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Introduction to Thermodynamics

I. Conservation of Energy

A. First Law of Thermodynamics = energy cannot be created or destroyed, only

converted between different forms

1. Example: CH4(g) + 2O2(g) CO2(g) + 2H2O + energy

a) Reaction gives off energy as heat

b) Potential energy stored in chemical bonds is lowered

c) Total energy is unchanged

2. Uses and Shortcomings

a) Lets us keep track of energy flow in processes

b) Does not tell us if or why a given process occurs

c) Does not tell us direction of a chemical reaction

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II. Entropy

A. The Heat Tax

1. Conversion of energy between forms is inefficient:

2. Usually, some energy is lost as heat

3. The fewer energy conversion, the better

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B. Direction of Processes and Reactions

1) Examples:

a) Ball at the top of a hill Ball at the bottom of a hill

b) Steel + H2O + O2 Rust

c) Gas in one part of a container Gas filling a container

d) Ice at 5 oC Water at 5 oC

2) What is in common?

a) Exothermic? Not ice melting or gas expandingb) Increased Disorder = Increased Entropy = +DS

3) Entropy = S = driving force of spontaneous reactions = disorder or random

a) Probability (likelihood): there are many ways for objects/molecules to

be disordered, but only a few to be ordered

b) Nature proceeds towards the most likely state

= state with greatest number of energetically

equivalent arrangements

4) Expansion of a Gas

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a) Possible arrangements of 4 gas molecules in a 2-bulb system

b) Microstates = possible configurations of a particular arrangement

c) Entropy selects most likely arrangement = 2 molecules in each bulb

MicrostatesABCD ---

ABC D

ABD C

ACD BBCD A

AB CD

CD AB

AC BD

BD AC

AD BC

BC AD

Order

1 microstate

Ordered

4 microstates

SomewhatDisordered

6 microstatesFully disordered

B

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5) Probabilities

Probability of all molecules in left bulb

Number of Molecules Probability

1

2 x =

3 x x = 1/8

5 1/25 = 1/32

10 1/210 = 1/1024

n 1/2n

6.022 x 1023 1 x 10-23

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6) Positional Entropy = entropy depending on configuration in space

a) Changes in state depend on positional entropy

b) Ssolid < Sliquid

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f) Example: Predict the sign of the entropy change for

i) Dissolving solid sugar into water

ii) Iodine vapor condensing to crystals on a surface

7) Second Law of Thermodynamics = in any spontaneous process, there isalways an increase in the entropy of the universe

a) Energy is conserved = constant

b) Entropy is always increasing

c) DSuniverse

= DSsystem

+ DSsurroundings

d) For a given process: ifDSuniverse = + the process is spontaneous

ifDSuniverse = - the process is not spontaneous

e) Life = constant battle against entropy

i) Large molecules are assembled from smaller ones

ii) Organizing a cell is DSsystem = - the process is not spontaneous

iii) Fortunately, it is DSuniverse that must be positive in a process

iii) DSsurroundings = large + for life to occur

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III. Temperature and Spontaneity

A. Change in state: 1 mol = 18 ml H2O(l) 1 mol = 31 L H2O(g)

1) DSsurrounding depends on flow of heat into or out of the system

a) Heat increases the motion (randomness) of particles

b) Exothermic reactions release heat to surroundings DSsurrounding = +

c) Endothermic reactions absorb heat from surroundings DSsurrounding = -

d) Vaporization of water is endothermic DSsurrounding = -

2) DSuniverse = DSsystem + DSsurrounding = (+) + (-) = +/- ?a) Depends on the temperature

b) If T > 100 oC, DSuniverse = + If T < 100oC, DSuniverse = -

B. Temperature Effects

1) DSsurroundings depends on heat flow

a) Exothermic reactions usually favors spontaneity

b) Spontaneity usually lowers the energy of the starting material as it

becomes product

c) The difference of these energies = heat released to surroundings

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2) Importance of Exothermicity ofDSuniverse depends on Temperature

a) Adding heat to hot surroundings has little effect

b) Adding heat to cold surroundings has a large effect

c) Heat transfer is more important at low temperatures

3) In Summary

a) Sign of DSsurrdepends on direction of heat transfer

b) Magnitude ofDSsurrdepends on T

c)

d) The (-) is there because DH is for the system, which is opposite of

DH of the surroundings

4) Example: Find DSsurrat 25oC

a) Sb2S3(s) + 3Fe(s) 2Sb(s) + 3FeS(s) DH = -125 kJ/mol

b) Sb4O6(s) + 6C(s) 4Sb(s) + 6CO(g) DH = +778 kJ/mol

T

H

e(K)Temperatur

Heat(J)Ssurr

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IV. Free Energy

A. Free Energy = G = HTS

1) DGprocess = DHTDS

2) Divide byT

3)

4) A process is spontaneous ifDG = -

5) Chemists use DG rather than DS because we only need to know systemB. Example: Predicting Spontaneity using DG

1) H2O(s) H2O(l) DHo = 6030 J/mol, DSo = 22.1 J/K mol

univsurrSSSS

T

H

T

G

T

GSuniv D

T T DHo DSo DSsurr DSuniv TDSo DGo

oC K J/mol J/Kmol J/Kmol J/K mol J/mol J/mol

-10 263 6030 22.1 -22.9 -0.8 5810 +220

0 273 6030 22.1 -22.1 0.0 6030 0

10 283 6030 22.1 -21.3 0.8 6250 -220

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2) Classifying Processes/Reactions based on DH and DS

3) Example: At what T is Br2(l) Br2(g) spontaneous (1 atm) given thatDH = 31.0 kJ/mol and DS = 93.0 J/Kmol

a) Spontaneous when DG = -

b) Set DG = 0 and solve for T

c) When T > 333K, TDS > DH and DG = - (Entropy controlled)

d) When T < 333K, TDS < DH and DG = + (Enthalpy controlled)

e) 333K is the boiling point of Br2(l)

333K

93J/Kmol

31000J/mol

S

HT

ST-H0

D

D

DD

ST-HG D