1. Lecture 8: Thermochemistry
Lecture 8 Topics Brown chapter 5
8.1: Kinetic vs. potential energy
8.2: Transferring energy as heat & work
Thermal energy
8.3: System vs. surroundings
Closed systems
8.4: First Law of Thermodynamics
Internal energy of chemical reactions
Energy diagrams
E, system & surroundings
8.5: Enthalpy
Exothermic vs. endothermic
Guidelines thermochemical equations 5.4
Hess’s Law
8.6: Calorimetry
Constant pressure calorimetry
8.7: Enthlapy of formation

2. Enthalpy of formation Enthalpy of formation: enthalpy to form 1 mole
Enthalpies of formation can be used to calculate enthalpy of reaction.
The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws.
It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made.
This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory.
While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

3. Enthalpy of formation (ΔHf)
There are many forms (or expressions) of enthalpy including Hvaporization, Hfusion, Hcombination & Hformation.
The most useful is standard enthalpy of formation (H°f): the enthalpy change that accompanies the formation of 1 mole of a compound from elements in their standard states.
The form of the element at 1 atm, 298 K/25°C gases - 1 atm
solutions - 1 M
standard states?
Na?
Hg? N2?
solid
liquid gas
The enthalpy of formation of any substance in its most stable form is always zero: Carbon as graphite; oxygen as O2 gas; copper as an elemental solid (metal).
Write the reactions for H°f for:
CH3CH2OH
FePO4
2C + 1/2O2 + 3H2  CH3CH2OH
Fe + P + 2O2  FePO4
see table 5.3, p. 177 & appendix C p

4. Example: ΔHf
For which reaction (at 25°C) would enthalpy change represent standard enthalpy of formation?
If not, how could you change the reaction conditions?
a. 2Na(s) + 1/2O2(g)  Na2O(s)
2K(l) + Cl2(g)  2KCl(s)
C6H12O6(s)  6C(diamond) + 6H2(g) + 3O2(g)
Yes - all in standard states & 1 mole product is formed
No - K should be (s) & 2 moles product are formed
No - this is a decomposition, not a formation; reverse it p

5. ΔHfs can be summed to calculate ΔHrxn
So this concept combines Hess’s Law with Hf .
If we know the Hf of all reactants & products, we can calculate Hrxn:
H°rxn = nH°f(products) - nH°f(reactants)
Propane (C3H8) is combusted to form CO2 and H2O under standard cond.
C3H8 + 5O2  3CO2 + 4H2O
(gas) (gas) (gas) (liquid)
H°rxn = -Hf [C3H8(g)] + 3Hf [CO2(g)] + 4Hf [H2O(l)]
= -( kJ) + 3( kJ) + 4( kJ)
= kJ
(moles)(kJ/mole)
Expressed via Hess’s Law:
C3H8 --> 3C + 4H2
H1 = -Hf [C3H8(g)]
reversed
3C + 3O2 --> 3CO2
H2 = 3Hf [CO2(g)]
stoichiometry
4H2 + 2O2 --> 4H2O
H3 = 4Hf [H2O(l)]
state !
C3H8 + 5O2 --> 3CO2 + 4H2O
H°rxn = H1 + H2 + H3 p

6. Examples: ΔHf
Calculate the H for combustion of 1 mole benzene (C6H6).
C6H6 + 15/2O2  6CO2 + 3H2O
(liquid) (gas) (gas) (liquid)
Hrxn = [6Hf(CO2) + 3Hf(H2O)] - [Hf(C6H6) + 15/2 Hf(O2)]
 products
 reactants
= [6( kJ) + 3( kJ)] - [(49.0 kJ) + 15/2(0 kJ)]
= ( ) kJ
= kJ
Use enthalpies of formation to calculate the Hrxn for: CaCO3(s) --> CaO(s) + CO2(g)
Hrxn = [Hf(CaO) + Hf(CO2)] - [Hf(CaCO3)]
 products
 reactants
= [( kJ) + ( kJ)] - [( kJ)] = kJ p

7. Enthalpies of combustion of fuels
Carbohydrates? Glucose = C6H12O6
C6H12O O2  6CO H2O Hrxn = kJ
Average fuel value for carbohydrates (& proteins) is ~17 kJ/g.
Fats? Triacylglycerol (body fat) = C57H110O6
2C57H110O O2 --> 114CO H2O Hrxn = - 75,520 kJ
Average fuel value for carbohydrates (& proteins) is ~38 kJ/g.
What about industrial fuels?
fuel value (kJ/g)
Coal
Oil 45
Natural gas 49
Gasoline 48
H2 142 p