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Chapter 17

Spontaneity, Entropy, �and Free Energy

Section 17.1

Spontaneous Processes and Entropy

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(17.1) Spontaneous processes and entropy
(17.2) Entropy and the second law of thermodynamics
(17.3) The effect of temperature on spontaneity
(17.4) Free energy
(17.5) Entropy changes in chemical reactions
(17.6) Free energy and chemical reactions
(17.7) The dependence of free energy on pressure
(17.8) Free energy and equilibrium
(17.9) Free energy and work

Chapter 17

Table of Contents

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Thermodynamics - An Introduction

First law of thermodynamics states that the energy of the universe is constant

Statement of law of conservation of energy

Spontaneous process: Occurs without external intervention

Can be fast or slow

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Section 17.1

Spontaneous Processes and Entropy

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Thermodynamics versus Kinetics

Domain of kinetics

Rate of a reaction depends on the pathway from reactants to products

Thermodynamics

Provides information on whether a reaction is spontaneous based only on the properties of the reactants and products

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Section 17.1

Spontaneous Processes and Entropy

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Entropy (S)

Thermodynamic function that describes the number of arrangements that are available to a system existing in a given state
Measure of molecular randomness or disorder

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Section 17.1

Spontaneous Processes and Entropy

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The Expansion of an Ideal Gas into an Evacuated Bulb

Nature spontaneously proceeds toward the states that have the highest probabilities of existing

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Section 17.1

Spontaneous Processes and Entropy

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Microstate

Each configuration that gives a particular arrangement
Probability of occurrence of a state depends on the number of microstates in which the arrangement can be achieved

Section 17.1

Spontaneous Processes and Entropy

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Table 17.1 - The Microstates That Give a Particular Arrangement (State)

Section 17.1

Spontaneous Processes and Entropy

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Table 17.1 - The Microstates That Give a Particular Arrangement (State) (Continued)

Section 17.1

Spontaneous Processes and Entropy

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Positional Probability

Depends on the number of configurations in space that yield a particular state
Gas expands into a vacuum to give a uniform distribution

Expanded state has the highest positional probability of the states available to the system

Section 17.1

Spontaneous Processes and Entropy

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Positional Probability and Changes of State

Positional entropy increases when going from solid to gaseous state

Section 17.1

Spontaneous Processes and Entropy

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Positional Entropy and Solutions

Entropy change when mixing two pure substances is expected to be positive

Result of the presence of more microstates for the mixed condition
Caused due to the increased volume available to a given particle after mixing occurs

Formation of solutions is favored by an increase in positional entropy that is associated with mixing

Section 17.1

Spontaneous Processes and Entropy

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Interactive Example 17.1 - Positional Entropy

For each of the following pairs, choose the substance with the higher positional entropy (per mole) at a given temperature

Solid CO₂ and gaseous CO₂
N₂ gas at 1 atm and N₂ gas at 1.0×10^–2 atm

Section 17.1

Spontaneous Processes and Entropy

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Interactive Example 17.1 - Solution (a)

Since a mole of gaseous CO₂ has the greater volume by far, the molecules have many more available positions than in a mole of solid CO₂

Thus, gaseous CO₂ has the higher positional entropy

Section 17.1

Spontaneous Processes and Entropy

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Interactive Example 17.1 - Solution (b)

A mole of N₂ gas at 1.0×10^–2 atm has a volume 100 times that (at a given temperature) of a mole of N₂ gas at 1 atm

Thus, N₂ gas at 1.0×10^–2 atm has the higher positional entropy

Section 17.1

Spontaneous Processes and Entropy

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Interactive Example 17.2 - Predicting Entropy Changes

Predict the sign of the entropy change for each of the following processes

Solid sugar is added to water to form a solution
Iodine vapor condenses on a cold surface to form crystals

Section 17.1

Spontaneous Processes and Entropy

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Interactive Example 17.2 - Solution (a)

Sugar molecules become randomly dispersed in the water when the solution forms and thus have access to a larger volume and a larger number of possible positions

Positional disorder is increased, and there will be an increase in entropy
ΔS is positive, since the final state has a larger entropy than the initial state, and ΔS = S_final – S_initial

Section 17.1

Spontaneous Processes and Entropy

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Interactive Example 17.2 - Solution (b)

Gaseous iodine is forming a solid

This process involves a change from a relatively large volume to a much smaller volume, which results in lower positional disorder
For this process ΔS is negative, implying that the entropy decreases

Section 17.1

Spontaneous Processes and Entropy

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Second Law of Thermodynamics

In any spontaneous process, there is always an increase in the entropy of the universe
First law of thermodynamics

Energy of the universe is constant
Energy is conserved, entropy is not

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Section 17.2

Entropy and the Second Law of Thermodynamics

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Changes in Entropy of the Universe

ΔS_univ is positive

Entropy of the universe increases
Process is spontaneous in the direction written

ΔS_univ is negative

Process is spontaneous in the opposite direction

ΔS_univ is zero

Process has no tendency to occur
System is at equilibrium

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Section 17.2

Entropy and the Second Law of Thermodynamics

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Critical Thinking

What if ΔS_univ was a state function?

How would the world be different?

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Section 17.2

Entropy and the Second Law of Thermodynamics

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Example 17.3 - The Second Law

In a living cell, large molecules are assembled from simple ones

Is this process consistent with the second law of thermodynamics?

Section 17.2

Entropy and the Second Law of Thermodynamics

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Example 17.3 - Solution

To reconcile the operation of an order-producing cell with the second law of thermodynamics, we must remember that ΔS_univ, not ΔS_sys, must be positive for a process to be spontaneous
A process for which ΔS_sys is negative can be spontaneous if the associated ΔS_surr is both larger and positive

The operation of a cell is such a process

Section 17.2

Entropy and the Second Law of Thermodynamics

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Entropy Changes in the Surroundings (ΔS_surr)

ΔS_surr is determined by flow of energy as heat
Exothermic process increases ΔS_surr

Important driving force for spontaneity

Endothermic process decreases ΔS_surr
Impact of transfer of energy as heat to or from the surroundings is greater at lower temperatures

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Section 17.3

The Effect of Temperature on Spontaneity

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Entropy Changes in the Surroundings (ΔS_surr) (Continued)

Sign of ΔS_surr depends on the direction of the heat flow

At constant temperature:

ΔS_surrfor exothermic processes is positive
ΔS_surrfor endothermic processes is negative

Magnitude of ΔS_surrdepends on the temperature

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Section 17.3

The Effect of Temperature on Spontaneity

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Characteristics of the Entropy Changes That Occur in the Surroundings - Summary

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Section 17.3

The Effect of Temperature on Spontaneity

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ΔS_surr and ΔH

ΔS_surr can be expressed in terms of ΔH for a process occurring at constant pressure

Heat flow = change in enthalpy = ΔH

Components of ΔH

Sign - Indicates the direction of flow

Determined in accordance with the reaction system

Number - Indicates the quantity of energy

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Section 17.3

The Effect of Temperature on Spontaneity

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ΔS_surr and ΔH (Continued)

Reaction takes place under conditions of constant temperature (in Kelvins) and pressure

If the reaction is exothermic:

ΔH has a negative sign
ΔS_surr is positive since heat flows into the surroundings

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Section 17.3

The Effect of Temperature on Spontaneity

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Table 17.3 - Interplay of ΔS_sys and ΔS_surr in Determining the Sign of ΔS_univ

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Section 17.3

The Effect of Temperature on Spontaneity

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Interactive Example 17.4 - Determining ΔS_surr

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

Section 17.3

The Effect of Temperature on Spontaneity

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Interactive Example 17.4 - Determining ΔS_surr(Continued)

Carbon is used as the reducing agent for oxide ores:

Calculate ΔS_surr for each of these reactions at 25°C and 1 atm

Section 17.3

The Effect of Temperature on Spontaneity

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Interactive Example 17.4 - Solution

For the sulfide ore reaction,

ΔS_surris positive since this reaction is exothermic, and heat flow occurs to the surroundings, increasing the randomness of the surroundings

Section 17.3

The Effect of Temperature on Spontaneity

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Interactive Example 17.4 - Solution (Continued)

For the oxide ore reaction,

In this case ΔS_surris negative because heat flow occurs from the surroundings to the system

Section 17.3

The Effect of Temperature on Spontaneity

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Free Energy (G)

H - Enthalpy
T - Temperature in K
S - Entropy

At constant temperature,

All quantities refer to the system

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Section 17.4

Free Energy

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Relationship between Free Energy (G) and Spontaneity

Divide both sides of the equation ΔG = ΔH – TΔS by –T

At constant temperature (T) and pressure (P),

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Section 17.4

Free Energy

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Relationship between Free Energy (G) and Spontaneity (Continued)

Processes that occur at constant T and P are spontaneous in the direction in which the free energy decreases

Negative ΔG means positive ΔS_univ

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Section 17.4

Free Energy

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Table 17.5 - Various Possible Combinations of ΔH and ΔS

Section 17.4

Free Energy

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Critical Thinking

Consider an ideal gas in a container fitted with a frictionless, massless piston

What if weight is added to the top of the piston?
We would expect the gas to be compressed at constant temperature
For this to be true, ΔS would be negative (since the gas is compressed) and ΔH would be zero (since the process is at constant temperature) �

Section 17.4

Free Energy

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Critical Thinking (Continued)

This would make ΔG positive
Does this mean the isothermal compression of the gas is not spontaneous?
Defend your answer�

Section 17.4

Free Energy

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Interactive Example 17.5 - Free Energy and Spontaneity

At what temperatures is the following process spontaneous at 1 atm?

What is the normal boiling point of liquid Br₂?

Section 17.4

Free Energy

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Interactive Example 17.5 - Solution

The vaporization process will be spontaneous at all temperatures where ΔG°is negative

Note that ΔS°favors the vaporization process because of the increase in positional entropy, and ΔH°favors the opposite process, which is exothermic
These opposite tendencies will exactly balance at the boiling point of liquid Br₂, since at this temperature liquid and gaseous Br₂ are in equilibrium (ΔG°= 0)

Section 17.4

Free Energy

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Interactive Example 17.5 - Solution (Continued 1)

We can find this temperature by setting ΔG°= 0 in the following equation:

Section 17.4

Free Energy

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Interactive Example 17.5 - Solution (Continued 2)

At temperatures above 333 K, TΔS°has a larger magnitude than ΔH°, and ΔG°is negative

Above 333 K, the vaporization process is spontaneous
The opposite process occurs spontaneously below this temperature

At 333 K, liquid and gaseous Br₂ coexist in equilibrium

Section 17.4

Free Energy

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Interactive Example 17.5 - Solution (Continued 3)

Summary of observations (the pressure is 1 atm in each case)

T > 333 K

The term ΔS°controls, and the increase in entropy when liquid Br₂ is vaporized is dominant

T < 333 K

The process is spontaneous in the direction in which it is exothermic, and the term ΔH°controls

Section 17.4

Free Energy

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Interactive Example 17.5 - Solution (Continued 4)

T = 333 K

The opposing driving forces are just balanced (ΔH°= 0), and the liquid and gaseous phases of bromine coexist
This is the normal boiling point

Section 17.4

Free Energy

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Exercise

Ethanethiol (C₂H₅SH; also called ethyl mercaptan) is commonly added to natural gas to provide the “rotten egg” smell of a gas leak

The boiling point of ethanethiol is 35°C and its heat of vaporization is 27.5 kJ/mol
What is the entropy of vaporization for this substance?

89.3 J/K·mol

Section 17.4

Free Energy

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Entropy Changes and Chemical Reactions

Positional probability determines the changes that occur in a chemical system
Fewer the molecules, fewer the possible configurations

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Section 17.5

Entropy Changes in Chemical Reactions

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Entropy Changes in Reactions That Involve Gaseous Molecules

Change in positional entropy is dominated by the relative numbers of molecules of gaseous reactants and products

If the number of product molecules is greater than the number of reactant molecules:

Positional entropy increases
ΔSis positive

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Section 17.5

Entropy Changes in Chemical Reactions

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Interactive Example 17.6 - Predicting the Sign of ΔS°

Predict the sign of ΔS°for each of the following reactions

Thermal decomposition of solid calcium carbonate

Oxidation of SO₂ in air

Section 17.5

Entropy Changes in Chemical Reactions

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Interactive Example 17.6 - Solution

Since in this reaction a gas is produced from a solid reactant, the positional entropy increases, and ΔS°is positive
Here three molecules of gaseous reactants become two molecules of gaseous products

Since the number of gas molecules decreases, positional entropy decreases, and ΔS°is negative

Section 17.5

Entropy Changes in Chemical Reactions

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Entropy Changes

Third law of thermodynamics

Entropy of a perfect crystal at 0 K is zero

Entropy of a substance increases with temperature

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Section 17.5

Entropy Changes in Chemical Reactions

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Entropy Values

Standard entropy values (S°) represent increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm

More complex the molecule, the higher the standard entropy value

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Section 17.5

Entropy Changes in Chemical Reactions

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Entropy Change for a Given Chemical Reaction

Entropy is a state function of a chemical system
Entropy changes can be calculated as follows:

Σ - Sum of all terms
n_r - Number of moles of a reactant
n_p- Number of moles of a product

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Section 17.5

Entropy Changes in Chemical Reactions

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Figure 17.6 - Entropy of Water

H₂O molecule can vibrate and rotate in several ways

Freedom of motion leads to a higher entropy for water

Section 17.5

Entropy Changes in Chemical Reactions

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Interactive Example 17.7 - Calculating ΔS°

Calculate ΔS°at 25°C for the following reaction:

The following information is given:

Section 17.5

Entropy Changes in Chemical Reactions

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Interactive Example 17.7 - Solution

Section 17.5

Entropy Changes in Chemical Reactions

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Interactive Example 17.7 - Solution (Continued)

We would expect ΔS°to be negative because the number of gaseous molecules decreases in this reaction

Section 17.5

Entropy Changes in Chemical Reactions

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Exercise

Predict the sign of ΔS^ₒ and then calculate ΔS°for each of the following reactions:

Negative

ΔS°= –186 J/K

Positive

ΔS°= 187 J/K

Hard to predict since Δn = 0

ΔS°= 138 J/K

Section 17.5

Entropy Changes in Chemical Reactions

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Standard Free Energy Change (ΔG°)

Change in G that will occur if the reactants in their standard states are converted to the products in their standard states

More negative the value of ΔG°, the further the reaction shifts to the right to attain equilibrium

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Section 17.6

Free Energy and Chemical Reactions

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Methods for Calculating ΔG°

Use the following formula

Treat free energy as a state function and use Hess’s law
Use standard free energy of formation

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Section 17.6

Free Energy and Chemical Reactions

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Interactive Example 17.9 - Calculating ΔH°, ΔS°, and ΔG°

Consider the following reaction carried out at 25°C and 1 atm:

Calculate ΔH^ₒ, ΔS^ₒ, and ΔG^ₒusing the following data:

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Section 17.6

Free Energy and Chemical Reactions

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Interactive Example 17.9 - Solution

The value of ΔH°can be calculated from the enthalpies of formation using the following formula:

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Section 17.6

Free Energy and Chemical Reactions

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Interactive Example 17.9 - Solution (Continued 1)

The value of ΔS°can be calculated using the following formula:

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Section 17.6

Free Energy and Chemical Reactions

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Interactive Example 17.9 - Solution (Continued 2)

We would expect ΔS°to be negative because three molecules of gaseous reactants give two molecules of gaseous products

The value of ΔG°can now be calculated

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Section 17.6

Free Energy and Chemical Reactions

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Interactive Example 17.10 - Calculating ΔG°

Use the following data (at 25°C):

Calculate ΔG°for the following reaction:

(17.5)

(17.6)

Section 17.6

Free Energy and Chemical Reactions

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Interactive Example 17.10 - Solution

Reverse equation (17.6) to make graphite a product, as required, and then add the new equation to equation (17.5)

Reversed Equation (17.6)

Section 17.6

Free Energy and Chemical Reactions

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Interactive Example 17.10 - Solution (Continued 1)

Since ΔG°is negative for this process, diamond should spontaneously change to graphite at 25°C and 1 atm

However, the reaction is so slow under these conditions that we do not observe the process
This is another example of kinetic rather than thermodynamic control of a reaction

Section 17.6

Free Energy and Chemical Reactions

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Interactive Example 17.10 - Solution (Continued 2)

We can say that diamond is kinetically stable with respect to graphite even though it is thermodynamically unstable

Section 17.6

Free Energy and Chemical Reactions

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Standard Free Energy of Formation ( )

Change in free energy that accompanies the formation of 1 mole of a substance from its constituent elements

All reactants and products are in their standard states
Used to calculate the free energy change for a reaction

of an element in its standard state = 0

Section 17.6

Free Energy and Chemical Reactions

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Interactive Example 17.11 - Calculating ΔG°

Methanol is a high-octane fuel used in high-performance racing engines

Calculate ΔG °for the following reaction:

The following energies of formation are provided:

Section 17.6

Free Energy and Chemical Reactions

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Interactive Example 17.11 - Solution

Use the following equation:

Section 17.6

Free Energy and Chemical Reactions

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Interactive Example 17.11 - Solution (Continued)

The large magnitude and the negative sign of ΔG°indicate that this reaction is very favourable thermodynamically

Section 17.6

Free Energy and Chemical Reactions

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Free Energy and Pressure

System under constant P and T proceeds spontaneously in the direction that lowers its free energy

Free energy of a reaction system changes as the reaction proceeds

Dependent on the pressure of a gas or on the concentration of species in solution

Equilibrium - Point where free energy value is at its lowest

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Section 17.7

The Dependence of Free Energy on Pressure

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Free Energy and Pressure (Continued 1)

For ideal gases:

Enthalpy is not pressure-dependent
Entropy depends on pressure due to its dependence on volume

At a given temperature for 1 mole of ideal gas:

S_{large volume} > S_{small volume}

Or,

S_{low pressure}> S_{high pressure}

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Section 17.7

The Dependence of Free Energy on Pressure

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Free Energy and Pressure (Continued 2)

G° - Free energy of a gas at 1 atm
G - Free energy of the gas at a pressure of P atm
R - Universal gas constant
T - Temperature in Kelvin

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Section 17.7

The Dependence of Free Energy on Pressure

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Free Energy and Pressure (Continued 3)

Q - Reaction quotient
T - Temperature in Kelvin
R - Universal gas constant (8.3145 J/K·mol)
ΔG°- Free energy change at 1 atm
ΔG - Free energy change at specified pressures

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Section 17.7

The Dependence of Free Energy on Pressure

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Interactive Example 17.13 - Calculating ΔG°

One method for synthesizing methanol (CH₃OH) involves reacting carbon monoxide and hydrogen gases

Calculate ΔGat 25°C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol

Section 17.7

The Dependence of Free Energy on Pressure

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Interactive Example 17.13 - Solution

To calculate ΔG for this process, use the following equation:

First compute ΔG^ₒ from standard free energies of formation

Section 17.7

The Dependence of Free Energy on Pressure

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Interactive Example 17.13 - Solution (Continued 1)

One might call this the value of ΔG°for one round of the reaction or for 1 mole of the reaction
Thus, the ΔG°value might better be written as –2.9×10⁴ J/mol of reaction, or –2.9×10⁴ J/mol rxn

Use this value to calculate the value of ΔG

Section 17.7

The Dependence of Free Energy on Pressure

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Interactive Example 17.13 - Solution (Continued 2)

ΔG°= – 2.9×10⁴ J/mol rxn
R = 8.3145 J/K·mol
T = 273 + 25 = 298 K

Note that the pure liquid methanol is not included in the calculation of Q

Section 17.7

The Dependence of Free Energy on Pressure

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Interactive Example 17.13 - Solution (Continued 3)

Section 17.7

The Dependence of Free Energy on Pressure

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Interactive Example 17.13 - Solution (Continued 4)

Note that ΔG is significantly more negative than ΔG°, implying that the reaction is more spontaneous at reactant pressures greater than 1 atm

This result can be expected from Le Châtelier’s principle

Section 17.7

The Dependence of Free Energy on Pressure

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The Meaning of ΔG for a Chemical Reaction

Even when value of ΔG provides information regarding whether the system is favored under a given set of conditions:

System may not proceed to pure products (if ΔG is negative)
System may not remain at pure reactants (if ΔG is positive)

A system will spontaneously seek equilibrium

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Section 17.7

The Dependence of Free Energy on Pressure

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The Meaning of ΔG for a Chemical Reaction (Continued)

System can achieve the lowest possible free energy by going to equilibrium, not by going to completion

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Section 17.7

The Dependence of Free Energy on Pressure

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Equilibrium Point

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Section 17.8

Free Energy and Equilibrium

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Figure 17.8 - Equilibrium Point

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Section 17.8

Free Energy and Equilibrium

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Equilibrium Point (Continued 1)

When substances undergo a chemical reaction, the reaction proceeds to the minimum free energy (equilibrium)

This corresponds to the point where:

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Section 17.8

Free Energy and Equilibrium

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Equilibrium Point (Continued 2)

Quantitative relationship between free energy and the value of the equilibrium constant is given by:

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Section 17.8

Free Energy and Equilibrium

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Figure 17.9 - Plot of Energy versus the Mole Fraction of the reaction of A(g)

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Section 17.8

Free Energy and Equilibrium

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Table 17.6 - Qualitative Relationship between the ΔG°and the K for a Given Reaction

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Section 17.8

Free Energy and Equilibrium

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Interactive Example 17.15 - Free Energy and Equilibrium II

Section 17.8

Free Energy and Equilibrium

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Interactive Example 17.15 - Solution

Calculate ΔG° from ΔG° = ΔH°– TΔS°

Section 17.8

Free Energy and Equilibrium

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Interactive Example 17.15 - Solution (Continued 1)

Therefore,

ΔG° = ΔH°– TΔS°

= (– 1.652×10⁶ J) – (298 K) (– 543 J/K)

= – 1.490 ×10⁶ J

Section 17.8

Free Energy and Equilibrium

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Interactive Example 17.15 - Solution (Continued 2)

Therefore,

K = e⁶⁰¹

This is a very large equilibrium constant

The rusting of iron is clearly very favourable from a thermodynamic point of view

Section 17.8

Free Energy and Equilibrium

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The Temperature Dependence of K

Quantitative dependence of K on temperature is given by:

y = mx + b

Section 17.8

Free Energy and Equilibrium

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The Temperature Dependence of K (Continued)

If the values of K are determined at various levels of T:

Plot of ln(K) versus 1/T will be linear
Slope = – ΔH°/R
Intercept = ΔS°/R

Assume that ΔH°and ΔS°are independent of T

Section 17.8

Free Energy and Equilibrium

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Relationship Between Free Energy and Work

Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy

Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway

Any real pathway wastes energy

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Section 17.9

Free Energy and Work

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Reversible and Irreversible Processes

Reversible process: Universe is exactly the same as it was before a cyclic process
Irreversible process: Universe is different after a cyclic process

All real processes are irreversible

Characteristics of a real cyclic process

Work is changed to heat
Entropy of the universe increases

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Section 17.9

Free Energy and Work

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Figure 17.11 - Reversible Process as Seen in a Battery

Section 17.9

Free Energy and Work

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Revisiting the Laws of Thermodynamics

First law

You can’t win; you can only break even

Second law

You can’t break even

As we use energy, we degrade its usefulness

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Section 17.9

Free Energy and Work

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Critical Thinking

What if the first law of thermodynamics was true but the second law was not?

How would the world be different?

Section 17.9

Free Energy and Work