Chapter 19: Chemical Thermodynamics
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Chapter 19: Chemical Thermodynamics. Chapter 19: Chemical Thermodynamics. Spontaneous processes. at 25 o C. H 2 O (s) H 2 O (l). Chapter 19: Chemical Thermodynamics. Spontaneous processes. at 25 o C. ‘dry ice’. at 25 o C.

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Chapter 19: Chemical Thermodynamics

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Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

Spontaneous processes

at 25oC

H2O (s) H2O (l)


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

Spontaneous processes

at 25oC

‘dry ice’

at 25oC

CO2 (s) → CO2 (g)


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

Spontaneous processes

at 25oC

Fe (s) + O2 (g)

Fe2O3 (s)

rust


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

Spontaneous processes

Burning Paper

at 25oC

C6H10O5 (s) + 6 O2 (g)

5 H2O (g) + 6 CO2 (g)


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

Spontaneous processes

…happen “on their own”

although they sometimes require an initial push to get going

…are “product favored”

occur in a definite direction: towards the formation of product

at 25oC

H2O (s) → H2O (l)

CO2 (s) → CO2 (g)

4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s)


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

The direction of a spontaneous processes

may depend on temperature

below 0oC

H2O (s) H2O (l)


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

  • At a given temperature and pressure, processes are

  • spontaneous only in one direction

  • If a processes is spontaneous in one direction it is non-spontaneous in the other direction

at 25oC

H2O (l) → H2O (s)

non-spontaneous

CO2 (s) → CO2 (g)

spontaneous

non-spontaneous

2 Fe2O3 (s) → 4 Fe (s) + 3 O2 (g)


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

Which reactions are spontaneous?

  • many spontaneous reactions are exothermic (DH < 0)

… but not all!

  • Some reactions are endothermic(DH > 0) and still spontaneous

NH4NO3 (s) → NH4+ (aq) + NO3- (aq)

DH > 0


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

Entropy can be thought of as a measure of disorder

Ludwig Boltzmann (1844-1906)

S = k log W

W = Wahrscheinlichkeit (probability)

k = 1.38 x 10-23 J / K


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

The change in entropy for any process is:

DS = Sfinal - Sinitial

What is the sign of DS for the following processes at 25oC ?

H2O (s) → H2O (l)

DS > 0 (positive)

DS < 0 (negative)

CO2 (g) → CO2 (s)


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

Second Law of Thermodynamics

For any spontaneous process,

the entropy of the universe increases

DSouniverse = DSosystem + DSosurroundings> 0


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

Of all phase states, gases have the highest entropy

Ssolid

< Sliquid

< Sgas

gas

solid

liquid

In a gas, molecules are more randomly distributed


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

Larger molecules/atoms generally have a larger entropy

Ssmall

< Smedium

< Slarge

Larger molecules have more internal motion


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

Often, dissolving a solid or liquid will increase the entropy

dissolves

lower entropy

more disordered arrangement

higher entropy


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

Dissolving a gas in a liquid decreases the entropy

dissolves

overall more disordered

arrangement:

higher entropy


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

The entropy of a substance increases with temperature


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

}

size of molecules increases

}

Sgas > Sliquid

}

dissolving a gas in a liquid is accompanied by

a lowering of the entropy

}

dissolving a liquid in another liquid is accompanied

by an increase in entropy


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

What is the sign of DS for the following reactions?

FeCl2 (s) + H2 (g) → Fe (s) + 2 HCl (g)

DS > 0

solid

gas

solid

gas

1 mol

2 mol

Ba(OH)2 (s) → BaO (s) + H2O (g)

DS > 0

solid

solid

gas

2 SO2 (g) + O2 (g) → 2 SO3 (g)

DS < 0

gas

gas

gas

2 mol

1 mol

2 mol

DS < 0

Ag+ (aq) + Cl-(aq) → AgCl (s)

in solution

insoluble


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

For each of the following pairs, which substance has a higher

molar entropy at 25oC ?

HCl (l) HCl (s)

Li (s) Cs (s)

C2H2 (g) C2H6 (g)

Pb2+ (aq) Pb (s)

O2 (g) O2 (aq)

HCl (l) HBr (l)

N2 (l) N2 (g)

CH3OH (l) CH3OH (aq)


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

If you know the standard molar entropies of reactants and

products, you can calculate

DS for a reaction:

DSorxn = Σ n So(products) – Σ m So(reactants)


Chapter 19 chemical thermodynamics

substanceSo (J/K-mol)

H2130.6

C2H4 (g)219.4

C2H6 (g)229.5

So for elements are NOT zero

Chapter 19: Chemical Thermodynamics

What is DSo for the following reaction?

C2H4 (g) + H2 (g) → C2H6 (g)

Do you expect DS to be positive or negative?

DSorxn = [229.5] – [219.4 + 130.6] =

-120.5 J/K


Chapter 19 chemical thermodynamics

DGo

= Gibbs free energy…

Chapter 19: Chemical Thermodynamics

Which reactions are spontaneous?

We need to know the magnitudes of both DS and DH !

DGo = DHo - TDSo

… is a measure of the amount of “useful work” a system can perform


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

DGo = DHo - TDSo

A reaction is spontaneous if DGo is negative

J. Willard Gibbs

(1839 – 1903)


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

The reaction of sodium metal with water:

2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g)

Is the reaction spontaneous?

Yes

What is the sign of DGo?

DGo = negative

What is the sign of DHo?

DHo = negative (exothermic!)

What is the sign of DSo?

DSo = positive


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

DGo< 0=> reaction is spontaneous

(“product favored”)

“exergonic”

DGo> 0=> reaction is non-spontaneous

“endergonic”

DGo= 0=> reaction is at equilibrium


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

DGo = DHo - TDSo

DGo

DHoDSo

+ -

+

- +

-

- -

sign depends on T !

low T => DGo is negative

high T => DGo is positve

+ +

sign depends on T !

low T => DGo is positive

high T => DGo is negative


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

e. I have no idea how to even start thinking about this

Spontaneity: DG must be negative - think about the signs of DH and DS

endothermic DH > 0


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

e. I have no idea how to even start thinking about this

Spontaneity: DG must be negative - think about the signs of DH and DS

endothermic DH > 0

number of gas molecules doubles DS > 0


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

DGo = DHo - TDSo

DGo

DHoDSo

+ -

+

- +

-

- -

sign depends on T !

low T => DGo is negative

high T => DGo is positve

+ +

sign depends on T !

low T => DGo is positive

high T => DGo is negative


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

e. I have no idea how to even start thinking about this

Spontaneity: DG must be negative - think about the signs of DH and DS

endothermic DH < 0

number gas molecules doubles DS > 0


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

Consider the following reaction:

2 H2 (g) + O2 (g) → 2 H2O (l)

DS = -326.3 J/K, DH = -571.7 kJ , and DG = -475.3 kJ at 25oC. Does the the decomposition of water ever become spontaneous?

If the decomposition of water is spontaneous, then

the formation of water is non-spontaneous:

DG > 0

2 H2 (g) + O2 (g) → 2 H2O (l)

i.e. is there a temperature at which DG becomes > 0 (non-spontaneous)?


Chapter 19 chemical thermodynamics

DH < 0

DS < 0

Chapter 19: Chemical Thermodynamics

Consider the following reaction:

2 H2 (g) + O2 (g) → 2 H2O (l)

DS = -326.3 J/K, DH = -571.7 kJ , and DG = -475.3 kJ at 25oC. Does the the decomposition of water ever become spontaneous?

DG = DH - TDS


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

DGo = DHo - TDSo

DGo

DHoDSo

+ -

+

- +

-

- -

sign depends on T !

low T => DGo is negative

high T => DGo is positve

+ +

sign depends on T !

low T => DGo is positive

high T => DGo is negative


Chapter 19 chemical thermodynamics

DH < 0

DS < 0

Chapter 19: Chemical Thermodynamics

Consider the following reaction:

2 H2 (g) + O2 (g) → 2 H2O (l)

DS = -326.3 J/K, DH = -571.7 kJ , and DG = -475.3 kJ at 25oC. Does the the decomposition of water ever become spontaneous?

DG = DH - TDS

The formation of water becomes non-spontaneous at high temperatures

The decomposition of water becomes spontaneous at high temperatures!


Chapter 19 chemical thermodynamics

substanceDHfo (kJ/mol)DGfo(kJ/mol)So (J/K-mol)

O200205.0

C (diamond, s)1.882.842.43

C (graphite, s) 005.69

CO (g)-110.5-137.2197.9

Chapter 19: Chemical Thermodynamics

A diamond left behind in a burning house reacts according to

2 C (s) + O2 (g) → 2 CO (g)

What is the value of DGo for the reaction at 298K ?


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

There are two possible ways to calculate DGo:

DGo = DHo - TDSo

I)

calculate DGo from DHo and So :

DGo = Σ n DGfo (products) – Σ m DGfo (reactants)

II)


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

2 C (s) + O2 (g) → 2 CO (g)

DGo = DHo - TDSo

I) calculate DGo from DHo and DSo :

substanceDHfo (kJ/mol)DGfo(kJ/mol)So (J/K-mol)

O200205.0

C (diamond, s)1.882.842.43

C (graphite, s) 005.69

CO (g)-110.5-137.2197.9

DHo =[2 x (-110.5] – [2 x 1.88 + 0]

= -224.8 kJ

DSo = [2 x 197.9] – [2 x 2.43 + 205.0]

= 185.9 J/K

DGo = -224.8 kJ -

= - 280.2 kJ

298K x 185.9 J/K


Chapter 19 chemical thermodynamics

Chapter 19: Chemical Thermodynamics

2 C (s) + O2 (g) → 2 CO (g)

DGo = Σ n DGfo (products) – Σ m DGfo (reactants)

II)

substanceDHfo (kJ/mol)DGfo(kJ/mol)So (J/K-mol)

O200205.0

C (diamond, s)1.882.842.43

C (graphite, s) 005.69

CO (g)-110.5-137.2197.9

DGo = [2 x (-137.2)] – [2 x 2.84 + 0]

= -280.1 kJ


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