1. Thermochemistry
CHAPTER 17

2. Thermochemistry Thermochemistry: study of energy changes
During chemical reactions
During phase changes

3. How does Heat differ from Temperature?
Temperature: average kinetic energy of particles
The faster the particles are moving, the higher the temperature.
Heat (q): transfer of kinetic energy
Heat flows from high temp  low temp
Units: Joules (SI unit)
or calories (kCal = Cal in foods)
1 calorie = J

4. Joule/Calorie Conversion Practice
Example: 9.6 calories = ______ Joules
450 Joules = ______ calories
53 calories = ______ Joules
0.9 Joules = _______ calories
9.6 calories Joules = Joules
1 calorie
107.5
221.8
0.22

5. Energy Changes in Chemical Reactions
Exothermic Reactions – give off energy
A + B  C + D + energy
Endothermic Reactions – absorb energy
energy + A + B  C + D

6. Endothermic or Exothermic?
Reactants  Products + Energy
Reactants + Energy  Products
2NaHCO3 (s) kJ  Na2CO3 (s) + H2O (g) + CO2 (g)
CaO (s) + H2O (l)  Ca(OH)2 (s) kJ
baking bread
campfire
burning fuel
photosynthesis
breaking down carbohydrates for energy
Electrolysis of water

7. Exothermic Reactions
Exothermic Reactions: net release of energy (heat)
“feels hot to the touch”
ex. combustion reactions
PE of Reactants > PE of Products (b/c energy is lost)
Activation Energy
Potential Energy (J)
Time
Reactants
Enthalpy
Products

8. Endothermic Reactions
Endothermic Reactions: net absorption of energy (heat)
“Feels cold to the touch”
ex. decomposition of water using electrolysis
PE of Reactants < PE of Products
Activation Energy
Potential Energy (J)
Time
Products
Enthalpy
Reactants

9. Activation Energy & Catalysts
Activation energy: The amount of energy needed to get a reaction started.
Required for BOTH endothermic and exothermic reactions.

10. How does a catalyst speed up a reaction?
Catalyst: speeds up the reaction, but is not used up during the reaction.
Activated Complex
Activation Energy
Potential Energy (J)
Time
Reactants
Catalyst Reduces Activation Energy
Products

11. Enthalpy
Enthalpy (ΔH): the difference in energy between the products and reactants in a chemical change
ΔHreaction = Hproducts - Hreactants
ΔHreaction is negative for exothermic reactions. Why?
ΔHreaction is positive for endothermic reactions. Why?
Nature favors lower energy  exothermic
Example:
2H2 (g) + O2 (g)  2H2O (l) ΔH = -572 kJ
2H2O (l)  2H2 (g) + O2 (g) ΔH = +572 kJ

12. Law of Conservation of Energy
Law of Conservation of Energy: the energy released when elements come together to form a compound is equal to the energy required to decompose the compound into its elements.
Example:
2H2 (g) + O2 (g)  2H2O (l) ΔH = -572 kJ
2H2O (l)  2H2 (g) + O2 (g) ΔH = +572 kJ

13. Heating Curve Energy changes also occur in physical processes. L  G
(vaporization)
Gas
boiling point
Liquid
**NOTE – Temp DOES NOT
Change during a phase change!!!!
BECAUSE – All the energy goes
into breaking the bonds…
NOT into speeding up the particles.
S  L
(melt)
melting
point
Solid

14. Cooling Curve

15. Endothermic or Exothermic?
Phase changes are physical changes (no new substance is formed)…
Melting
Freezing
Boiling/Evaporation/Vaporization
Condensation

16. Specific Heat
Specific Heat (C): amount of energy (in Joules) needed to increase the temperature of 1 gram of a substance by 1oC.
See Reference Tables
q = mCΔT (+q = endothermic, -q = exothermic)
q = heat (joules or calories)
m = mass (grams)
C = specific heat 
ΔT = change in Temperature = Tf - Ti J
goC

17. Remember!!!
Temperature does not change during a phase change! (Energy goes into pulling molecules farther apart.)
Therefore, we cannot use this equation for phase changes: q = mCΔT

18. Latent Heats
Heat of Vaporization (Hv): amount of energy absorbed when one gram of a liquid is changed into a vapor (or, the amount of energy released when 1 gram of a vapor is changed into a liquid).
q = mHv
Hv = heat of vaporization (J/g)
Hv of water = 2260 J/g

19. Heat of Fusion
Heat of Fusion (Hf): amount of energy needed to convert 1 gram of a solid into a liquid.
(Or the heat released when 1 gram of liquid  solid.)
q = mHf
Hf = heat of fusion (J/g)
Hf of water = 334 J/g

20. Example Problems – Level One
1. How much heat is added if 100 grams of liquid water increases in temperature from 30oC to 70oC? 2. How much heat is absorbed if 200 g of ice increases in temperature from -15oC to -5oC?
q = mCΔT
q = (100 g)(4.18 J/goC)(70o – 30o)
q = 16,720 J
q = (200 g)(2.05 J/goC)(-5o – -15oC)
q = 4100 J

21. (cont’d)
3. How much heat is released if 80 grams of water vapor is decreases in temperature from 150oC to 125oC? 4. How much heat is absorbed when 30 g of ice is changed into liquid water at 0oC?
q = (80 g)(2.02 J/goC)(125o – 150o)
q = J
q = mHf
q = (30 g)(334 J/g)
q = 10,020 J

22. Continued…
5. How much heat is released when 50 grams of water vapor is changed into liquid water at 100oC?
q = mHv
q = (50 g)(2260 J/g)
q = 113,000 J  -113,000 J

23. Example Problems – Level Two
6. How much heat is absorbed if 30 grams of ice at -10oC is converted into liquid water? 2 steps: (1) heat the ice and (2) melt ice  liquid (1) q = mCΔT q = (30g)(2.05 J/goC)(0o – -10o) q = 615J (2) q = mHf q = (30g)(334J/g) q = 10,020J
Add both steps together!
Total heat = 615J + 10,020J
Total heat = 10,635 J

24. Continued…
7. How much heat is absorbed if 45 grams of water at 80oC is converted into steam at 105oC?
3 steps: (1) heat water, (2) water  steam, (3) heat steam
(1) q1 = mCΔT
q1 = (45g)(4.18J/goC)(100o – 80o)
q1 = 3,762 J
Add all 3 together!
qtotal = q1 + q2 + q3
qtotal = 3,762J + 101,700J J
qtotal = 105,916.5 J
q2 = mHv
q2 = (45g)(2260J/g)
q2 = 101,700 J
(3) q3 = mCΔT
q3 = (45g)(2.02J/goC)(105o – 100o)
q3 = J

25. (cont’d)
8. How much heat is released if 500 grams of vapor at 120oC changes to liquid water 70oC? 3 steps: cool the vapor, condensation, cool the liquid
(1) q1 = mcpΔT
q1 = (500g)(2.02J/goC)(100o – 120o)
q1 = -20,200J
q2 = mHv
q2 = (500g)(2260J/g)
q2 = 1,130,000J  - 1,130,000J
Add together: ,200J
-1,130,000J
+ -62,700J
-1,212,900J
(3) q3 = mcpΔT
q3 = (500g)(4.18J/goC)(70o – 100o)
q3 = -62,700J

26. Calorimetry
Calorimetry: the precise measurement of the heat flow out of a system for chemical and physical processes.

27. Calorimeter Example Problems
1. How much heat is released by the reaction if 10 grams of water is heated from 20oC to 60oC. q = mCΔT q = (10g)(4.18J/goC)(60o – 20o) q = 1,672J  amount absorbed by water. Therefore, -1,672J was released by the reaction. 2. What is the mass of the water if a release of 650 J of heat causes the water to increase in temperature by 15oC? (Water absorbed 650J of heat from the reaction.) 650J = m(4.18J/goC)(15oC) m = 10.4g
Joules (released by reaction)
2. 43 kg
3. 15 degrees Celsius

28. Example Problems
3. What is the final temperature of 15 g of water if the water starts at 20oC and loses 300 J of heat? **Losing heat means q is negative! -300J = (15g)(4.18J/goC)(Tf – 20o) Tf = 15.2oC 4. Determine the substance if a sample of 20 grams increases in temperature from 10oC to 28oC when it absorbs Joules of energy J = 20g(C)(28o – 10o) C = 4.5 J/goC  Magnesium (see ref.pack)

29. Entropy
Entropy (S): a measure of the disorder of the particles in a system.
Systems tend to become more disordered.
Entropy increases when:
solid  liquid  gas
(melting)
# moles increase in a reaction
(1A + 2B  3C + 3D)
Temperature increases

30. Predicting Changes in Entropy
Predict whether ΔSsystem will be positive or negative:
1. H2O (l)  H2O (g)
[positive]
2. CH3OH (s)  CH3OH (l)
3. Increase the temperature of a substance

31. (cont’d)
4. Dissolving of a gas in a solvent CO2 (g)  CO2 (aq) [negative] 5. What happens when the number of gaseous product particles is greater than the number of gaseous reactant particles? 2SO3 (g)  2SO2 (g) + O2 (g) [positive] 6. Generally, when a solid or liquid dissolves to form a solution? NaCl (s)  Na+ (aq) + Cl- (aq)

33. Spontaneous Reactions
A reaction is spontaneous if:
Energy decreases (energy is released = exothermic) AND
Disorder increases (positive entropy).
Ex. Wood burning