1. THERMODYNAMICS
Chapter 19

2. SPONTANEOUS PROCESS
 A process that occurs without ongoing outside intervention.
Examples
Nails rusting outdoors

3. Ice melting at room temperature
Expansion of gas into an evacuated space
Formation of water from O2(g) and H2(g):
2H2(g) + O2(g) H2O(g)

4. Why are some processes spontaneous and others not?
We know that temperature has an effect on the spontaneity of a process.
e.g. T>0oC  ice melts
 spontaneous at this temp.
H2O (s)  H2O (l)
T<0oC  water freezes
 spontaneous at this temp.
H2O (l)  H2O (s)
T=0oC  water and ice in equilibrium
H2O (l) H2O (s)

5. Exothermic processes tend to be spontaneous.
Example
Rusting of nail - SPONTANEOUS!
4Fe(s) + 3O2(g)  Fe2O3(s)
H = kJ.mol-1
Formation of water - SPONTANEOUS!
2H2(g) + O2(g) H2O(l)
H = kJ.mol-1
Pt cat

6. However, the dissolution of ammonium nitrate is also spontaneous, but it is also endothermic.
NH4NO3(s)  NH4+(aq) + NO3-(aq)
H = kJ.mol-1
So is:
2N2O5(s)  4NO2(g) + 2O2(g)
H = kJ.mol-1
 a process does not have to be exothermic to be spontaneous.
 something else besides sign of H must contribute to determining whether a process is spontaneous or not.

7. ENTROPY (S) That something else is:  extent of disorder!
More disordered  larger entropy
Entropy is a state function
S = Sfinal - Sinitial
Units: J K-1mol-1

8. Entropy’s effects on the mind

9. Examples of spontaneous processes where entropy increases:
Dissolution of ammonium nitrate:
NH4NO3(s)  NH4+(aq) + NO3-(aq)
H = kJ.mol-1
Decomposition of dinitrogen pentoxide:
2N2O5(s)  4NO2(g) + 2O2(g)
H = kJ.mol-1

10. However, entropy does not always increase for a spontaneous process
At room temperature:
Spontaneous
Non-spontaneous

11. SECOND LAW OF THERMODYNAMICS
The entropy of the universe increases in any spontaneous process.
Suniverse = Ssystem + Ssurroundings
Spontaneous process: Suniverse > 0
Process at equilibrium: Suniverse = 0
Thus Suniv is continually increasing!
Suniv must increase during a spontaneous process, even if Ssyst decreases.

13. Q: What is the connection between sausages and the second law of thermo?
A: Because of the 2nd law, you can put a pig into a machine and get sausage, but you can't put sausage into the machine and get the pig back.

14. Rusting nail = spontaneous process
Suniv>0
For example:
Rusting nail = spontaneous process
4Fe(s) + 3O2(g)  2Fe2O3(s) Ssyst<0
BUT reaction is exothermic,
 entropy of surroundings increases as heat is evolved by the system thereby increasing motion of molecules in the surroundings.
 Ssurr>0
+ve
-ve
+ve
Thus for Suniv = Ssyst + Ssurr >0
Ssurr > Ssyst

15. Special circumstance = Isolated system:
Does not exchange energy nor matter with surroundings Ssurr = 0
Spontaneous process: Ssyst > 0
Process at equilibrium: Ssyst = 0
Spontaneous  Thermodynamically favourable
(Not necessarily occur at observable rate.)
Thermodynamics  direction and extent of reaction, not speed.

16. EXAMPLE
State whether the processes below are spontaneous, non-spontaneous or in equilibrium:
CO2 decomposes to form diamond and O2(g)
Water boiling at 100oC to produce steam in a closed container
Sodium chloride dissolves in water
NON-SPONTANEOUS
EQUILIBRIUM
SPONTANEOUS

17. MOLECULAR INTEPRETATION OF S
Decrease in number of gaseous molecules  decrease in S
e.g. 2NO(g) + O2(g)  2NO2(g)
3 moles gas
2 moles gas

18. Molecules have 3 types of motion:
Translational motion - Entire molecule moves in a direction (gas > liquid > solid)
Vibrational motion – within a molecule
Rotational motion – “spinning”
Greater the number of degrees of freedom
 greater entropy

19. Decrease in temperature
 decrease in thermal energy
 decrease in translational, vibrational and rotational motion
 decrease in entropy
As the temperature keeps decreasing, these motions “shut down”  reaches a point of perfect order.

20. EXAMPLE
Which substance has the great entropy in each pair? Explain.
C2H5OH(l) or C2H5OH(g)
2 moles of NO(g) or 1.5 moles of NO(g)
1 mole O2(g) at STP or 1 mole NO2(g) at STP

21. THIRD LAW OF THERMODYNAMICS
The entropy of a pure crystalline substance at absolute zero is zero.
S(0 K) = 0  perfect order

22. Ginsberg's Theorem
(The modern statement of the three laws of thermodynamics)
1. You can't win.
2. You can't even break even.
3. You can't get out of the game.

23. Entropy increases for s  l  g

24. EXAMPLE
Predict whether the entropy change of the system in each reaction is positive or negative.
CaCO3(s)  CaO(s) + CO2(g)
2SO2(g) + O2(g)  2SO3(g)
N2(g) + O2(g)  2NO(g)
H2O(l) at 25oC  H2O(l) at 55oC
+ve
-ve
3 mol gas  2 mol gas ?
Can’t predict, but it is close to zero
2 mol gas  2 mol gas
+ve
Increase thermal energy

25. Standard molar entropy (So)
= molar entropy for substances in their standard state
NOTE
So  0 for elements in their standard state
So(gas) > So(liquid) > So(solid)
So generally increases with increasing molar mass
So generally increases with increasing number of atoms in the formula of the substance

26. Calculation of S for a reaction
Stoichiometric coefficients
(So from tabulated data)

28. EXAMPLE
Calculate So for the synthesis of ammonia from N2(g) and H2(g):
N2(g) + 3H2(g)  2NH3(g)
So/J.K-1.mol-1
N2(g)
H2(g)
NH3(g)

29. Calculation of S for the surroundings
For a process that occurs at constant temperature and pressure, the entropy change of the surroundings is:
(T &P constant)

30. GIBB’S FREE ENERGY (G)

31.  - TSuniv = - TSsys + Hsys
Defined as: G = H – TS
- state function
- extensive property
Suniv = Ssys + Ssurr
At constant T and P:
Suniv = Ssys -
Hsys T
 - TSuniv = - TSsys + Hsys
(at constant T & P)
G = H – TS

32. We know:
Spontaneous process: Suniv > 0
Process at equilibrium: Suniv = 0
Therefore:
Spontaneous process:
Process at equilibrium:
< 0
-TSuniv
-TSuniv
= 0
Gsyst = -TSuniv

33. G = H – TS
Spontaneity involves S H T
Spontaneity is favoured by increasing S and H is large and negative.

34. G allows us to predict whether a process is spontaneous or not (under constant temperature and pressure conditions):
G < 0  spontaneous in forward direction
G > 0  non-spontaneous in forward direction/spontaneous in reverse direction
G = 0  at equilibrium

35. But nothing about rate

36. Standard free energy (Go)
Go = Ho – TSo
Standard states:
Gas - 1 atm
Solid - pure substance
Liquid - pure liquid
Solution - Concentration = 1M
Gfo = 0 kJ/mol for elements in their standard states

37. Tabulated data of Gfo can be used to calculate standard free energy change for a reaction as follows:
Stoichiometric coefficients

38. Substance
H (kJ mol-1)
S (J K-1 mol-1)
G (kJ mol-1)
S (J K-1 mol-1)
AgS
+42.6
I2(s)
+116.1
AgCl(s)
127.1
+96.2
-109.8
I2(g)
+62.4
+260.6
+19.4
Al(s)
28.32
MgO(s)
-601.5
+27.0
-569.2
AlCl3(s)
-704.2
+110.7
-628.8
MnO2(s)
-520.0
+53.1
-465.2
Al2O3(s)
+51.0
N2(g)
+191.5
Br2()
+152.2
N2O4(g)
+9.3
+304.2
+97.8
BrF3(g)
-255.6
+292.4
-229.5
Na(s)
+51.3
C(g)
+716.7
+158.0
+671.3
NaF(s)
-569.0
-546.3
C(graphite)
+5.8
NaCl(s)
-411.1
+72.4
-384.3
C(diamond)
+1.9
+2.4
+2.9
NaBr(s)
-361.1
+87.2
-349.1
CO(g)
-110.5
+197.6
-137.2
NaI(s)
-287.8
+98.5
-282.4
CO2(g)
-393.5
+213.7
-394.4
NaOH(s)
-425.6
+64.5
-379.5
CH4(g)
-74.5
+186.1
-50.8
NH3(g)
-46.2
+192.7
-16.4
C3H8(g)
-103.8
+269.9
-23.4
N2H4()
+50.6
+121.2
+149.2
Ca(s)
41.4
NO(g)
+90.3
+210.6
+86.6
CaO(s)
-635.1
+38.1
-603.5
NO2(g)
+33.2
+240.0
CaCO3(s)(calcite)
+92.9
HNO3()
-174.1
+155.6
-80.8
Cl2(g)
+223.0
O2(g)
+205.0
Cu(s)
O3(g)
+142.7
+238.8
+163.2
F2(g)
+202.7
P(s)(white)
+41.1
Fe(s)
+27.3
P4O10(s)
+231.0
Fe2O3(s)(hematite)
-824.2
+87.4
-742.2
PCl3(g)
-287.0
+311.7
-267.8
H(g)
+218.0
+114.6
+203.3
PCl5(g)
-374.9
+364.5
-305.0
H2(g)
+130.6
PbO2(s)
-277.4
+68.6
-217.4
HCl(g)
-92.3
+186.8
-95.3
S(s)(orthorhombic)
+32.0
HF(g)
-271.1
+173.8
-273.2
H2S(g)
-20.6
+205.6
-33.4
HI(g)
+26.4
+206.5
+1.6
SiO2(s)(quartz)
-910.7
+41.5
-856.3
HBr(g)
-36.4
+198.6
-53.5
SiCl4()
-687.0
+239.7
-619.9
HCN(g)
+135.1
+201.7
+124.7
SO2(g)
-296.8
+248.1
-300.2
H2O(g)
-241.8
+188.7
-228.6
SO3(g)
-395.7
+256.6
-371.1
H2O()
-285.8
+70.0
-237.2
Zn(s)
+41.6
H2O2()
-187.8
+109.6
-120.4
ZnO(s)
-350.5
+43.6
-320.5
Hg()
+75.9

39. EXAMPLE
The combustion of propane gas occurs as follows:
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
Using thermodynamic data for Go, calculate the standard free energy change for the reaction at 298 K.
Gfo/kJ.mol-1
C3H8(g)
CO2(g)
H2O(g)
H2O(l)

40. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
Gfo/kJ.mol-1
C3H8(g)
CO2(g)
H2O(g)
H2O(l)
Go = [3(-394.4) + 4( )] – [(-23.47) – 5(0)]
Go = kJ

41. Free Energy and Temperature
How is change in free energy affected by change in temperature?
G = H – TS H S
-TS
G = H - TS + - +
+ at all temp - + -
- at all temp
- at high temp + at low temp + + -
+ at high temp - at low temp - - - +

42. Note:
For a spontaneous process the maximum useful work that can be done by the system:
wmax = G
“free energy” = energy available to do work

43. EXAMPLE
The combustion of propane gas occurs as follows at 298K:
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
Ho = kJ.mol-1
a) Without using thermodynamic data tables, predict whether Go, for this reaction is more or less negative than Ho.
b) Given that So = J.K-1.mol-1 at 298 K for the above reaction, calculate Go. Was your prediction correct?

44.  Ho – TSo will be less negative than Ho
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
Ho = kJ.mol-1
a) Without using thermodynamic data tables, predict whether Go, for this reaction is more or less negative than Ho.
Go = Ho – TSo
-ve
-ve
6 moles gas  3 moles gas
 – TSo > 0
 Ho – TSo will be less negative than Ho
i.e. Go will be less negative than Ho

45. Go = (-2220 kJ) – (298 K)(-374.46x10-3 kJ.K-1.mol-1)
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
Ho = kJ.mol-1
b) Given that So = J.K-1.mol-1 at 298 K for the above reaction, calculate Go. Was your prediction correct?
Go = Ho – TSo
Go = (-2220 kJ) – (298 K)( x10-3 kJ.K-1.mol-1)
Go = kJ.mol-1
 Prediction was correct.

46. EXAMPLE (TUT no. 5a)
At what temperature is the reaction below spontaneous?
AI2O3(s) + 2Fe(s)  2AI(s) + Fe2O3(s)
Ho = kJ; So = 38.5 J K-1

47. Assume H and S do not vary that much with temperature.
At what temperature is the reaction below spontaneous?
AI2O3(s) + 2Fe(s)  2AI(s) + Fe2O3(s)
Ho = kJ; So = 38.5 J K-1
Go = Ho – TSo
Assume H and S do not vary that much with temperature.
Set G = 0
equilibrium
0 = (851.5 kJ) – T(38.5x10-3 kJ.K-1)
T = K
at equilibrium
For spontaneous reaction: G < 0
 T > K

48. Free Energy and the equilibrium constant
Recall: G = Change in Gibb’s free energy under standard conditions.
G can be calculated from tabulated values.
BUT most reactions do not occur under standard conditions.
Calculate G under non-standard conditions:
Q = reaction quotient
R = gas constant = J.K-1.mol-1

49. Under standard conditions: (1 M, 1 atm) Q = 1  ln Q = 0  G = Go
At equilibrium: G = and Q = Keq 
If Go <  ln Keq >  Keq > i.e. the more negative Go, the larger K etc.
Go <  Keq > 1 Go >  Keq < 1 Go =  Keq = 1
Also

50. Calculate K for the following reaction at 25oC: 2H2O(l) 2H2(g) + O2(g)
EXAMPLE
Calculate K for the following reaction at 25oC:
2H2O(l) H2(g) + O2(g)
Gfo/kJ.mol-1
H2O(g)
H2O(l)
Go = [2(0) + (0)] – [2( )]
Go = kJ.mol-1
474.26x103 J.mol-1 = -(8.314 J.K-1.mol-1)(298 K) lnK
lnK =
K = 7.36x10-84