1

ThermodynamicsThermodynamically favored

reactions (“spontaneous”)

Enthalpy

Entropy

Free energy

2

Thermodynamically

Favored ProcessesWater flows downhill.

Sugar dissolves in coffee.

Heat flows from hot object to

cold object.

A car fender rusts

Why not the reverse?

3

Thermodynamically

Favored Processes

Enthalpy decrease

and/or

Entropy increase

Driven by two factors:

4

reactantsproductsP

ote

ntia

l E

ne

rgy

time

DH

Ea

Enthalpy: Many exothermic

processes are spontaneous.

Lower

PE

5

Enthalpy

“Forget about it.”

2H2O 2H2 + O2

DH = +570 kJ

spontaneous

(caution rate may be slow)

CH4 + 2O2 CO2 + H2O

DH = -890 kJ

Enthalpy

6

Recall enthalpy changes are related

to bond energies & IM forces of

reactants and products.

Molecules with strong bonds (high

bond energies) are relatively stable

and have low potential energy.

e.g. H2O and CO2 are often reaction

products due to their low PE.

7

EnthalpyYou might think only

exothermic reactions are

thermodynamically favored.

at 25 oC DH = +6.0 kJ

Ice melts, but is endothermic!

but H2O(s) H2O(l)

8

Entropy (S)

“Entropy” also helps determines

whether a reaction is spontaneous.

What is it ?

9

Gedanken

Gas Y

At 1 atm

Gas B

At 1 atm

What happens when the

valve is opened? Why?

10

EntropyThe gases mix because of

Entropy is actually the dispersal

of matter and energy.

11

If you drop a glass, what happens?

If you drop the broken pieces,

do they reform the glass?

Entropy

low Shigh S

12

Changing Entropy: #1

solid liquid gas

Increasing entropy (+DS)

increased dispersal of matter

13

NaCl(s) Na+(aq) + Cl-(aq)H2O

Dissolution:

1.Ionization: increases entropy

2.Mixing: increases entropy

3.Hydration: decreases entropy

Changing Entropy: #2

Changing Entropy: #3

14

Entropy increases in reactions where

the number of product molecules is

greater than reactant molecules.

(matter becomes more dispersed)

4H3PO4 P4O10 + 6H2O

15

Entropy increases greatly in reactions

where moles of gas increase.

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

Changing Entropy: #4

Gases have high entropy due

to the dispersal of matter.

0 mol gas 1 mol gas

16

Increasing temperature raises

the kinetic energy which

increases the disorder (entropy).

Changing Entropy: #5

Raising T

disperses KE

17

Entropy Prediction

Is DS positive or negative?

CO2 sublimes

Br2(l) freezes

Sugar is dissolved in water

18

Predict DS

Is DS positive or negative?

2H2(g) + O2(g) 2H2O(g)

19

3rd Law of Thermodynamics

The entropy of a perfect crystalline

substance at 0 K is zero.

Unlike enthalpy, entropy values

are absolute.

(no arbitrary reference point)

20

Calculating Entropy

aA + bB cC + dD

For the reaction:

Calculate DS from table of

standard entropies.

(appendix 3; 25oC and 1 atm)

DSo = SnSo(prod) –SnSo(react)

21

Calculating Entropy

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

So 192 131 193(J/mol.K)

DSo = 2(193J/K) – (1)192J/K – 3(131J/K)

= -199 J/molrxn.K

Decrease in entropy as expected.

units

22

Calculating Entropy

Your turn.

H2(g) + Cl2(g) 2HCl(g)

So(J/mol.K) 131 223 187

Does entropy go up or down?

23

2nd Law of Thermodynamics

The entropy of the universe increases

in a spontaneous reaction and is

unchanged in a equilibrium process.

24

2nd Law of ThermodynamicsA simple function can be derived

from the 2nd law that describes

whether a reaction is spontaneous.

G = Gibb’s Free Energy

25

Gibb’s Free Energy

DGo = DHo –TDSo

DGo < 0 reaction spontaneous

(“exergonic”)

DGo > 0 reaction nonspontaneous

(spontaneous in reverse)

(“endergonic”)

DGo = 0 system at equilibrium

26

Standards for DGo

Gas 1 atm

Liquid pure substance

Solid pure substance

Elements DGo = 0

Solution 1 M concentration

Temp. 25oC

27

Calculating Free Energy

1. From “standard” DGfo values

(appendix 3; 1 mole,1atm, 25oC).

This works just like DHfo and DSo.

2. From DHo and DSo and T using

the free energy equation.

DGo = DHo –TDSo

28

Standard DGo

H2(g) + Br2(l) 2HBr(g)

DGfo kJ/mol 0 0 -53.2

most stable form of element

DGo = 2(-53.2kJ) –1(0) –1(0)

= -106.4 kJ/molrxn (at 25oC)

Spontaneous !!!

29

DH is +

(endo)

DH is –

(exo)

DS is +(more random)

DS is –(less random)

DG = DH –TDS

Is a Reaction Spontaneous?

30

Spontaneous? Try It!CaCO3(s) CaO(s) + CO2(g)

1.Use appendix 3 data (above)

to calculate DHrxno & DSrxn

o.

DHfo (kJ/mol) -1207 -636 -394

DSo (J/mol.K) 92.9 39.8 213.6

Be careful with units.

31

Spontaneous? Try It!

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

2. Use DHrxno & DSrxn

o to calc.

DGrxno at 25oC.

3. Is this reaction spontaneous

at 25oC?

4. At what T does this reaction

become spontaneous?

32

Free Energy: Interpretation

CaCO3(s)

CaO(s) + CO2(g)

Temperature (oC)

P(C

O2)(a

tm)

3

2

1

0

CO2 hits std.

pressure (1atm)

Coupled Reactions

33

Although some reactions are not

thermodynamically favored (+DG),

they can be made to happen by

coupling to a thermodynamically

favored second reaction.

Coupled Reactions

34

S(s) + O2(g) SO2(g) -300kJ

Write the net reaction and calculate

DGorxn when this reaction is coupled with

the oxidation of sulfur to SO2.

PbS(s) Pb(s) + S(s) +99kJ

The production of lead from PbS ore by

heating is nonspontaneous.

DGrxno

External Energy

35

External energy can be used to drive

reactions where DG is positive.

• thermal energy

• electrical energy

• light energy

e.g. photosynthesis DG = 2880 kJ/molrxn

hn + 6CO2 + 6H2O C6H12O6 + 6O2

36

Free Energy: Reminders

1.Be careful of units (kJ vs. J, & T)

DHo and DGo are kJ/molrxn

DSo = J/molrxn.K

2. AP units for chemical reactions:

2Al(s) + 3Cl2(g) 2AlCl3(s)

Free Energy: Reminders

37

3. DHo & DSo are only exact at

25oC, but using them to

calculate DGo at other

temperatures is “good enough”.

(DHo & DSo don’t change

much with temperature.)

Free Energy: Reminders

38

4. Spontaneous reactions

(thermodynamically favored)

may appear not to occur if the

reaction rate is slow due to

high activation energy.

“kinetically controlled reaction”

39

Controlled by Thermodynamics

Controlled

by kinetics

Kinetically Controlled

Reaction

40

Graphite is the most stable form of

carbon under standard conditions

Cdiamond(s) Cgraphite(s)

DGo = -3 kJ/molrxn (spontaneous)

However it is too slow to be

observed due to high Ea.

Free Energy: Reminders

41

5. External energy can be used to

drive reactions that are not

thermodynamically favored.

• 2H2O(l) 2H2(g) + O2(g)

can be driven by electrical energy

• Photosynthesis, initiated by

photon absorption

42

Phase Transitions(e.g. ice water at 0oC)

System is in equilibrium, so:

DGo = 0 = DHo -TDSo

DSo =DHo

T

43

Melt Ice at 0 oC

=6010 J

273 K

= 22 J/mol.K

DSo =DHo

T

DHfusT

=

Entropy increases while

melting, as expected

44

Phase Change: Try It !!!

1.What is the entropy change for

freezing benzene at its freezing

point of 5.5oC?

(DHfus is 10.9 kJ)?

2.Is this process spontaneous?

45

Free Energy & Equilibrium

aA(aq) + bB(aq) cC(aq) + dD(aq)

If reactants start in standard states

(1 M), as reaction proceeds, the

chemicals are no longer in their

standard states.

46

Free Energy & Equilibrium

As the reaction proceeds, the

free energy change (DG) is

related to standard free energy

change (DGo) by:

DG = DGo + RT lnQ

“reaction quotient”, Q, remember?

47

Free Energy & Equilibrium

DG = DGo + RT lnQ

= DGo + RT ln[C]c[D]d

[A]a[B]b

aA + bB cC + dD

R = 8.314 J/mol.K

48

Try It

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

What is DG at these initial pressures

and 25oC? (DGfo = -17 kJ/mol)

Predict the reaction direction.

Po (atm) 0.87 0.25 12.9

49

Free Energy & Equilibrium

DG = DGo + RT lnQ

At equilibrium:

and DGo = -RT lnK

and the net reaction stops.

0 K

50

Free Energy & Equilibrium

DG = DGo + RT lnQ

Case 1. If reaction is spontaneous

(DGo negative), reaction will proceed

toward products to a great extent

until the RT lnQ equals DGo .

negative positivezero

51

Free Energy & Equilibrium

DG = DGo + RT lnQ

Case 2. If reaction is non- spont.

(DGo positive), reaction will proceed

toward products to only a small

extent until the RT lnQ equals DGo .

positive negativezero

52

Free Energy & EquilibriumAt equilibrium:

DGo = -RT lnK

(Kp for gases and Kc for solutions)

Note: the larger the K, the more

negative DGo (reaction goes more

toward products).

53

Free Energy & Equilibrium

K lnK DGo

> 1 + -Products favored

& spontaneous

< 1 - +Reactants favored

& nonspontaneous

DGo = -RT ln K

Free Energy & Equilibrium

54

DGo = -RT ln K

Rearranges to: lnK = - DGo

RT

and: K = e -DGo/RT

RT at room temperature ~ 2.4kJ/mol.

So if DG ~ 2.4kJ/mol, K ~ 1, otherwise

K will be very large or very small.

55

Free Energy & Equilibrium

DGo = -RT lnK

Measuring DGo (via DHo & DSo )

is a convenient way to measure K

for many reactions.

56

Calculating K from DGo

2H2O(l) 2H2(g) + O2(g)

Gfo kJ –237 0 0

Calculate Kp at 25oC:

Try It !!!

Does your answer make sense?

57

Calculating DGo from K

Calculate DGo for the aqueous

dissolution of AgCl at 25oC.

Ksp = 1.6 x 10-10

AgCl(s) Ag+(aq) + Cl-(aq)

Does your answer make sense?

58

One More !!!

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

What is DG at these initial pressures?

Predict reaction direction.

(Kp = 6.59 x 105 at 25oC)

Po (atm) 1.21 0.43 10.9

59

The End

60

What is the pH when 25 mL of 0.12 M HI is mixed with 32 mLof 0.058 M Ba(OH)2?

Warm-up

Warm-up

61

Salt water is evaporated to dryness.

What happens to the entropy of:

Na+ ions and Cl- ions?

water ?

Warm-up

62

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

So(J/K) 192 131 193

Ho(kJ) -46.3

Under standard conditions,

what temperature is this a

spontaneous reaction?

Warm-up

63

What is the sign of DG when the

equilibrium constant is very small?

Is the reaction spontaneous

(Explain using the equation

relating these two variables.)

Warm-up

64

What is the difference between

DG & DGo? For a given reaction,

when does DGrxn = DGo?

What is DGo and DSo when water

condenses at 100oC?

(DHvap = 40.6 kJ/mol)