1. Chpt 6 - Thermochemistry
Energy and framework (definitions)
Internal energy, heat & work
Enthalpy & Calorimetry
Hess’s Law
Standard Enthalpies
Applications?
HW: Chpt 6 - pg , #s 14, 16, 24, 27, 28, 32, 34, 38, 41, 42, 44, 49, 54, 62, 72, 73, 80, 84, Due Wed Oct. 7

2. Thermodynamic Framework
Energy - capacity to do work or produce heat
Law of conservation of energy
PE (generally chemical potential energy) and KE
Heat involves a transfer of energy - Very different from temperature (a measure of motions of particles) Heat is NOT a substance
Work is force acting over a distance

3. State Functions Energy = work + heat
Ball rolling down hill example - heat and work are different depending on pathway

4. State Functions - con’t
State function or state property only depends on current state not pathway
Energy is state function
Heat and work are not state functions
Another example, Elevation vs. Distance

5. Chemical Energy System and surroundings Exothermic and endothermic
Chemical PE <--> thermal energy
Recall energy reaction diagrams
Only concerned with Energy of Reactants and Products (not pathway) Internal Energy
E = q + w q=heat, w=work
1st Law of thermodynamics
Energy of Universe is constant

6. Universe Energy is Constant
Exothermic or
Endothermic?

7. Chemical Energy - con’t
Quantities have sign and magnitude
System’s perspective for sign, thus endo is flow into system so q is positive (gaining heat)
E <0 exothermic, E>0 endothermic
Work done on system is positive, w>0
Work done by system is negative, w<0

8. PV Work
Common types of work are expansion by a gas and compression on a gas
PV work,
w = - PV if volume is expanding
w = PV if volume is compressing
V = Vfinal - Vinitial,

9. PV Work - derivation Example that shows Pressure=force/area = F/A
so F = P x A
Work is force x dist = F x h
so W = P x A x h
volume of cylinder = A x h
Thus W = PV
The sign is (-) for expanding gas, since work is done by system on surroundings

10. Enthalpy, H
H = E + PV, since E, P and V are state functions H is also a state function
At constant pressure H = qp (qp is heat at const p) In general, for open laboratory chemical reactions pressure is constant, so the change in enthalpy is used interchangeably with the heat of a reaction.
For a chemical reaction
H = Hproducts -  Hreactants
Exothermic means enthalpy, H < 0
Endothermic means enthalpy, H > 0

11. Calorimetry
The science of measuring heat. Substances absorb heat differently; heat capacity, C, measures this
C = heat absorbed / increase in temp
Specific heat capacity is per gram substance
A calorimeter measures heat change.
q = C x m x T our text uses ‘s’ for C
The AP test will use C, actually Cp

12. Calorimeter Simple styrofoam cup
calorimeter is easily used for lab measurements and constant pressure measurements.

13. Bomb Calorimeter
A “Bomb” calorimeter schematic - it only looks like a bomb. It is used when constant volume measurements are needed.

14. Hess’s Law
Since enthalpy is a state function, heat of reactions can be calculated from a known set of simple chemical rxns combined together to get the final rxn.
One step:
N2(g) + 2O2(g) --> 2NO2(g) H = 68kJ
Two distinct steps
N2(g) + O2(g) --> 2NO(g) H = 180kJ
2NO(g) + O2(g) --> 2NO2(g) H = -112kJ
Total these reactions

15. Hess’s Law schematic
The overall reaction enthalpy is independent of pathway

16. Hess’s Law rules Characteristics of H for a rxn
If a reaction is reversed, the sign of H is also reversed.
The magnitude of H is directly proportional to the quantities of reactants and products in the rxn
i.e. if the coefficients are multiplied by an integer, H is multiplied by same integer.

17. Example 6.8 pg. 254
We want to calculate the H for the synthesis of diborane, B2H6, from its elements.
2B(s) + 3H2(g) --> B2H6 (g) H = ?
Use the following data:
2B(s) + 3/2O2(g) --> B2O3(s) H = kJ
B2H6(g) + 3O2(g) --> B2O3(s) +3H2O(g) H = kJ
H2(g) + 1/2O2(g) --> H2O(l) H = kJ
H2O(l) --> H2O(g) H = kJ
Hints - 1) work backward from the required/desired reaction, 2) reverse any reaction as needed to align reactants and products, 3)multiply any reaction as necessary to get correct coefficients.
Recall H is a state function (independent of pathway)

18. Standard Enthalpies of Formation
Hfo of a substance is the change in enthalpy that accompanies the formation of 1 mole of the substance from its elements in their standard states.
Standard state: gas is 1 atm, liquid or solid is pure substance, solutions are 1M. Elements are state at 1atm and 25oC.

19. CalculatingH from Hfo
Since enthalpies are state functions independent of pathway, in chemical reactions, the reactants can be taken apart into their elements and the products can be constructed from their elements.
For a chemical reaction
H = npHfo (products) -  nrHfo (reactants)
Data found in Appendix pg A19 - A22

20. Example H from Hfo 4 NH3(g) + 7 02(g) --> 4 NO2(g) + 6 H2O(l)
NH3(g) -46 kJ/mol NO2(g) 34kJ/mol
H2O(l) kJ/mol O2(g) 0 kJ/mol
H = Products - Reactants…
4x34kJ/mol + 6x(-286kJ/mol) x(-46kJ/mol)
H = -1396kJ/mol

21. Greenhouse Effect