# CHEM C5: Entropy

Today we’ll be covering:

• What is Entropy?
• Standard entropy
• Entropy changes
• State & Temperature change
• Reaction with a change in the number of gaseous molecules
• Calculating the entropy change for a reaction
• What is a spontaneous/feasible reaction?

Let’s go!

What is Entropy?
A measure of dispersal of energy at a specific temperature.
It is also a measure of disorder of a system.

The higher the disorder or the more spread out the energy of a substance, the higher the entropy.

It is denoted as a capital ‘S’.

It is measured as:
Energy transferred in a reaction per Kelvin of temperature per mole of substance
= Joules per Kelvin per Mole
= JK-1mol-1

Why does entropy tend to increase?
This is all due to probability.

 Consider the different possible arrangements for particles in a container. From OpenText BC: Chemistry There are more ways of arrangement that are spread out than orderly. Compare the orderly arrangements (red) with the dispersed arrangements (yellow). Thus it is more probable for particles to be in a disordered/dispersed position than in an orderly one. In systems of billions of particles, the probability of disorder is much larger than the probability of order. It can be said that entropy is a measure of the number of disordered ways. Thus, entropy tends to increase in a system unless external energy is put in to make it more ordered. However, the process of providing external energy itself causes entropy in another system. Thus, the total entropy in the Universe cannot decrease. This is Newton’s 2nd Law of Thermodynamics.

A video explanation: Jeff Phillips: What is Entropy? (TED Ed)

What is Standard Entropy?
The measure of entropy of a certain substance (compound or element) at standard conditions.

It is denoted as S.

Just as standard enthalpies are listed out, standard entropies can also be listed for each element, like so:

Note that all values of entropy are positive.
There is no such thing as negative entropy, unlike enthalpy which CAN be negative.

What are the Factors affecting Entropy?

 State The more possible movement a state has, the higher its entropy. Generally*, gases have higher entropy than liquids, which have higher entropy than solids. This is because particles can move more freely in a gas than in a liquid, etc.   S of Η2Ο (g) > S of Η2Ο (l) > S of Η2Ο (s) Temperature The higher the kinetic energy of a substance, the higher its entropy. Substances at a higher temperature have a higher entropy than those at a lower temperature.   S of Η2Ο (l) at 350 K > S of Η2Ο (l) at 320 K Number of particles The more particles there are, the more possible arrangements, thus the higher the entropy.   S of 2 moles of Η2Ο (l) > S of 1 mole of Η2Ο (l) Number of atoms The more atoms there are in a molecule, the more possible arrangements for the energy between bonds, thus the higher the entropy.   S of CaCO3 > S of CaO Hardness of solids The softer the solid, the more possible arrangements for particles to have, thus the higher the entropy. Harder substances have a lower entropy than soft substances.   S of C (graphite) > S of C (diamond)

*These factors do have exceptions!

What is an Entropy Change (ΔS)?
The change in the dispersal of energy between reactants & products in a reaction.

We will look at a few types of processes & their corresponding entropy changes.

 Temperature & State changes When substances melt/vaporise, their entropy increases. Here’s a graph of entropy against temperature, as ice is heated, melted, & then boiled: From Annenberg Learner: Chemistry Entropy gradually increases as temperature increases, since particles have higher kinetic energy. Entropy has sudden jumps between states because state changes occur at constant temperatures (the melting & boiling points). Reaction with a change in the number of gas particles In a reaction, the reactants’ elements reshuffle & form new compounds as products. We can say entropy INCREASES when the products have a higher number of moles of gases than the reactants.  For example: CaCO3 (s) → CaO (s) + CO2 (g) Here there are 0 moles of gaseous reactants, but 1 mole of gaseous products. Thus, entropy increases. N2 (g) + 3H2 (g) → 2NH3 (g) Here there are 4 moles of gaseous reactants, but 2 moles of gaseous products. Thus, entropy decreases.

How do you calculate entropy changes in a reaction?
First, we need to be aware that every reaction can be divided into 2 parts:

• System: reactants & products
• Surroundings: not involved in the reaction, but can transfer energy to/from the system

Let’s look at each part:

 System Entropy change in a system = difference between entropies of products & reactants. ΔSsystem = ΣSproducts – ΣSreactants ΔS⦵system = ΣS⦵products – ΣS⦵reactantsSince the exact values of entropy for given substances are known, all we need to do is to substitute the values into the equation! For example: Calculate standard entropy change for the system: 2Ca (s) + O2 (g) → 2CaO (s) Given: S⦵ Ca (s) = 41.4 JK-1mol-1 S⦵ O2 (g) = 205.0 JK-1mol-1 S⦵ CaO (s) = 39.7 JK-1mol-1 ΔS⦵system = ΣS⦵products – ΣS⦵reactants ΔS⦵system = [2S⦵ CaO (s)]  – [2S⦵ Ca (s) + S⦵ O2 (g)] ΔS⦵system = -208.4 JK-1mol-1 Surroundings Entropy change in surroundings = energy transferred to surroundings divided by temperature in Kelvin (at standard conditions, T = 298K) ΔSsurroundings = –ΔHreaction/T ΔS⦵surroundings = –ΔH⦵reaction/TNote that the negative sign is part of the equation. No matter whether ΔHreaction was negative or positive, you still have to multiply it by -1. For example: Calculate standard entropy change for the surroundings for: 2Ca (s) + O2 (g) → 2CaO (s) Given: ΔH⦵r = – 1270.2 kJmol-1 Converting to Jmol-1: ΔH⦵r = – 1270200 Jmol-1 ΔS⦵surroundings = –ΔH⦵reaction/T ΔS⦵surroundings = – (–1270200)/298 ΔS⦵surroundings = +4262.4 JK-1mol-1 Total Total entropy change = entropy change of system + entropy change of surroundings ΔStotal = ΔSsystem + ΔSsurroundings ΔS⦵total = ΔS⦵system + ΔS⦵surroundingsFor example: Calculate total entropy change for: 2Ca (s) + O2 (g) → 2CaO (s) ΔS⦵total = ΔS⦵system + ΔS⦵surroundings ΔS⦵total = -208.4 JK-1mol-1 + +4262.4 JK-1mol-1 ΔS⦵total = +4054.0 JK-1mol-1

NOTE:
When a question asks to “calculate the entropy change of a reaction”, they usually mean the entropy change of the system, UNLESS you are specifically asked to calculate the total.
A clue:

• If the S of each species is given, you can calculate ΔSsystem
• If the ΔΗ of the reaction is given, you can calculate ΔSsurroundings
• You need both S & ΔΗ to calculate ΔStotal
• If only Sis given, the question’s asking for ΔSsystem

What is a feasible/spontaneous reaction?
A reaction that can occur without continuous supply of external energy.

This means that when the reaction is started (which may require a little activation energy), it can continue on its own without external work put in.

A spontaneous reaction MUST have a POSITIVE total entropy change (i.e. the net entropy of the system + surroundings must increase).

If [ΔSsystem + ΔSsurroundings] > 0, a reaction is spontaneous.

For example:

• Neutralisation is feasible:
an acid & an alkali will react when their solutions are in contact
• Rusting is feasible:
iron will rust as long as it is in contact with oxygen & water
• Combustion of petrol at room temperature is feasible:
you only need a spark to start it burning, but it will continue to burn on its own
• Charging a battery is NOT feasible:
You need to continuously supply electrical current

There are a few ways to know if a reaction is feasible:

• Calculating Total Entropy Change (as above)
If Total Entropy Change is POSITIVE, it is feasible
• Calculating Gibbs Free Energy Change (see here)
If Gibbs Free Energy Change is NEGATIVE, it is feasible