1. Big Idea 5
Thermodynamics

2. Energy Transfer
Basically if you have a hot object and cold object- energy is transferred until both objects are the same temperature.
T= 100oC
T= 0oC

3. Standard state conditions
O - stands for standard state
DH =DHo
DS =DSo
DG =DGo
If DHo is negative then exothermic
If DHo is positive then endothermic
Standard state conditions
All gases at 1 atm
All liquids are pure
All solids are pure
All solutions are 1M
DHof of elemental state = 0

4. Bond Energy Find DHo for following reaction 2H2(g) + O2(g) 2H2O(g)
DHo = S Bond energies of bonds broken - S Bond energies of bonds formed
Find DHo for following reaction
2H2(g) + O2(g) H2O(g)
DHo = -481kJ/mol
Bond Bond energy (kJ/mol)
H-H 436
O-O 499
O-H 463

5. Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ
CH4  C + 2H kJ
Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

6. Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ
CH4  C + 2H kJ
C + O2  CO kJ
Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

7. Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ
CH4  C + 2H kJ
C + O2  CO kJ
2H2 + O2  2 H2O kJ
Step #3: Multiply reaction #3 by 2

8. Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ
CH4  C + 2H kJ
C + O2  CO kJ
2H2 + O2  2 H2O kJ
CH4 + 2O2  CO2 + 2H2O kJ
Step #4: Sum up reaction and H

9. Short cut for Hess’s Law Heat of Formation- DHof
DHo = S DHof products - S DHof reactants
Remember to multiply by # moles
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H= [3( kJ) + 4( kJ)] – [1( kJ) + 5(0 kJ)]
DHof Table
C3H8 (g) = kJ
CO2 (g) = kJ
H2O (l) = kJ

10. Phase changes
Heat of fusion- energy required for solid to melt
q= nDHofusion
Heat of vaporization = energy required to turn liquid into gas
q= n DHovaporization
Where DHofusion and DHovaporization
are physical constants dependent on material and must be given
Solid to liquid = melting Liquid to solid = freezing Liquid to gas = vaporization Gas to liquid = condensation Solid to gas = sublimation Gas to solid = deposition

11. Calorimetry q = m * c* T q = Joules of heat
m = total mass in calorimeter
(If a solution, use m = D * V to obtain the mass)
Water has a density of 1.00 g/mL
Most water solutions have a density of 1.02 g/mL
c = specific heat (for water, 4.18 J/g-K)
DT = Tf - Ti
© 2009, Prentice-Hall, Inc.

12. Entropy- S Entropy = disorder DSo = S DSoproducts - S DSo reactants
Liquids have higher disorder than solids
Gases have higher disorder than liquids
Particles in solution have more disorder than just solids
2 moles of substances have more disorder than 1 mol

13. Gibbs Free Energy -G DGo = S DGoproducts - S DGo reactants
Measure of whether or not reaction will proceed forward w/o outside energy added, it will occur spontaneously (old wording), or is thermodynamically favored (new wording)
If DGo is negative - the reaction is thermodynamically favored
If DGo is positive - the reaction is thermodynamically UNfavored
If DGo =0 - the reaction is at equilibrium

14. DGo = DHo - TDSo T must be in Kelvin
So to be thermodynamically favored DGo must be NEGATIVE
DHo
DSo T
DGo
Reaction - +
Low
High
Always favored
Never favored
Not favored at low temp
Favored at hi temp
Favored at low temp
Not favored at hi temp

15. DGo = -nF Eo Calculate DGo for below reaction
DGo = Gibbs free energy
n = moles
F= faradays constant coulomb/mol (that is 1 mole of electrons has a charge of coulombs)
Eo = standard reaction potential in volts
Calculate DGo for below reaction
Zn(s) + 2Ag+ Zn Ag(s) Eo= +1.56V
(How many electrons are transferred?)
-301,000J/mol