Contents and Concepts - Bakersfield College Spring_10/Chap_13... · 1 1 Contents and Concepts 1....

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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 Entropy

A spontaneous process is one that occurs by itself. As we will see, the entropy of the system

increases in a spontaneous process.

2

Free-Energy Concept

The quantity ∆H – T∆S 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 – T∆S, so the free energy gives us a thermodynamic criterion of spontaneity.

4. Free Energy and Spontaneity

5. Interpretation of Free Energy

3

Free Energy and Equilibrium Constants

The 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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• 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 -P∆V.

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 - T∆S 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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Chapter 13 Chemical Thermodynamics

1. Spontaneous Chemical and Physical Processes

2. Entropy and Disorder

3. Entropy and the Second Law of Thermodynamics

4. Standard-State Entropies of Reaction

5. The Third Law of Thermodynamics

6. Calculating Entropy Changes for Chemical Reactions

7. Gibbs Free Energy

8. The Effect of Temperature on the Free Energy of a Reaction

9. Beware of Oversimplification

10. Stand-State Free Energies of Reaction

11. Equilibria Expressed in Partial Pressures

12. Interpreting Stand-State Free Energy of Reaction Data

13. Relationship Between Free Energy and Equilibrium Constants

14. Temperature Dependence of Equilibrium Constants

15. Gibbs Free Energies of Formation and Absolute Entropies

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Thermodynamics and Equilibrium

Thermodynamics: Define

The study of the relationship between

heat and other forms of energy

involved in a chemical or physical

process

How do we use this knowledge in chemistry?

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2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)

Consider this Reaction

Concerning this reaction:

1. Does this reaction naturally occur as written?

2. Will the reaction mixture contain sufficient

amount of product at equilibrium?

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We can answer these questions with

heat measurements only!!!

1. We can predict the natural direction.

2. We can determine the composition of the

mixture at equilibrium.

How?

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Consider the Laws of Thermodynamics

1st Law: The change in internal energy of a

system ∆U, equals q + w

∆U = q + w

2nd Law: The total entropy of a system and its

surroundings increases for a spontaneous

process.

3rd Law: A substance that is perfectly crystalline

at 0 K has an entropy of zero.

You can’t win

You cant break even.

You can’t quit the game

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1st Law: The change in internal energy of a

system ∆U, equals q + w

∆U = q + w

∆U = Internal energy = Sum of kinetic energy and

potential energy of the

particle making up the system.

Kinetic Energy = Energy of motion of electrons, protons

and molecules.

Potential Energy = Energy from chemical bonding of

atoms and from attraction between

molecules.

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∆U is a state function:

Most often we are interested in the change:

∆U = Uf - Ui

q = Energy that moves in and out of a system

w = Force x distance = F x d = P∆V

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Heat vs. Work

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w = - F x h

= - F x ∆V/A

= -F/A x ∆V

= -P∆V

Showing work is P∆V

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q = 165 J

P∆V = -92 J

∆U = q + w = (+165) + (-92) = +73 J

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Heat of Reaction and Internal Energy

Zn (s) + HCl (l) → ZnCl2 (aq) + H2 (g)

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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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q = - w = -

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w = -P∆V

= -(1.01 x 105 pa)(24.5 l)

= -1.01 x 105 pa)(24.5 x 10-3m3)

= -2.47 x 103 J

= -2.47 kJ

q = -152 kJ

∆U = -152 kJ + (-2.47 kJ) = -154.9 kJ

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Diagram and explain the change in internal energy

for the following reaction.

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)

q = -890.2 kJ

w = P∆V = +(1.01 x 105 pa)(24.5 l)(2)

= +(1.01 x 105 pa)(2.45 x 10-3m3)(2)

= +4.95 kJ

∆U = -890.2 kJ + (+4.95 kJ) = 885.2 kJ

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Enthalpy and Enthalpy Change

Enthalpy is defined as qp

Enthalpy = H = U + PV

All are state functions.

∆H = Hf - Hi

∆H = (Uf + PVf) – (Ui + PVi) = (Uf –Ui) + P(Vf –Vi)

∆U = qp - P∆V

∆H = (qp - P∆V) + P∆V = qp

∆Hof = Σn∆Ho

f (products) - Σm∆Hof (reactants)

∆Hof = Standard enthalpy change (25oC)

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∆Hof = Σn∆Ho

f (products) - Σm∆Hof (reactants)


Calculate ∆Hof for the reaction in slide 15

∆Hof = for NH3 (g) = -45.9 kJ

= for CO2 (g) = -393.5 kJ

= for NH2CONH2 = -319.2 kJ

= for H2O (l) = -285.8 kJ

∆Ho = [(-319.2 – 285.8) – (-2 x 45.9 – 393.50)] kJ = -119.7

Since ∆Ho has a negative sign, heat is evolved

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Spontaneity and Entropy

Definition of Spontaneous Process:

Physical or chemical process that occurs by itself.

Give several spontaneous processes:

Still cannot predict spontaneity…..

Why??

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Entropy and the 2nd Law of Thermodynamics



process.

Entropy = S = A thermodynamic quantity that is a

measure of how dispersed the energy

is among the different possible ways

that a system can contain energy.

Consider:

1. A hot cup of coffee on the table

2. Rock rolling down the side of a hill.

3. Gas expanding.

4. Stretching a rubber band.

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Flask connected to an evacuated flask by a

valve or stopcock

∆S = Sf - Si

H2O (s) → H2O (l)

∆S = (63 – 41) J/K = 22 J/K

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Concept Check: You have a sample of solid iodine

at room temperature. Later you notice

that the iodine has sublimed. What

can you say about the entropy change

of the iodine?

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process.

Process occurs naturally as a result of energy dispersal

In the system.

∆S = entropy created + q/T

∆S > q/T

For a spontaneous process at a given temperature, the

change in entropy of the system is greater than the heat

divided by the absolute temperature.

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Entropy and Molecular Disorder

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37

Entropy Change for a Phase Transition

∆S > q/T (at equilibrium)

What processes can occur under phase

change at equilibrium?

Solid to liquid

Liquid to gas

Solid to gas

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Was thought that in order to be spontaneous

A reaction had to be exothermic.

∆H < 0

For water undergoing fusion.

∆S = ∆Hfus/T = 6.0 x 103 J/273 K

∆S = 22J/K

The heat of vaporization, ∆Hvap, of carbon tetrachloride,

CCl4, at 25 oC is 39.4 kJ/mol.

CCl4 (l) → CCl4 (g) ∆Hvap = 39.4 kJ/mol

If 1 mol of liquid carbon tetrachloride has an entropy of

216 J/K at 25 oC, what is the entropy of 1 mol of vapor

at equilibrium with the liquid at this temperature?

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Solution:

∆S = ∆Hvap/T = (39.4 x 103 J/mol)/ 298 K

= 132 j/(mol·K)

Entropy of Vapor = (216 + 132) J/(mol·K)

= 348 J/(mol·K)

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Figure 13.2

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Criterion for a Spontaneous Reaction


Is this reaction spontaneous?

That is, does it go left to right as written?

∆S > q/T

qp = ∆H at constant pressure

If we know ∆S and ∆H we can make prediction.

∆S > qp/T = ∆H/T ∆H/T - ∆S < 0

∆H - T∆S < 0 (spontaneous rxn, constant T and P)

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Standard Entropies and the Third Law of Thermodynamics

To determine the entropy of a substance you first

measure the heat absorbed by the substance by

warming it at various temperatures (heat capacity

at different temperatures)

Determination of entropy is based on the 3rd Law.

3rd Law: A substance that is perfectly crystalline

at 0 K has an entropy of zero.

See page 742 – Ebbing and Gammon Houghton Mifflin,

9th Edition

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Standard entropy of methyl chloride, CHCl3, at various

Temperatures (approximate schematic graph)

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Entropy Change for a Reaction

∆So may be positive for reactions with the following:

1. The reaction is one in which a molecule is

broken into two or more smaller molecules.

2. The reaction is one in which there is an increase

in moles of gas.

3. The process is one in which a solid changes to

a liquid or a liquid changes to a gas.

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Standard Entropy (Absolute Entropy): Entropy

value for the standard state of a species.

See Table B-13 p. B-17 and Table B-16 p. B-28

Entropy values of substances must be positive.

So must be >0 but Ho can be plus or minus

(Why?)

How about ionic species?

So for H3O+ is set at zero.

46

Predict The Entropy sign for the following reactions:

a. C6H12O11 (s) → 2CO2 (g)+ C2H5OH (l)

b. 2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)

c. CO (g) + H2O (g) → CO2 (g) + H2 (g)

Exercise 13.2 p 578

d. Stretching a rubber band

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Calculating ∆So for a reaction

∆So = Σn∆So (products) - Σm∆So (reactants)

Calculate the entropy change for the following reaction

At 25 oC.


So 2 x 193 214 174 70

∆So = Σn∆So (products) - Σm∆So (reactants)

∆So= [(174 + 70) - (2x 193 + 214)] J/K= -356 J/K

See exercise 13.3 p 584 and problems 8-12 and 23-26

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Free Energy Concept

∆Ho – T∆So Can Serve as a criteria for Spontaneity


∆Ho = -119.7 kJ ∆So = -365 J/K = -0.365 kJ/K

∆Ho – T∆So = (-119.7 kJ) – (298 K) x (-0.365kJ/K

= -13.6 kJ

∆Ho – T∆So is a negativity quantity, from which we can

conclude that the reaction is spontaneous

under standard conditions.

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49

Free Energy and Spontaneity

Free Energy: Thermodynamic quantity defined by

the equation G = H - TS

∆G = ∆H - T∆S

If you can show that ∆G for a reaction at a given

temperature and pressure is negative, you can

predict that the reaction will be spontaneous…

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Standard Free Energy Changes

Standard Conditions:

1 atm pressure

1 atm partial pressure

1 M concentration

Temperature of 25 oC or 298 K

Standard free energy is free-energy change that takes

place when reactants in their standard states are

converted to products in their standard states.

∆Go = ∆Ho - T∆So

51

Calculate ∆Go from ∆Ho and ∆So

What is the standard free energy change ∆Go, for the

following reaction at 25 oC?

N2 (g) + 3 H2 (g) → 2 NH3 (g)

Use values of ∆Hfo and ∆So from tables 6.1 and 18.1

∆Hfo : 0 0 2 x (-45.9) kJ

N2 (g) + 3 H2 (g) → 2 NH3 (g)

So : 191.6 3 x 130.6 2 x 192.7 J/K

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∆Ho = Σn∆Ho (products) - Σm∆Ho (reactants)

= [2 x (-45.9) – 0] = -91.8 kJ

∆So = ΣnHo (products) - ΣmHo (reactants)

= [2 x 192.7 – (191.6 + 3 x 130.60] J/K = -198.0 J/K


∆Go = −91.8 kJ - (298 K)(-0.1980 kJ/K) = -32.8 kJ

See exercise 13.4-6 and problems 30-36

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Standard Free Energies of Formation

∆Gof = The free energy change that occurs when

1 mol of substance is formed from its

elements in their most stable states at 1 atm

and at a specific temperature (normally 25 oC )

∆Go = Σn∆Gfo (products) - Σm∆Gf

o (reactants)

½ N2 (g) + 3/2 H2 (g) → NH3 (g)

∆Go for 2 mols of NH3 = -32.8 kJ

∆Go for 1 mol of NH3 = -32.8 kJ/2 mol = -16.4 kJ

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Calculation of ∆Go from Standard Free Energies of Formation

Calculate ∆Go for the following reaction:

C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (g)

∆Go = Σn∆Gfo (products) - Σm∆Gf

o (reactants)

∆Go = [2(-394.4) + 3(-228.6) – (- 174.9)] = -1299.7kJ

∆Gfo = -174.9 0 2(-394.4) 3(-228.6)

Calculate ∆Go for the following reaction:

CaCO3 (s) → CaO (s) + CO2 (g) ∆Go = 130.9 kJ

See problems 45-46Do Exercise 13.7-8

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∆Go as a Criterion for Spontaneity


1. When ∆Go is a large negative number (more

negative than about – 10 kJ), the reaction is spontaneous

as written, and reactants transform almost entirely into

products when equilibrium is reached.

2. When ∆Go is a large positive number (more

positive than about + 10 kJ), the reaction is not

spontaneous as written, and reactants transform almost

entirely into products when equilibrium is reached.

3. When ∆Go has a small positive or negative value(less

than about 10 kJ), the reaction mixture gives an

equilibrium mixture with significant amounts of both

reactants and products..

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Interpreting the Sign of ∆Go

Calculate ∆Ho and ∆Go for the following reaction

2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)

Interpret the signs of ∆Ho and ∆Go

∆Hfo are as follows: KClO3 (s) = -397.7 kJ/mol

KCl (s) = -436.7 kJ/mol

O2 (g) = 0

∆Gfo are as follows: KClO3 (s) = -296.3 kJ/mol

KCl (s) = -408.8 kJ/mol

O2 (g) = 0

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2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)

∆Hfo 2 x (-397.70) 2 x (-436.7) 0 kJ

∆Gfo 2 x (-296.3) 2 x (-408.8) 0 kJ

Then:

∆Ho = [2 x (-436.7) – 2 x (-397.7)] kJ = -78 kJ

∆Go = [2 x (-408.8) – 2 x (-296.3)] kJ = -225 kJ

The reaction is exothermic, liberating 78 kJ of heat. The

large negative value of ∆Go indicates that the equilibrium

is mostly KCl and O2.

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∆Gfo = 86.60 kJ/mol

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Interpretation of Free Energy

Theoretically, spontaneous reactions can be used to

obtain useful work.

Combustion of gasoline

Battery generating electricity

Biochemical reaction in muscle tissue

Often reaction are not carried out in in a way that does

useful work.

Reactants simply poured together..

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Interpretation of Free Energy

In principle, if a reaction is carried out to obtain the

maximum useful work, no entropy is produced.

It can be shown that maximum useful work, wmax,

for a spontaneous reaction is ∆G,

The free energy change is the maximum energy

available, or free to do useful work.

The sign of w (work) is defined so that a negative

value means work (energy is subtracted from the

system) obtained from the system). You can obtain

work from a reaction if its ∆G is negative.

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Free energy change during reaction

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Free energy change during reaction

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Free Energy and Equilibrium Constant

Very important relation is the relation between

free energy and the equilibrium constant.

Thermodynamic Equilibrium Constant- the equilibrium

constant in which the concentration of gases are

expressed in partial pressures in atmospheres,

whereas the concentration of solutes in liquid

are expressed in molarities.

K = Kc for reactions involving only liquid solutions

K = Kp for reactions involving only gases

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Write Kp and Kc

• Consider again

N2(g) + 3H2(g) ⇄ 2 NH3(g)

Kc = [NH3]

2

[N2][H2]3

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• In terms of partial pressures,

N2(g) + 3H2(g) ⇄ 2 NH3(g)

Kp = [PNH3]

2

[PN2][PH2]3

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• Kc and Kp are related.

• PV = nRT

• P = [gas]RT

• Kp=Kc × (RT)∆n

• ∆n = (number of moles of product gas) –(number of moles of reactant gas)

• For this reaction, ∆n = -2.

N2(g) + 3H2(g) ⇄ 2 NH3(g)

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• For this reaction, ∆G° = -33.0 kJ.

N2(g) + 3H2(g) ⇄ 2 NH3(g)

• What does this value of ∆G° tell us

about the reaction?

Reaction should be spontaneous.

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Writing the Expression for a Thermodynamic

Equilibrium Constant

a. 2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)

K = [NH2CONH2]

P2NH3PCO2

b. AgCl (s) → Ag+ (aq) + Cl- (aq)

K = [Ag+][Cl-]

Exercises 13.9-10 and problems 58-62

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Table 13.2

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Relating ∆Go to the Equilibrium Constant

∆G = ∆Go + RT ln Q

∆G0 can be calculated from thermodynamic data

If you want ∆G for a none standard state use the

above equation.

At ∆G = 0 the reaction is at equilibrium

0 = ∆Go + RT ln K

And Q = K the equilibrium constant

∆Go = -RT ln K

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Problem: Find the equilibrium constant for the following

reaction:

∆Go for the reaction is -13.6 kJ

using ∆Go = ∆Ho - T∆So

ln K = =∆Go

-RT

-13.6 kJ

-8.31 J/(K·mol) x 298 K

K = e5.49 = 2.42 x 102

Do Exercise 13.11-13 and problems 72-74

0 = ∆Go + RT ln K

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• What is the relationship between ∆G°

and K?

∆G° = -RT lnK

• ∆G° = -RT lnK

– Using tabulated ∆G°ac values, ∆G°reaction

is calculated

– With ∆G°reaction (= ∆G°), K can be

calculated

K = e-∆G°/RT

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• ∆G° = -RT lnK

• ∆G° =∆H° – T∆S°

• K must depend on temperature.

– We have already seen qualitatively how Kchanges with temperature when we discussed Le Châtelier’s principle in Chapter 10.

• Consider this equilibrium

– 2NO2(g) ⇄ N2O4(g)

– NO2 is brown, N2O4 is colorless.

• When this equilibrium is cooled, the system becomes colorless.

• When this equilibrium is heated, the system turns dark brown.

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Table 13.3

2NO2NO22(g) (g) ⇄ N2O4(g)

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• ∆G° can increase with temperature

(∆H°>0 and ∆S°<0) with K also

increasing with temperature.

– Normally ∆G° decreases when K

increases.

• This is illustrated with this equilibrium

2CH4(g) ⇄ C2H6(g) + H2(g).

Table 13.4

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Understanding ∆Go as a criterion for Spontaneity

∆Go = -RT ln K

When K is <1, ln is neg, ∆Go is positive

When K is >1, ln is pos, ∆Go is negative

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a. No change

b. Increases

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Change of Free Energy with Temperature

∆GTo = ∆Ho – T∆So (approximation for ∆GT

o)

Measuring ∆Go and K at different temperatures is difficult

Spontaneity and Temperature Change

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79

80

Increase in temperature – increase in NO2

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Consider the decomposition of dinitrogen tetroxide,N2O4,

to nitrogen dioxide, NO2:

N2O4 (g) → 2 NO2 (g)

How would you expect the spontaneity of the reaction

to behave with temperature change?

Increase in temperature – increase in NO2

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• Instead of using ∆G°ac to calculate ∆G°, you can use ∆G°f, the Gibbs free energy of

formation.

• Gibbs free energy of formation values are

tabulated in Appendix B.16.

aA + bB ⇄ cC + dD

( ) ( ) ( ) ( ) ][][ 00000

BfAfDfCf GbGaGdGcG ∆+∆−∆+∆=∆

• Entropies of reaction can be calculated from tabulated values of absolute entropies. These are found in Appendix B.16.

aA + bB ⇄ cC + dD

][][ 00000

BADC bSaSdScSS +−+=∆

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Calculation of ∆∆∆∆Go at Various Temperatures

Consider the following reaction:

CaCO3 (s) → CaO (s) + CO2 (g)

At 25 oC ∆Go = +130.9 and Kp = 1.1 x 10-23 atm

What do these values tell you about CaCO3?

What happens when the reaction is carried out at a

higher temperature?

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Calculating ∆∆∆∆Go and K at Various Temperatures

a. What is ∆Go at 1000oC for the calcium carbonate reaction?


Is this reaction spontaneous at 1000oC and 1 atm?

b. What is the value of Kp at 1000oC for this reaction?

What is the partial pressure of CO2?

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Strategy for solution…

a. Calculate ∆Ho and ∆So at 25 oC using standard

enthalpies of formation and standard entropies.

Then substitute into the equation for ∆Gfo.

b. Use the ∆Gfo value to find K (=Kp ) as in

Exercise 13.16.

86

a. From Table B-16 you have the following:


∆Hfo: -1206.9 -635.1 -393.5 kJ

So: 92.9 38.2 213.7 J/K

∆Ho = [(-635.1 – 393.5) – (-1206.9)] = 178.3 kJ

∆So = [(38.2 + 213.7) – (92.9)] = 159.0 J/K

∆GTo = ∆Ho –T∆So

= 178.3kJ – (1273 K)(0.1590kJ/K) = -24.1 kJ

∆Go is negative – reaction is spontaneous

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b. Substitute the values of ∆Go at 1273 K, which

equals -24.1 x 103 J, into the equation relating

ln K and ∆Go.

ln K = ∆Go

-RT=

-24.1 x 103

-8.31 x 1273= 2.278

K = Kp = e2.278 = 9.76

Kp = PCO2 = 9.76 atm

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Where does the reaction change from spontaneous

to non-spontaneous?

∆Go = 0 = ∆Ho - T∆So

Solve for T;

T = ∆Ho

∆So =178.3 kJ

0.1590 kJ/K

T = 1121 K = 848 oC

89

That’s it for Chapter 13

Quiz 2.

Explain how you are able to determine whether

a reaction is spontaneous.

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