chapter 19 chemical thermodynamics - yonsei...

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Chapter 19

Chemical Thermodynamics

Lecture Presentation

Yonsei University

© 2012 Pearson Education, Inc.

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19.1 Spontaneous Processes

• Chemical thermodynamics is concerned with energy relationships in chemical reactions. – enthalpy

– entropy

ChemicalThermodynamics

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First Law of Thermodynamics

• Eenergy cannot be created or destroyed.(in chap. 5)

• Therefore, the total energy of the universe is a constant.

• Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa.

E= q + w© 2012 Pearson Education, Inc.


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What is Spontaneous Process ?


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Spontaneous Processes

• Spontaneous processesare those that can proceed without any outside intervention.

• The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously return to vessel B.



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Spontaneous Processes

Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.



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Temperature dependence

• Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures.

• Above 0 C, it is spontaneous for ice to melt.

• Below 0 C, the reverse process is spontaneous.



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Seeking a Criterion for Spontaneity

• To understand why some processes are spontaneous we must look at the ways in which the state of a system might change.– Temperature, internal energy, and enthalpy are state

functions.• internal energy = Kinetic & potential energy of molecule +

chemical energy

– Heat transferred between a system and the surroundings, as well as work done on or by a system, are not state functions.


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Reversible Processes

In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process.



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Irreversible Processes

• Spontaneous processes are irreversible.



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reversibe vs. irreversible

• For a system at equilibrium, reactants and products can interconvert reversibly.

• For a spontaneous process, the path between reactants and products is irreversible.


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19.2 Entropy and the Second Law of Thermodynamics

• Entropy (S), Rudolph Clausius (19C)

• Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered, .

– Entropy can be thought of as a measure of the randomness of a system.

– The more disordered, or random, the system is, the larger the value of S.

– It is related to the various modes of motion in molecules.

qT



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Entropy Change

• Like total energy, E, and enthalpy, H, entropy is a state function.

• Therefore, S = Sfinal Sinitial


• For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature:

• ∆S for Phase Changes : S =

qrev

T

T

H

T

qS fusionrev

fusion


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Sample Exercise 19.2 Calculating ∆S for a Phase Change

Elemental mercury is a silver liquid at room temperature. Its normal freezing point is –38.9 °C, and its molar enthalpy of fusion is ∆Hfusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point?

–38.9 °C = 1–38.9 + 273.152 K = 234.3 K

Solution

Practice ExerciseThe normal boiling point of ethanol, C2H5OH, is 78.3 °C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy in the system when 68.3 g of C2H5OH(g) at 1 atm condenses to liquid at the normal boiling point?Answer: –163 J/K


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Second Law of Thermodynamics

The second law of thermodynamics states that the entropy of the universe increases for spontaneous processes, and the entropy of the universe does not change for reversible processes.


For reversible processes:

Suniv = Ssystem + Ssurroundings = 0

For irreversible processes:

Suniv = Ssystem + Ssurroundings > 0

These last truths mean that as a result of all spontaneous processes, the entropy of the universe increases.


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19.3 Molecular Interpretation of Entropy

• Gas expansion into a vacuum is a spontaneous process. – Consider two flasks connected by a stopcock.

– The probability of finding molecules (n) in the left flask : (1/2)N

– For one mole, N = 6.02 1023!


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Boltzmann’s Equation and Microstates• Ludwig Boltzmann described the concept of entropy

on the molecular level.

S = k lnW

• Temperature is a measure of the average kinetic energy of the molecules in a sample.



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Molecular Motions and Energy

• Molecules exhibit several types of motion:– Translational: Movement of the entire molecule from

one place to another.

– Vibrational: Periodic motion of atoms within a molecule.

– Rotational: Rotation of the molecule about an axis or rotation about bonds.



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Molecular Motions and Energy

• Boltzmann envisioned the motions of a sample of molecules at a particular instant in time.

• He referred to this sampling as a microstate of the thermodynamic system.

• Each thermodynamic state has a specific number of microstates, W, associated with it.

• Entropy is S = k ln W

where k is the Boltzmann constant, 1.38 1023 J/K.



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Entropy on the Molecular Scale

• The change in entropy for a process, then, is

S = k ln Wfinal k ln Winitial

Wfinal

WinitialS = k ln

• Entropy increases with the number of microstates in the system.



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Microstates vs.Macrostate

42( ) ( ) ( ) 16

1f f N

i i

W V

W V

P : 1/16 4/16 6/16 4/16 1/16


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Probability Distribution


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Entropy on the Molecular Scale

• The number of microstates and, therefore, the entropy, tends to increase with increases in– Temperature (EK)

– Volume

– The number of independently moving molecules.

W(E, V, N)=gVNE(3N/2)



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Entropy and Physical States

• Entropy increases with the freedom of motion of molecules.

• Therefore,S(g) > S(l) > S(s)



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Entropy


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Solutions

Generally, when a solid is dissolved in a solvent, entropy increases.



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Entropy Changes

• In general, entropy increases when– Gases are formed from

liquids and solids;

– Liquids or solutions are formed from solids;

– The number of gas molecules increases;

– The number of moles increases.



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Third Law of Thermodynamics

The entropy of a pure crystalline substance at absolute zero is 0.

– In a perfect crystal at 0 K there is no translation, rotation, or vibration of molecules. Therefore, this is a state of perfect order.



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Predict whether is positive or negative for each process, assuming each occurs at constant ∆S temperature:(a) H2O(l) H2O(g)(b) Ag+(aq) + Cl–(aq) AgCl(s)(c) 4 Fe(s) + 3 O2(g) 2 Fe2O3(s)(d) N2(g) + O2(g) 2 NO(g)

Sample Exercise 19.3 Predicting the Sign of ∆S

Solution

(a) ∆S is positive.

(b) ∆S is negative.

(c) ∆S is negative.

(d) ∆S will be close to zero.

Practice ExerciseIndicate whether each process produces an increase or decrease in the entropy of the system:

(a) CO2(s) CO2(g)

(b) CaO(s) + CO2(g) CaCO3(s)

(c) HCl(g) + NH3(g) NH4Cl(s)

(d) 2 SO2(g) + O2(g) 2 SO3(g)Answer: (a) increase, (b) decrease, (c) decrease, (d) decrease


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In each pair, choose the system that has greater entropy and explain your choice: (a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25 °C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25 °C, (c)1 mol of HCl(g) or 1 mol of Ar(g) at 298 K.

Sample Exercise 19.4 Predicting Relative Entropies

Solution(a)HCl(g) :gases have more freedom of motion than the particles in solids.

(b)When these two systems are at the same pressure, the sample containing 2 mol of HCl has twice the number of molecules as the sample containing 1 mol. Thus, the 2–mol sample has twice the number of microstates and twice the entropy.

(c)The HCl system has the higher entropy because the number of ways in which an HCl molecule can store energy is greater than the number of ways in which an Ar atom can store energy. (Molecules can rotate and vibrate; atoms cannot.)

Practice ExerciseChoose the system with the greater entropy in each case: (a) 1 mol of H2(g) at STP or 1 mol of H2(g) at 100 °C and 0.5 atm, (b) 1 mol of H2O(s) at 0 °C or 1 mol of H2O(l) at 25 °C, (c) 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP, (d) 1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP.

Answers: (a) 1 mol of H2(g) at 100 °C and 0.5 atm, (b) 1 mol of H2O(l) at 25 °C, (c) 1 mol of SO2(g) at STP, (d) 2 mol of NO2(g) at STP


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19.4 Entropy Changes in Chemical Reactions

• These are molar entropy values of substances in their standard states.

• Standard entropies tend to increase with increasing molar mass.



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Standard Entropies

Larger and more complex molecules have greater entropies.



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Entropy Changes

Entropy changes for a reaction can be estimated in a manner analogous to that by which H is estimated:

S = nS(products) — mS(reactants)

where n and m are the coefficients in the balanced chemical equation.



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Sample Exercise 19.5 Calculating ∆S° from Tabulated Entropies

Calculate the change in the standard entropy of the system, ∆S° , for the synthesis of ammonia from N2(g) and H2(g) at 298 K:

N2(g) + 3 H2(g) 2 NH3(g)

Solution∆S° = 2S°(NH3)–[S°(N2) + 3S°(H2)]∆S°= (2 mol)(192.5 J/mol–K)–

[(1 mol)(191.5 J/mol–K) + (3 mol)(130.6 J/mol–K)]= –198.3 J/K

Practice ExerciseUsing the standard molar entropies in Appendix C, calculate the standard entropy change, ∆S°, for the following reaction at 298 K:

Al2O3(s) + 3 H2(g) 2 Al(s) + 3 H2O(g)Answer: 180.39 J/K


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Entropy Changes in Surroundings

• Heat that flows into or out of the system changes the entropy of the surroundings.

• For an isothermal process:

Ssurr =qsys

T

• At constant pressure, qsys is simply Hfor the system.



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Entropy Change in the Universe

• The universe is composed of the system and the surroundings.

• Therefore,Suniverse = Ssystem + Ssurroundings

• For spontaneous processes

Suniverse > 0



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Entropy Change in the Universe• Since

Ssurroundings =

and

qsystem = Hsystem

This becomes:

Suniverse = Ssystem +

Multiplying both sides by T, we get

TSuniverse = Hsystem TSsystem

Hsystem

T

qsystem

T



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19.5 Gibbs Free Energy

• TSuniverse is defined as the Gibbs free energy, G.

• When Suniverse is positive, G is negative.

• Therefore, when G is negative, a process is spontaneous.



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Gibbs Free Energy

1. If G is negative, the forward reaction is spontaneous.

2. If G is 0, the system is at equilibrium.

3. If G is positive, the reaction is spontaneous in the reverse direction.



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Standard Free Energy ChangesAnalogous to standard enthalpies of formation are standard free energies of formation, G:f

G = nG (products) mG (reactants)f f

where n and m are the stoichiometric coefficients.


At temperatures other than 25 C,

G = H TSHow does G change with temperature?


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Free Energy Changes


• There are two parts to the free energy equation:H — the enthalpy term

– TS — the entropy term

• The temperature dependence of free energy then comes from the entropy term.


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Sample Exercise 19.6 Calculating Free–Energy Change from ∆H°, T, and ∆S°

Calculate the standard free–energy change for the formation of NO(g) from N2(g) and O2(g) at 298 K:N2(g) + O2(g) 2 NO(g)

given that ∆H° = 180.7 kJ and ∆S° = 24.7 J/K. Is the reaction spontaneous under these conditions?

Solution

Practice ExerciseCalculate ∆G° for a reaction for which ∆H° = 24.6 kJ and ∆S° = 132 J/K at 298 K. Is the reaction spontaneous under these conditions?

Answer: ∆G° = –14.7 kJ; the reaction is spontaneous.


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Sample Exercise 19.7 Calculating Standard Free–Energy Change from Free Energies of Formation

(a) Use data from Appendix C to calculate the standard free–energy change for the reaction P4(g) + 6 Cl2 (g) 4 PCl3 (g) run at 298 K. (b) What is ∆G° for the reverse of this reaction?

Solution(a)

(b) 4 PCl3(g) P4(g) + 6 Cl2(g) ∆G° = +1102.8 kJ

Practice ExerciseUse data from Appendix C to calculate ∆G° at 298 K for the combustion of methane: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g).

Answer: –800.7 kJ


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Solution(a) ∆G° is less negative than ∆H°.

Sample Exercise 19.8 Estimating and Calculating ∆G°

In Section 5.7 we used Hess’s law to calculate ∆H° for the combustion of propane gas at 298 K:C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) ∆H° = –2220 kJ

(a) Without using data from Appendix C, predict whether ∆G° for this reaction is more negative or less negative than ∆H°. (b) Use data from Appendix C to calculate ∆G° for the reaction at 298 K. Is your prediction from part (a) correct?

(b)

Practice Exercise

For the combustion of propane at 298 K, C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g), do you expect ∆G° to be more negative or less negative than ∆H°?

Answers: more negative


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Sample Exercise 19.9 Determining the Effect of Temperature on Spontaneity

The Haber process for the production of ammonia involves the equilibriumAssume that ∆H° and ∆S° for this reaction do not change with temperature. (a) Predict the direction in which ∆G° for the reaction changes with increasing temperature. (b) Calculate ∆G° at 25 °C and 500 °C.

Solution(a) ∆G° becomes less negative (or more positive) with increasing temperature. Thus, the driving force for the production of NH3 becomes smaller with increasing temperature.

(b) We calculated ∆H° for this reaction in Sample Exercise 15.14 and ∆S° in Sample Exercise 19.5: ∆H° = –92.38 kJ and ∆S° = –198.3 J/K. If we assume that these values do not change with temperature, we can calculate ∆G° at any temperature by using Equation 19.12.

At T = 25 °C = 298 K, we have

At T = 500 °C = 773 K, we have


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Sample Exercise 19.9 Determining the Effect of Temperature on Spontaneity

Continued

Practice Exercise(a) Using standard enthalpies of formation and standard entropies in Appendix C, calculate ∆H° and ∆S° at 298 K for the reaction 2 SO2(g) + O2(g) 2 SO3(g). (b) Use your values from part (a) to estimate ∆G° at 400 K.

Answers: (a) ∆H° = –196.6 kJ, ∆S° = –189.6 J/K; (b) ∆G° = –120.8 kJ


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19.7 Free Energy and Equilibrium Constant

Under any conditions, standard or nonstandard, the free energy change can be found this way:

G = G + RT ln Q

(Under standard conditions, all concentrations are 1 M, so Q = 1 and ln Q = 0; the last term drops out.)



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Free Energy and Equilibrium

• At equilibrium, Q = K, and G = 0.

• The equation becomes

0 = G + RT ln K

• Rearranging, this becomes

G = RT ln K

or

K = e


G/RT


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• If G° < 0, then K > 1.

• If G° = 0, then K = 1.

• If G° > 0, then K < 1.

RTGeK /o


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Sample Exercise 19.11 Calculating the Free–Energy Change under Nonstandard Conditions

Calculate at 298 K for a mixture of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3 being used in the Haber process:

Solution

Practice Exercise

Calculate at 298 K for the Haber reaction if the reaction mixture consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3.

Answer: –26.0 kJ/mol


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Sample Exercise 19.12 Calculating an Equilibrium Constant from ∆G°

The standard free–energy change for the Haber process at was obtained in Sample Exercise 19.9 for the Haber reaction: Use this value of ∆G° to calculate the equilibrium constant for the process at 25 °C.

Solution

K=6.9x105


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Driving Nonspontaneous Reactions:coupled reactions

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l) G° = –2880 kJ.

– used to convert low energy ADP and inorganic phosphate into ATP


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Problems

• 30, 42, 54, 72, 86, 94,102,108

chapter 19 chemical thermodynamics - yonsei...

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