chapter 18 thermodynamics and equilibrium 18 | 2 contents and concepts 1.first law of thermodynamics...

Chapter 18Thermodynamics and Equilibrium

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Contents and Concepts

1. First Law of Thermodynamics

2. Entropy and the Second Law of Thermodynamics

3. Standard Entropies and the Third Law of Thermodynamics

Spontaneous Processes and EntropyA spontaneous process is one that occurs by itself. As we will see, the entropy of the system increases in a spontaneous process.

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Free-Energy ConceptThe quantity H – TS can function as a criterion for the spontaneity of a reaction at constant temperature, T, and pressure, P. By defining a quantity called the free energy, G = H – TS, we find that G equals the quantity H – TS, so the free energy gives us a thermodynamic criterion of spontaneity.

4. Free Energy and Spontaneity

5. Interpretation of Free Energy

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Free Energy and Equilibrium ConstantsThe total free energy of the substances in a reaction mixture decreases as the reaction proceeds. As we discuss, the standard free-energy change for a reaction is related to its equilibrium constant.

6. Relating G° to the Equilibrium Constant

7. Change of Free Energy with Temperature

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1. First Law of Thermodynamics; Enthalpy

a. Define internal energy, state function, work, and first law of thermodynamics.

b. Explain why the work done by the system as a result of expansion or contraction during a chemical reaction is -PV.

Learning Objectives

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1. First Law of Thermodynamics; Enthalpy (cont)

c. Relate the change of internal energy, U, and heat of reaction, q.

d. Define enthalpy, H.

e. Show how heat of reaction at constant pressure, qp, equals the change of enthalpy, H.

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Spontaneous Processes and Entropy

2. Entropy and the Second Law of Thermodynamics

a. Define spontaneous process.

b. Define entropy.

c. Relate entropy to disorder in a molecular system (energy dispersal).

d. State the second law of thermodynamics in terms of system plus surroundings.

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2. Entropy and the Second Law of Thermodynamics (cont)

e. State the second law of thermodynamics in terms of the system only.

f. Calculate the entropy change for a phase transition.

g. Describe how H - TS functions as a criterion of a spontaneous reaction.

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3. Standard Entropies and the Third Law of Thermodynamics

a. State the third law of thermodynamics.

b. Define standard entropy (absolute entropy).

c. State the situations in which the entropy usually increases.

d. Predict the sign of the entropy change of a reaction.

e. Express the standard change of entropy of a reaction in terms of standard entropies of products and reactants.

f. Calculate So for a reaction.

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Free-Energy Concept

4. Free Energy and Spontaneity

a. Define free energy, G.

b. Define the standard free-energy change.

c. Calculate Go from Ho and So.

d. Define the standard free energy of formation, Go.

e. Calculate Go from standard free energies of formation.

f. State the rules for using Go as a criterion for spontaneity

g. Interpret the sign of Go.

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5. Interpretation of Free Energy

a. Relate the free-energy change to maximum useful work.

b. Describe how the free energy changes during a chemical reaction.

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Free Energy and Equilibrium Constants

6. Relating Go to the Equilibrium Constant

a. Define the thermodynamic equilibrium constant, K.

b. Write the expression for a thermodynamic equilibrium constant.

c. Indicate how the free-energy change of a reaction and the reaction quotient are related.

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6. Relating Go to the Equilibrium Constant (cont)

d. Relate the standard free-energy change to the thermodynamic equilibrium constant.

e. Calculate K from the standard free-energy change (molecular equation).

f. Calculate K from the standard free-energy change (net ionic equation).

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7. Change of Free Energy with Temperature

a. Describe how Go at a given temperature (Go

T) is approximately related to Ho and So at that temperature.

b. Describe how the spontaneity or nonspontaneity of a reaction is related to each of the four possible combinations of signs of Ho and So.

c. Calculate Go and K at various temperatures.

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Thermodynamics is the study of heat and other forms of energy involved in chemical or physical processes.

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First Law of Thermodynamics

The first law of thermodynamics is essentially the law of conservation of energy applied to a thermodynamic system.

Recall that the internal energy, U, is the sum of the kinetic and potential energies of the particles making up the system:

U = Uf – Ui

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Exchanges of energy between the system and its surroundings are of two types: heat, q, and work, w.

Putting this in an equation gives us the first law of thermodynamics.

U = q + w

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Sign Convention for q

When heat is evolved by the system, q is negative. This decreases the internal energy of the system.

When heat is absorbed by the system, q is positive. This increases the internal energy of the system.

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Sign Convention for w

Recall from Chapter 6 that w = –PV.

When the system expands, V is positive, so w is negative. The system does work on the surroundings, which decreases the internal energy of the system.

When the system contracts, V is negative, so w is positive. The surroundings do work on the system, which increases the internal energy of the system.

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Here the system expands and evolves heat from A to B.

Zn2+(aq) + 2Cl-(aq) + H2(g)

V is positive, so work is negative.

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At constant pressure: qP = H

The first law of thermodynamics can now be expressed as follows:

U = H – PV

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To understand why a chemical reaction goes in a particular direction, we need to study spontaneous processes.

A spontaneous process is a physical or chemical change that occurs by itself. It does not require any outside force, and it continues until equilibrium is reached.

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A ball will roll downhill spontaneously. It will eventually reach equilibrium at the bottom.

A ball will not roll uphill spontaneously. It requires work.

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The first law of thermodynamics cannot help us to determine whether a reaction is spontaneous as written.

We need a new quantity—entropy.

Entropy, S, is a thermodynamic quantity that is a measure of how dispersed the energy of a system is among the different possible ways that system can contain energy.

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Examining some spontaneous processes will clarify this definition.

First, heat energy from a cup of hot coffee spontaneously flows to its surroundings—the table top, the air around the cup, or your hand holding the cup. The entropy of the system (the cup of hot coffee) and its surroundings has increased.

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The rock rolling down the hill is a bit more complicated. As it rolls down, the rock’s potential energy is converted to kinetic energy. As it collides with other rocks on the way down, it transfers energy to them. The entropy of the system (the rock) and its surroundings has increased.

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Now consider a gas in a flask connected to an equal-sized flask that is evacuated. When the stopcock is open, the gas will flow into the evacuated flask. The kinetic energy has spread out, and the entropy of the system has increased.

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In each of the preceding examples, energy has been dispersed (that is, spread out).

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Entropy is a state function. It depends on variables, such as temperature and pressure, that determine the state of the substance.

Entropy is an extensive property. It depends on the amount of substance present.

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Entropy is measured in units of J/K.

Entropy change is calculated as follows:

S = Sf – Si

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The iodine has spread out, so its entropy has increased.

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Second Law of Thermodynamics

The total entropy of a system and its surroundings always increases for a spontaneous process.

Note: Entropy is a measure of energy dispersal, not a measure of energy itself.

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For a spontaneous process at a constant temperature, we can state the second law of thermodynamics in terms of only the system:

S = entropy created +

For a spontaneous process:

S >

T

q

T

q

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A pendulum is put in motion, with all of its molecules moving in the same direction, as shown in Figures A and B.

As it moves, the pendulum collides with air molecules.

When it comes to rest in Figure C, the pendulum has dispersed its energy. This is a spontaneous process.

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Now consider the reverse process, which is not spontaneous. While this process does not violate the first law of thermodynamics, it does violate the second law because the dispersed energy becomes more concentrated as the molecules move together.

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Entropy and Molecular Disorder

Entropy is essentially related to energy dispersal. The entropy of a molecular system may be concentrated in a few energy states and later dispersed among many more energy states. The energy of such a system increases.

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In the case of the cup of hot coffee, as heat moves from the hot coffee, molecular motion becomes more disordered. In becoming more disordered, the energy is more dispersed.

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Likewise, when the gas moves from one container into the evacuated container, molecules become more disordered because they are dispersed over a larger volume. The energy of the system is dispersed over a larger volume.

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When ice melts, the molecules become more disordered, again dispersing energy more widely.

When one molecule decomposes to give two, as in

N2O4(g) 2NO2(g)

more disorder is created because the two molecules produced can move independently of each other. Energy is more dispersed.

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In each of these cases, molecular disorder increases, as does entropy.

Note: This understanding of entropy as disorder applies only to molecular situations in which increasing disorder increases the dispersion of energy. It cannot be applied to situations that are not molecular—such as a desk.

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Entropy Change for a Phase Transition

In a phase transition process that occurs very close to equilibrium, heat is very slowly absorbed or evolved. Under these conditions, no significant new entropy is created.

S =

This concept can be applied to melting using Hfus for q and to vaporization using Hvap for q.

T

q

?

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Acetone, CH3COCH3, is a volatile liquid solvent; it is used in nail polish, for example. The standard enthalpy of formation, Hf°, of the liquid at 25°C is –247.6 kJ/mol; the same quantity for the vapor is –216.6 kJ/mol.

What is the entropy change when 1.00 mol liquid acetone vaporizes at 25°C?

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CH3COCH3(l) CH3COCH3(g)

Hf° –247.6 kJ/mol –216.6 kJ/mol

n 1 mol 1 mol

nHf° –247.6 kJ –216.6 kJ

H° = 31.0 kJ

T

HS

Δ

K 298

J10 31.0Δ

3S

S = 104 J/K

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Criterion for a Spontaneous Reaction

The criterion is that the entropy of the system and its surroundings must increase.

HST ΔΔ

T

HS

ΔΔ

0ΔΔor ΔΔ0 ST-HST -H

T

qSΔ

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Third Law of Thermodynamics

A substance that is perfectly crystalline at zero Kelvin (0 K) has an entropy of zero.

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The standard entropy of a substance—its absolute entropy, S°—is the entropy value for the standard state of the species. The standard state is indicated with the superscript degree sign.

For a pure substance, its standard state is 1 atm pressure. For a substance in solution, its standard state is a 1 M solution.

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Standard Entropy of Bromine, Br2, at Various Temperatures

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Entropy Change for a Reaction

Entropy usually increases in three situations:

1. A reaction in which a molecule is broken into two or more smaller molecules.

2. A reaction in which there is an increase in the number of moles of a gas.

3. A process in which a solid changes to a liquid or gas or a liquid changes to a gas.

?

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The opening of Chapter 6, on thermo-chemistry, describes the endothermic reaction of solid barium hydroxide octahydrate and solid ammonium nitrate:

Ba(OH)2 8H2O(s) + 2NH4NO3(s) 2NH3(g) + 10H2O(l) + Ba(NO3)2(aq)

Predict the sign of S° for this reaction.

3 moles of reactants produces 13 moles of products. Solid reactants produce gaseous, liquid, and aqueous products.

S° is positive.

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To compute S° where n = moles:

)(reactants(products)Δ nSnSS

?

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When wine is exposed to air in the presence of certain bacteria, the ethyl alcohol is oxidized to acetic acid, giving vinegar. Calculate the entropy change at 25°C for the following similar reaction:

CH3CH2OH(l) + O2(g) CH3COOH(l) + H2O(l)

The standard entropies, S°, of the substances in J/(K mol) at 25°C are CH3CH2OH(l),161; O2(g), 205; CH3COOH(l), 160; H2O(l), 69.9.

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CH3CH2OH(l) + O2(g) CH3COOH(l) + H2O(l)

S = 229.9 J/K – 366 J/K

S = –136 J/K

S° J/(mol K) 161 205 160 69.9

n mol 1 1 1 1

nS° J/K 161 205 160 69.9

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Free Energy and Spontaneity

Physicist J. Willard Gibbs introduced the concept of free energy, G. Free energy is a thermodynamic quantity defined as follows:

G = H – TS

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As a reaction proceeds, G changes:

G = H – TS

Standard free energy change:

G° = H° – TS°

?

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Using standard enthalpies of formation, Hf° and the value of S° from the previous problem, calculate G° for the oxidation of ethyl alcohol to acetic acid.

CH3CH2OH(l) + O2(g) CH3COOH(l) + H2O(l)

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CH3CH2OH(l) + O2(g) CH3COOH(l) + H2O(l)

Hf° kJ/mol –277.6 0 –487.0 –285.8

n mol 1 1 1 1

nHf° kJ –277.6 0 –487.0 –285.8 –277.6 kJ –772.8 kJ

H° = –495.2 kJ

S° = –136 J/K

T = 298 K

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H° = –495.2 kJ

S° = –136 J/K = –0.136 kJ/K

T = 298 K

G° = H° – TS°

G° = –495.2 kJ – (298 K)(–0.136 kJ/K)

G° = –495.2 kJ + 40.5 kJ

G° = –454.7 kJ

The reaction is spontaneous.

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Standard Free Energies of Formation, Gf°

The standard free energy of formation is the free-energy change that occurs when 1 mol of substance is formed from its elements in their standard states at 1 atm and at a specified temperature, usually 25°C.

The corresponding reaction for the standard free energy of formation is the same as that for standard enthalpy of formation, Hf°.

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To find the standard free energy change for a reaction where n = moles:

)(reactantsΔ(products)ΔΔ GnGnG

?

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Calculate the free-energy change, G°, for the oxidation of ethyl alcohol to acetic acid using standard free energies of formation.

CH3CH2OH(l) + O2(g) CH3COOH(l) + H2O(l)

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CH3CH2OH(l) + O2(g) CH3COOH(l) + H2O(l)

Gf°, kJ/mol –174.8 0 –392.5 –237.2

n, mol 1 1 1 1

nGf°, kJ –174.8 0 –392.5 –237.2 –174.8 kJ –629.7 kJ

G° = –454.9 kJ

G° = –629.7 – (–174.8)

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G° as a Criterion for Spontaneity

The spontaneity of a reaction can now be determined by the sign of G°.

G° < –10 kJ: spontaneous

G° > +10 kJ: nonspontaneous

G° = very small or zero (< +10 kJ and > –10 kJ): at equilibrium

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For the reaction as written, G° = 173.20 kJ.

For 1 mol NO(g), Gf° = 86.60 kJ/mol.

The reaction is nonspontaneous.

; G° = 173.2 kJ

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Interpreting Free Energy

Theoretically, spontaneous reactions can be used to perform useful work. In fact, we use reactions such as the combustion of gasoline to move a vehicle.

We can also use spontaneous reactions to provide the energy needed for a nonspontaneous reaction.

The maximum useful work is

wmax = G

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The thermodynamic equilibrium constant is the equilibrium constant in which the concentrations of gases are expressed as partial pressures in atmospheres and the concentrations of solutes in solutions are expressed in molarities.

If only gases are present, K = Kp.

If only solutes in liquid solution are present, K = Kc.

?

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Write the expression for the thermodynamic equilibrium constant for these reactions:

a. N2O4(g) 2NO2(g)

b. Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)

a.

b.

42

2

ON

2NO

P

PK

2

2

][H

][Zn2H

P

K

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Standard free energy change is related to the thermodynamic equilibrium constant, K, at equilibrium:

G = G° + RT ln Q

At equilibrium:

G = 0 and Q = K

G° = –RT ln K

?

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Calculate the value of the thermodynamic equilibrium constant at 25°C for the reaction

N2O4(g) 2NO2(g)

The standard free energy of formation at 25°C is 51.30 kJ/mol for NO2 and 97.82 kJ/mol for N2O4(g).

G° = 2 mol(51.30 kJ/mol) – 1 mol(97.82 kJ/mol)

G° = 102.60 kJ – 97.82 kJ

G° = 4.78 kJ

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G = –RT ln K

RT

GK

ln

K) (298K mol

J 8.315

)mol

J 10 (4.78

ln

3K

929.1 ln K

0.145K

?

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Sodium carbonate, Na2CO3, can be prepared by heating sodium hydrogen carbonate, NaHCO3:

2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)

Estimate the temperature at which the reaction proceeds spontaneously at 1 atm.

See Appendix C for data.

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2NaHCO3(s) Na2CO3(s) + H2O(g) + CO(g)Hf°, kJ/mol –947.7 –1130.8 –241.8 –393.5n, mol 2 1 1 1nHf°, kJ –1895.4 –1130.8 –241.8 –393.5

–1895.4 kJ –1766.1 kJH° = 129.3 kJ

Sf°, J/mol K 102 139 188.7 213.7n, mol 2 1 1 1nSf°, J/K 204 139 188.7 213.7

204 J/K 541.4 J/KS° = 337.4 J/K

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S

HT

Δ

Δ

KJ

337.4

J 10 129.3 3T

K 383T

C110T