1. Chapter 17. Thermodynamics: Spontaniety, Entropy and Free Energy
First Law of Thermodynamics Enthalpy -DH, DHf Second Law of Thermodynamics Third Law of Thermodynamics Entropy DS, Free Energy DG Spontaneity

2. Thermochemistry Heat changes during chemical reactions
Thermochemical equation. eg H2 (g) + O2 (g) ---> 2H2O(l) DH = kJ; DH is called the enthalpy of reaction.
if DH is + reaction is called endothermic
if DH is - reaction is called exothermic

3. Why is it necessary to divide Universe into System and Surrounding
Universe = System + Surrounding

4. Why is it necessary to divide Universe into System and Surrounding
Universe = System + Surrounding

5. What is the internal energy change (DU) of a system?
DU is associated with changes in atoms, molecules and subatomic particles
Etotal = Eke + E pe + DU
DU = heat (q) + w (work)
DU = q + w
DU = q -P DV; w =- P DV

6. What forms of energy are found in the Universe?
mechanical
thermal
electrical
nuclear
mass: E = mc2
others yet to discover

7. What is 1st Law of Thermodynamics
Eenergy is conserved in the universe
All forms of energy are inter-convertible and conserved
Energy is neither created nor destroyed.

8. What exactly is DH? Heat measured at constant pressure qp
Chemical reactions exposed to atmosphere and are held at a constant pressure.
Volume of materials or gases produced can change. ie: work = -PDV
DU = qp + w; DU = qp -PDV
qp = DU + PDV; w = -PDV
DH = DU + PDV; qp = DH(enthalpy )

9. How do you measure DU? Heat measured at constant volume qv
Chemical reactions take place inside a bomb.
Volume of materials or gases produced can not change. ie: work = -PDV= 0
DU = qv + w
qv = DU + o; w = 0
DU = qv = DU(internal energy )

10. What is Hess's Law of Summation of Heat?
To heat of reaction for new reactions.
Two methods?
1st method: new DH is calculated by adding DHs of other reactions.
2nd method: Where DHf ( DH of formation) of reactants and products are used to calculate DH of a reaction.

11. Method 1: Calculate DH for the reaction:
SO2(g) + 1/2 O2(g) + H2O(g) -----> H2SO4(l) DH = ? Other reactions: SO2(g) > S(s) + O2(g) DH = 297kJ H2SO4(l)------> H2(g) + S(s) + 2O2(g) DH = kJ H2(g) +1/2O2(g) -----> H2O(g) DH = -242 kJ

12. SO2(g) ------> S(s) + O2(g);DH1 = 297 kJ - 1
H2(g) + S(s) + 2O2(g) > H2SO4(l) DH2 = kJ - 2
H2O(g) >H2(g) + 1/2 O2(g) DH3 = kJ - 3
______________________________________ SO2(g) + 1/2 O2(g) + H2O(g) -----> H2SO4(l) ªH = DH1 + DH DH3
ªH = ªH = kJ

13. Calculate Heat (enthalpy) of Combustion: 2nd method
C7H16(l) O2(g) > CO2(g) + 8 H2O(l) ; DH = ?
DHf (C7H16) = kJ/mol
DHf (CO2) = kJ/mol
DHf (H2O) = kJ/mol
DHf O2(g) = 0 (zero)
What method?
2nd method

14. DH = [Sn ( DHof) Products] - [Sn (DHof) reactants]
= [ ] - [-198.8] =
= kJ = kJ

15. Why is DHof of elements is zero?
DHof, Heat formations are for compounds
Note: DHof of elements is zero

16. What is 2nd Law of Thermodynamics
Entropy (DS) of the Universe increases during spontaneous process.
What is entropy
Entropy (DS) : A measure of randomness or disorder of a system.
Spontaneous Process: A process that occurs without outside intervention.

17. Definitions The Universe: The sum of all parts under consideration.
System: Part of the Universe we are interested in and a change is taking place.
Surrounding: Part of the Universe we are not interested in or can not observe.

18. Entropy DS DSuniv = entropy of the Universe
DSsys = entropy of the System
DSsurr = entropy of the Surrounding
DSuniv = DSsys + DSsurr

19. DSuniv = DSsys + DSsurr
DSuniv DSsys DSsurr
+ +(DSsys>DSsurr)
(DSsurr>DSsys)

20. Standard entropies at 25oC
Substance So (J/K.mol) Substance So (J/K.mol)
C (diamond) HBr (g)
C (graphite) HCl (g)
CaO (s) HF (g)
CaCO3 (s) HI (g)
C2H2 (g) H2O (l)
C2H4 (g) H2O (g)
C2H6 (g) NaCl (s)
CH3OH (l) O2 (g)
CH3OH (g) SO2 (g)
CO (g) SO3 (g)

21. Entropy Change
Entropy (DS) normally increase (+) for the following changes:
i) Solid ---> liquid (melting)
ii) Liquid ---> gas
iii) Solid ----> gas
iv) Increase in temperature
v) Increasing in pressure(constant volume, and temperature)
vi) Increase in volume ( at constant temperature and pressure)

22. Free energy
Gibbs free energy (G) can be used to describe the energy changes of a system.
As usual, it is the changes in free energy that are of interest - DG.
At constant temperature and pressure, DG is equal to:
DG = DH - TDS
T is the Kelvin temperature.

23. What is DG ?
Free Energy
DG = - TDSuni.
DG = DH - TDS.

24. How do you get: DG = DH - TDS.
Dsuniv= Dssys +Dssurr
Dssurr = -DHsys/T)
DSuniv = DSsys -DHsys/T
DSuniv x T = TDSsys -DHsys : x T
-DSuniv x T = -TDSsys +DHsys : x -1
-DSuniv x T = DHsys-TDSsys
-DSuniv x T = DG;
DG = DHsys-TDSsys or DG = DH - TDS.

25. What DG means If DG is - for a change it will take place spontaneously
If DG is + for a change it will not take place
If DG is 0 for a change it will be in equilibrium

26. Free energy
The sign of DG indicates whether a reaction will occur spontaneously.
+ Not spontaneous
0 At equilibrium
- Spontaneous
The fact that the effect of DS will vary as a function of temperature is important. This can result in changing the sign of DG.

27. Predict DG at different DH, DS, T
DG = DH T DS.
all
all
- / high/low +
-/ low/high -

28. Predict the spontaneity of the following processes from DH and DS at various temperatures. a)DH = 30 kJ, DS = 6 kJ, T = 300 K b)DH = 15 kJ,DS = -45 kJ,T = 200 K

29. DH = 30 kJ DS = 6 kJ T = 300 K DG = DHsys-TDSsys or DG = DH - TDS. DH = 30 kJ DS = 6 kJ T = 300 K DG = 30 kJ - (300 x 6 kJ) = kJ DG = kJ b) DH = 15 kJ DS = -45 kJ T = 200 K DG = DHsys-TDSsys or DG = DH - TDS. DH = 15 kJ DS = -45 kJ T = 200 K DG = 15 kJ -[200 (-45 kJ)] = 15 kJ -(-9000) kJ DG = kJ = 9015 kJ

30. Calculation of DGo
We can calculate DGo values from DHo and DSo values at a constant temperature and pressure.
Example.
Determine DGo for the following reaction at 25oC
Equation N2 (g) H2 (g) NH3 (g)
DHfo, kJ/mol
So, J/K.mol

31. Qualitative prediction of DS of Chemical Reactions
Look for (l) or (s) --> (g)
If all are gases: calculate
Dn = Sn(gaseous prod.) -S n(gaseous reac.)
N2 (g) H2 (g) > 2 NH3 (g)
Dn = = -2
If Dn is - DS is negative (decrease in S)
If Dn is + DS is positive (increase in S)

32. Predict DS! 2 C2H6(g) + 7 O2(g)--> 4 CO2(g) + 6H2O(g).
2 CO(g) + O2(g)-->2 CO2(g).
HCl(g) + NH3(g)-->NH4Cl(s).
H2(g) + Br2(l) --> 2 HBr(g).

33. Calculating DS of reactions
Based on Hess’s Law second method:
DSo=SDSo(prod.) - SDS o(react.)

34. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
Calculation of standard entropy changes
Example.
Calculate the DSorxn at 25 oC for the following reaction.
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
Substance So (J/K.mol)
CH4 (g)
O2 (g)
CO2 (g)
H2O (g)

35. Calculate the DS for the following reactions using: D So = 3 D So (products) D S o(reactants) a) 2SO2 (g) + O2 (g) > 2SO 3(g) D So [SO2(g)] = J/K mole ; D So [O2(g)] = J/K mole; D So [SO3(g)] = 257 J/K mole b) 2NH 3 (g) N2O (g) > 4N2 (g) + 3 H2O (l) D So[ NH3(g)] = J/K mole ; D So [N2(g)] = J/K mole; D So [N2O(g)] = J/K mole; D S[ H2O(l)] = 70 J/K mole

36. a) 2SO2 (g) + O2 (g) > 2SO 3(g) D So [SO2(g)] = J/K mole ; D So [O2(g)] = J/K mole; D So [SO3(g)] = 257 J/K mole DSo DSo = 3 DSo (products) DS o(reactants) DSo = [514] - [ ] DSo = DSo = J/K mole

37. How do you calculate DG There are two ways to calculate DG
for chemical reactions.
i) DG = DH - TDS.
ii) DGo = S DGof (products)
- S DG of (reactants)

38. Standard free energy of formation
DGfo
Free energy change that results when one mole of a substance if formed from its elements will all substances in their standard states.
DG values can then be calculated from:
DGo = SnpDGfo products - SnpDGfo reactants

39. Standard free energy of formation
Substance DGfo Substance DGfo
C (diamond) HBr (g)
CaO (s) HF (g)
CaCO3 (s) HI (g)
C2H2 (g) H2O (l)
C2H4 (g) H2O (g)
C2H6 (g) NaCl (s)
CH3OH (l) O (g)
CH3OH (g) SO2 (g)
CO (g) SO3 (g)
All have units of kJ/mol and are for 25 oC

40. Calculate the DG value for the following reactions using: D Go = SD Gof (products) - S D Gof (reactants) N2O5 (g) H2O(l) > HNO3(l) ; DGo = ? D Gfo[ N2O5 (g) ] = kJ/mole ; D Gfo [H2O(g)] = -237 kJ/mole; DGfo[ HNO3(l) ] = kJ/mole N2O5 (g) H2O(l) > HNO3(l) ; DGo = ? DGfo 1 x x (-237) (-81) DGo = DGof (products) DGof (reactants) DGo = [-162] - [134 + (-237)] DGo = DGo = -59 kJ/mole The reaction have a negative DG and the reaction is spontaneous or will take place as written.

41. How do you calculate DG at different T and P
DG = DGo + RT ln Q
Q = reaction quotient
at equilibrium DG = 0
0 = DGo + RT ln K
DGo = - RT ln K
If you know DGo you could calculate K

42. Free energy and equilibrium
For gases, the equilibrium constant for a reaction can be related to DGo by:
DGo = -RT lnK
For our earlier example,
N2 (g) + 3H2 (g) NH3 (g)
At 25oC, DGo was kJ so K would be:
ln K = = =
ln K = ; K = 5.8 x 105
DGo
-RT kJ
-( kJ.K-1mol-1)(298.2K)

43. Calculate the D G for the following equilibrium reaction and predict the direction of the change using the equation: DG = D Go + RT ln Q ; [D Gfo[ NH3(g) ] = -17 kJ/mole N2 (g) H2 (g) W 2 NH3 (g); D G = ? at 300 K, PN2 = 300, PNH3 = 75 and PH2 = N2 (g) H2 (g) W 2 NH3 (g); DG = ?

44. To calculate DGo Using DGo = DGof (products) - 3 DGof (reactants)
To calculate DGo Using DGo = DGof (products) DGof (reactants) DGfo[ N2(g) ] = 0 kJ/mole; DGfo[ H2(g) ] = 0 kJ/mole; DGfo[ NH3(g) ] = -17 kJ/mole Notice elements have DGfo = 0.00 similar to DHfo N2 (g) H2 (g) W 2 NH3 (g); DG = ? DGfo x (-17) DGo = DGof (products) DGof (reactants) DGo = [-34] - [0 +0] DGo = -34 DGo = -34 kJ/mole

45. To calculate Q Equilibrium expression for the reaction in terms of partial pressure: N2 (g) H2 (g) W 2 NH3 (g) p2NH3 K = _________ pN2 p3H p2NH3 Q = _________ ; pN2 p3H2 Q is when initial concentration is substituted into the equilibrium expression Q = _________ ; p2NH3= 752; pN2 =300; p3H2= x Q = x 10-7

46. To calculate DGo DG = DGo + RT ln Q DGo= -34 kJ/mole R = 8
To calculate DGo DG = DGo + RT ln Q DGo= -34 kJ/mole R = J/K mole or x 10-3kJ/Kmole T = 300 K Q= x 10-7 DG = (-34 kJ/mole) + ( x 10-3 kJ/K mole) (300 K) ( ln x 10-7) DG = ln x 10-7 DG = x (-14.18) DG = DG = kJ/mole