

Spontaneous Process
A spontaneous process happens by itself without any outside help. Once it starts, it keeps going on its own.
  1. 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.
  2. 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

Nonspontaneous Process
A nonspontaneous process does not happen by itself. It requires continuous input of energy or effort to occur.
  1. 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.
  2. 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.

Entropy
Entropy, denoted by the symbol S, is a measure of molecular randomness or disorder.

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?

  1. 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.
  2. 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-1mol-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) + O2(g) → 2CaO(s)

Solution:

The standard entropy values are:
Sϴ [Ca(s)] = 41.40 J K–1 mol–1
Sϴ [O2(g)] = 205.0 J K–1 mol–1
Sϴ [CaO(s)] = 39.70 JK–1 mol–1
ΔSϴsystem = Sϴproducts – Sϴreactants

= 2 × Sϴ [CaO(s)] – {2 × Sϴ [Ca(s)] + Sϴ [O2(g)]}
= 2 × 39.70 – ((2 × 41.40) + 205.0)
= 79.40 – 287.8
ΔSϴsystem = –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.

ΔStotal = ΔSsurroundings + ΔSsystem

ΔSsurroundings =  - \frac{\Delta H}{T}

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:

Sb2S3(s) + 3Fe(s) → 2Sb(s) + 3FeS(s) ΔH = -125 kJ

Calculate ΔSsurroundings for the reaction.

Solution:

We use

ΔSsurroundings =  - \frac{\Delta H}{T}

where

T = 25 + 273 = 298 K

ΔSsurroundings =  - \frac{-125 kJ}{298 K} = 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).

Gibb's Free Energy
Gibbs’s free energy is the energy associated with a chemical reaction that can be used to do work. It combines enthalpy (𝐻), entropy (𝑆), and temperature (𝑇) in a way that helps predict the direction of spontaneous processes.

The formula is G = HTS.

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).
  • ΔS 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).
Applying the equation ΔG = ΔH – TΔS

Example

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

ZnCO3(s)→ ZnO(s) + CO2(g) ΔHr = +71.0 kJ mol1

(Values for S in J K–1 mol–1: CO2(g) = +213.6,
ZnCO3(s) = +82.4, ZnO(s) = +43.6)

Solution

Step 1 Convert the value of ΔHr to J mol–1:
+71.0 × 1000 = 71 000 J mol–1

Step 2 Calculate ΔSsystem:
ΔSsystem = Sproducts – Sreactants
                = S[ZnO(s)] + S[CO2(g)] – S[ZnCO3(s)]                = 43.6 + 213.6 – 82.4
ΔSsystem = +174.8 J K–1 mol–1

Step 3 Calculate ΔG:
ΔG = ΔHreaction – TΔSsystem
      = 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.