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Entropy and�The Second Law of Thermodynamics

Reaction Direction

Why does a chemical reaction go naturally in a particular direction and not the other?
When asking this question, we are asking why a reaction is spontaneous in one direction but nonspontaneous in the other direction under certain conditions.
To answer this question, we will have to understand the second law of thermodynamics.

+

Urea reacts spontaneously with water to produce ammonia and carbon dioxide at 25°C and 100 kPa.

The reverse reaction is nonspontaneous and requires an input of energy to occur.

+

Spontaneous Processes

A spontaneous process is a physical or chemical change that occurs by itself – it requires no continuing outside energy to make it happen.

Ball → Rolls downhill
Heat → Flows from hot to cold
Iron → Rusts to iron(III) oxide

A nonspontaneous process requires an input of energy to occur.

Ball → Rolls uphill
Heat → Flows from cold to hot
Iron → Forms from iron(III) oxide

Spontaneous

Nonspontaneous

Hot

Cold

4Fe_(s) + 3O_2(g)

2Fe₂O_3(s)

32

30

32

30

⇀

↽

Spontaneity Versus Rate

Spontaneity tells you which direction a system tends to change, not how fast the change will occur – that is the rate and is determined by other factors.
For example, diamond turns into graphite spontaneously but at an incredibly slow rate.

Diamond

C_(s)

Graphite

C_(s)

spontaneous

rate = 10⁶⁰ years!

Entropy

To explain the spontaneity of a reaction, we will have to understand a quantity called entropy.
Entropy (S) is a thermodynamic quantity that measures how dispersed (i.e., spread out) energy is within a system or surroundings.
When energy is concentrated, the entropy is low.
When the energy is dispersed, the entropy is high.
In spontaneous processes, the entropy of the system plus its surroundings increases.

Energy

Low Entropy

High Entropy

J

Spontaneous

Nonspontaneous

J

Entropy – The Dispersing of Energy

Energy

Entropy

q

J

Entropy – The Dispersing of Energy

Energy

J

Entropy

q

Entropy – The Dispersing of Energy

Energy

Entropy

J

Entropy – The Dispersing of Energy

Energy

Entropy

J

Entropy – The Dispersing of Energy

Energy

Entropy

J

Entropy – The Dispersing of Energy

Energy

Entropy

J

Entropy – The Dispersing of Energy

Energy

Entropy

J

Entropy – The Dispersing of Energy

Energy

J

q

J

Entropy

Entropy – The Dispersing of Energy

Energy

Entropy

q

J

Entropy – The Dispersing of Energy

Entropy increases when:

energy is dispersed among more particles.
energy is dispersed among more spatial locations.
energy is dispersed among more energy modes (kinetic and potential).

There are other ways for energy to be dispersed within a system and surroundings, but this is a good place to start.
The energy in a mole of gas can be dispersed e different ways!

particles

spaces

energy modes

10²⁵

Increasing Entropy

J

Energy is concentrated in the coffee cup.

Increasing Entropy

J

As a cup of coffee cools down, the energy disperses, increasing entropy.

Increasing Entropy

Vacuum

Kinetic energy is concentrated

on left side

Increasing Entropy

Kinetic energy is dispersed

on both sides, increasing entropy.

Increasing Entropy

N₂O_4(g) → 2NO_2(g)

Increasing Entropy

N₂O_4(g) → 2NO_2(g)

The kinetic energy is dispersed among two particles instead of one, so entropy increases.

Increasing Entropy

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When a solute dissolves, the energy disperses in the system, increasing entropy.

The Maxwell-Boltzmann Distribution

Number of Molecules

Kinetic Energy

The Maxwell-Boltzmann Distribution

Number of Molecules

Kinetic Energy

Energy disperses among more kinetic energy modes, increasing entropy.

Increasing Entropy

At 5 °C and 100 kPa, the melting of ice is spontaneous.

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Increasing Entropy

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At 5 °C and 100 kPa, the melting of ice is spontaneous.

This must mean that entropy has increased and, therefore, energy has been dispersed.
In what sense has energy been dispersed when liquid water has the same number of particles and takes up less space than ice?
Entropy also measures how dispersed energy is among energy modes, not just just how dispersed it is among particles and in space.

Increasing Entropy

Kinetic Energy Modes

Vibrational

Rotational

Translational

Kinetic Energy Modes

Vibrational

Rotational

Translational

J

States of Matter and Entropy

As a rule, gases have higher entropy than liquids, and liquids have a higher entropy than solids.

This makes sense as gases have more freedom of movement than liquids, which have more freedom of movement than solids, so there are more ways to disperse energy in the system.

solid

liquid

gas

Increasing Entropy

S_(gas)> S_(liquid)>S_(solid)

Standard Entropies for Elements Across Period 2
Element	Li	Be	B	C	N₂	O₂	F₂	Ne
State	solid	solid	solid	solid	gas	gas	gas	gas
S°(JK^-1mol^-1)	29	10	6	6	192	205	203	146

States of Matter and Entropy Practice

You open a bottle of iodine at room temperature and later notice that the iodine has sublimed.
What can you say about the change in entropy of the iodine?

Iodine (I₂)

253.81 g/mol

!

Causes serious eye irritation

Calculating Entropy Changes

Kinetic Energy Modes

Vibrational

Rotational

Translational

Kinetic Energy Modes

Vibrational

Rotational

Translational

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Entropy = 41.2 JK^-1mol^-1

The SI unit of entropy is joules per kelvin (J/K).
In chemistry, entropy is often expressed as molar quantities at 298 K and 100 kPa – these are called standard entropy values (S°) with the units JK^-1mol^-1.
The entropy change, ΔS, is the final entropy (S_f) minus the initial entropy (S_i).

Entropy = 63.3 K^-1mol^-1

J

ΔS = S_f – S_i

ΔS = 63.3 – 41.2

ΔS = +22.1 JK^-1mol^-1

J

Entropy Increases

H₂O_(s)

H₂O_(l)

5 °C and 100 kPa

Standard Entropy Change

Calculate the standard entropy change (ΔS°) when dry ice sublimates.

We can use standard entropy (also called absolute entropy) values to calculate the standard entropy change (ΔS°) of a system using the following equation.

ΔS° = S_f° + ΔS_i°

Standard Entropy Table
Substance	S° (JK^-1mol^-1)
CO_2(s)	41.1
CO_2(l)	120
CO_2(g)	213.7

CO_2(s) → CO_2(g)

Positive and Negative Heat (q) of System

Heat (q) flowing into the system is given a positive sign.
Heat (q) flowing out of the system is given a negative sign.
The former is endothermic; the latter is exothermic.

Surroundings

System

+q

– q

exothermic

endothermic

Positive and Negative Heat (q) of Surroundings

From the surroundings perspective, the flow of heat is opposite.

Surroundings

System

Surrounding View

Surroundings

System

System View

q_sys = – q_surr

+q

– q

System View

Surrounding View

Heat Flow and Entropy Change

The flow of heat (q) into or out of a system or surroundings results in a change of entropy.
The change in entropy for the surroundings (ΔS_surr) can be calculated using the equation below, where ‘T’ is absolute temperature (K).

Surroundings

System

+q

ΔS_surr

=

q_surr

T_surr

System

– q

Entropy Change and Temperature

The equation to the right shows us that when heat is added to the surroundings, its change in entropy (ΔS_surr) is inversely proportional to the temperature.

ΔS_surr

=

q_surr

T_surr

In other words, adding heat (+q_surr) to cold surroundings increases its entropy more than adding heat hotter surroundings.
Similarly, removing heat (– q_surr) from cold surroundings decreases its entropy more than removing heat from hotter surroundings.

Cold Surroundings

(273 K)

System

+6000 J

ΔS_surr

=

273 K

6000 J

= 22 J/K

Hot Surroundings

(373 K)

System

+6000 J

ΔS_surr

=

373 K

6000 J

= 16 J/K

The Second Law of Thermodynamics

With our understanding of entropy, we can now state the second law of thermodynamics as follows:

In a spontaneous process, the total entropy of the system plus its surroundings always increases.

The total entropy change during a process is the sum of the entropy change of the system plus the entropy change of the surroundings.

ΔS_tot = ΔS_sys + ΔS_surr

This can be stated mathematically as follows:

Spontaneous Process
ΔS_tot > 0

Nonspontaneous Process
ΔS_tot < 0

Practice Problem

An 80 °C cup of coffee cools to room temperature (22 °C).
It loses 60 kJ of heat and its entropy decreases by 188 J/K.
Is this process spontaneous?

Surroundings

(22 °C or 295 K)

System

(80 °C or 353 K)

q

ΔS_tot = ΔS_sys + ΔS_surr

ΔS_sys

=

-188 J/K

ΔS_surr

=

295 K

ΔS_tot =

+

ΔS_tot > 0

spontaneous

-188 J/K

203 J/K

60,000 J

Practice Problem

Is the melting of one mole of ice at -5.0 °C spontaneous or nonspontaneous?

H₂O_(s) → H₂O_(l) ΔH= 6010 Jmol^-1
H₂O_(s) → H₂O_(l) ΔS = 22 JK^-1mol^-1

H₂O_(l)

Spontaneous?

H₂O_(s)

-5.0 °C

ΔS_tot = ΔS_sys + ΔS_surr

q

6010 J

Surroundings

Predicting the Sign of ΔS in a Chemical Reaction

Because gases have significantly higher entropy than solids and liquids, they are the most important factor for determining ΔS of a chemical reaction.

An increase in the number of moles of gas results in an increase in entropy (ΔS > 0).
A decrease in the number of moles of gas results in a decrease in entropy (ΔS < 0).
If the number of moles of gas is unchanged, entropy is also approximately unchanged (ΔS = 0).

Standard Entropies for Elements Across Period 2
Element	Li	Be	B	C	N₂	O₂	F₂	Ne
State	solid	solid	solid	solid	gas	gas	gas	gas
S°(JK^-1mol^-1)	29	10	6	6	192	205	203	146

N_2(g) + 3H_2(g) → 2NH_3(g)

CaCO_3(s)→ CaO_(s) + CO_2(g)

4 mol gas

2 mol gas

0 mol gas

2 mol gas

ΔS < 0

ΔS > 0

Predicting the Sign of ΔS Practice

Predict the sign of ΔS in the reactions below.

Reaction	ΔS
CH_4(s) + 2O_2(g) → CO_2(g) + 2H₂O_(l)
C₂H_4(g) + H_2(g) → C₂H_6(g)
2KClO_3(s) → 2KCl_(s) + 3O_2(g)
2NO_2(g) → N₂O_4(g)
2SO_3(g) → 2SO_2(g) + O_2(g)
COCl_2(g) → CO_(g) + Cl_2(g)
2C_(s) + O_2(g) → 2CO_(g)

Calculating the ΔS° of a Chemical Reaction

The standard entropy change (ΔS°) of a system can be calculated by subtracting the initial entropy value (S_i°) from the final entropy value (S_f°).

ΔS° = S_f° – S_i°

During a chemical change, the system is the chemical reaction – this means the S_f° is the entropy of the products and S_i° is the entropy of the reactants.
We can rewrite the equation as follows:

ΔS° = ∑S°_(products) – ∑S°_(reactants)

We can use the equation above along with standard entropy tables to calculate the standard entropy change for a chemical reaction.

Standard Entropy Values
Substance	S° (JK^-1mol^-1)
N₂	192
H₂	131
NH₃	193

Calculating the ΔS° of a Chemical Reaction

Calculate the standard entropy change for the following reaction.

ΔS° = ∑S°_(products) – ∑S°_(reactants)

Standard Entropy Values
Substance	S° (JK^-1mol^-1)
N₂	192
H₂	131
NH₃	193

N_2(g) + 3H_2(g) → 2NH_3(g)

ΔS°=

2

(193)

–

[

(192)

+

3

(131)

]

ΔS°=

Calculating the ΔS° Practice

Calculate the standard entropy change for the following reactions.

ΔS° = ∑S°_(products₎ – ∑S°_(reactants₎

Standard Entropy Values
Substance	S° (JK^-1mol^-1)
CH_4(g)	186
O_2(g)	103
CO_2(g)	214
H₂O_(l)	70
Cu(NO₃)_2(s)	193
CuO_(s)	43
NO_2(g)	240
BCl_3(l)	206
SF_4(g)	292
BF_3(g)	254
SCl_2(g)	282
Cl_2(g)	83

CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(l)

2Cu(NO₃)_2(s) → 2CuO_(s) + 4NO_2(g) + O_2(g)

4BCl_3(l) + 3SF_4(g) → 4BF_3(g) + 3SCl_2(g) + 3Cl_2(g)

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Surroundings

System

+q

– q

Resource Slide

Surroundings

System

– q

+q

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Kinetic Energy Modes

Vibrational

Rotational

Translational

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