1. Chapter 6
Thermochemistry

2. Thermochemistry Deals with the energy changes Chemical reactions
Determines whether a reaction is exothermic or endothermic
Physical Changes
Energy changes

3. Vocabulary Energy - The ability to do work (E) in J
Potential Energy – Energy of position (PE)
Kinetic Energy – Energy of motion (KE)
Temperature – Random motion of particles
Heat – Transfer of energy (q) in J
Work – Force x Distance (W) in J
Enthalpy – Energy Change for reactions
(ΔH) in J

4. Vocabulary Pathway – How energy is transferred
State Function – Properties that depend on current state. (Independent of pathway)
Energy is a state function
Temperature, Volume, Pressure
Heat and Work are not state functions

5. Example of Energy Change

6. Endothermic Reactions
Energy is put into a reaction
Product bonds are weaker than reactant bonds

7. Exothermic Reactions Energy is released from a reaction
Product bonds are stronger than reactant bonds

8. Signs for Exo vs. Endo.

9. Calorimetry Measuring heat associated with a reaction
Uses a calorimeter
Coffee Cup Calorimeter
At constant pressure q = -ΔH
q=m*s* ΔT
q = heat, m = mass, ΔT = Change in Temp
s=specific heat capacity

11. Specific Heat Capacity
Energy required to heat one gram of a substance 1 degree Celcius
Units J/g*ºC
Molar Heat Capacity = Just per mole
Metals have low heat capacities

12. Example
How much energy is required to heat grams of Aluminum metal from 22.0ºC to ºC? Specific heat of water is J/g*ºC the specific heat of aluminum is 0.89 J/g*ºC

13. Example
#46 p. 282

14. Homework
P. 281 #’s 32, 37,40, 43, 48

15. Hess’s Law
In going from a particular set of reactants to a particular set of products, the change in enthalpy (H) is the same whether the reaction takes place in one step or in a series of steps
Energy is a state function

16. Comparison One step: N2(g) + 2O2(g)  2NO2(g) H1 = 68kJ Two step
2NO(g) + O2(g)  2NO2(g) H3 = -112kJ
N2(g) + 2O2(g)  2NO2 (g) H2 + H3 = 68kJ

17. Hess’s Law Rules
Start with the end reaction in mind and work backward rearrange your equations
Work to cancel like terms
When reversing the equation the sign of ΔH must be changed
When multiplying an equation by a coefficient ΔH must be multiplied by the same coefficient

18. Example #54 p. 283
NH3 ½ N2 + 3/2 H2 H = 46 kJ
2H2 + O2  2H2O H = -486 kJ
2N2 + 6H2O  3O2 + 4NH3 H = ?
Based on enthalpy is this a useful synthesis?

19. Example #52 p. 283
C4H4 + 5O2  4CO2 + 2H2O H = -2341kJ
C4H8 + 6O2  4CO2 + 4H2O H = -2755kJ
H2 + ½ O2  H2O H = -286kJ
C4H4 + 2H2  C4H8 H = ?

20. Homework
P. 283 #’s 51,53,55,57

21. Calculate how much energy (kJ) is required to heat 50
Calculate how much energy (kJ) is required to heat 50.0 L of tap water from 10.0 ºC to 45.0 ºC. Assume the density of water is 1.00 g/mL and specific heat of water is 4.18 J/g* ºC
7320 kJ
A McDonalds hamburger contains Calories (that’s 2.50x105 calories). How many burgers would it take to supply the same amount of energy? calorie = 4.18 J
7.00 burgers

22. Enthalpy of Formation
The change in enthalpy that accompanies the formation of one mole of a compound from its elements with all elements in their standard state
Symbol = ΔHfº
f is formation
Degree symbol means standard states

23. Standard States Standard States Temperature is 25 ºC
Pressure = 1.00 atm
Concentration = 1.00 M
The standard state is how something exists at these conditions
O2(g), Na(s), Br2(l)

24. Reactions 4C(s) + 2H2(g)  CH4(g) ΔHfº = -75 kJ/mol
H2(g) + ½ O2(g)  H2O(l) ΔHfº = -286 kJ/mol
Write the enthalpy of formation reaction for gaseous carbon dioxide. Don’t worry about the value
C(s) + O2(g)  CO2(g)
Values are in Appendix 4 (A21)
ΔHfº = kJ/mol

25. Enthalpy of Formation Values
Most values are negative
Compounds are more stable than element
All elements are zero
Even diatomic elements!
States matter
Liquid water is -286
Gaseous water is -242
Use this information to calculate enthalpy of the reaction!

27. Standard Enthalpy of Reaction
Standard Enthalpy of Reaction is the energy change at standard conditions
Symbol = ΔHº
ΔHº = Σ ΔHfº(Products) - Σ ΔHfº(Reactants)
Remember to include coefficients
Remember elements are zero!

28. Al(s) + Fe2O3(s)  Al2O3(s) + Fe(s)
Calculate the standard enthalpy of reaction for the following
Al(s) + Fe2O3(s)  Al2O3(s) + Fe(s)
Al = 0, Fe = 0, Fe2O3 = -826, Al2O3 = -1676

29. Homework
P. 284 #63