1. AP Chem
Tests have been graded; you may start coming in to do test corrections.
There was a typo in the FRQ section so I’ll add in the points back for everyone…
You will have a quest (so not a big test) for this unit before your midterm & winter break
Official Midterm Date: Tues 12/19…we can talk about moving it earlier…

2. Thermodynamics
The study of energy changes in physical and chemical processes.

3. Definitions
System
includes the molecules we want to study (here, the hydrogen and oxygen molecules).
Surroundings
everything else (here, the cylinder and piston).

4. Energy Energy the ability to do work or transfer heat Types of Energy:
1. Potential energy
energy an object possesses by virtue of its position or chemical composition.
2. Kinetic energy
energy an object possesses by virtue of its motion.

5. Definitions
Temperature=measure of the average kinetic energy of a sample of a substance.
Heat=is measured as the energy that is transferred from one object to another because of a difference in temperature. The direction of heat flow is always from the hotter object to the colder object.
Specific Heat Capacity: The amount of energy needed to raise the temperature of 1 gram of a substance by 1°C

6. Answer practice questions
Heat Capacity Questions

7. Answer questions 1-6 based on the “Heating Curve”

8. Heat = Energy transferred due to a difference in temperatures
Heat = Energy transferred due to a difference in temperatures. The amount of heat absorbed or released in a physical or chemical reaction can be calculated using the equation q = mCΔT

9. Example
How much heat is required to raise the temperature of 5 grams of water from 20°C to 100°C? The specific heat of water is 4.18 J/g°C.

10. Answer practice questions

11. Why can’t we use this formula when there is a phase change
Why can’t we use this formula when there is a phase change? Consider the problem “how many joules are required to melt 100 grams of ice at 0°C?”
If we actually tried to use q=mCΔT to solve this problem…
q = 100 g x 4.18 J/g°C x 0°C
q = 0 Joules….
But we know that you need heat energy to melt something…!

12. Heat of Fusion: amount of heat energy required to change a substance from a solid to a liquid; Heat energy required to melt a substance: q=mHf
Note: The amount of energy released when a substance changes from liquid to solid is equal to the Heat of Fusion
q=mHf
q=-mHf

13. Heat of Vaporization: amount of heat energy required to change a substance from a liquid to a gas; heat energy required to boil a substance: q=mHv
Note: the amount of energy released when a substance changes from a gas to a liquid is equal to the Heat of Vaporization
q=mHv
q=-mHv

14. q=mHf and q=mHv Problems:
1. How many joules are required to melt 100 grams of ice at 0°C? The heat of fusion for water is 334 J/g q= m= Hf =
q = 100 g x 334 J/g
100 g
q = 33,400 J
334 J/g

15. How many joules are absorbed by a 50 gram sample of H2O that is boiling? The heat of vaporization of water is 2260 J/g q= m=
Hv =
q = 50 g x 2260 J/g
50 g
2260 J/g
q = 113,000 J

16. Enthalpy A measurement for the heat flow in a system
H > 0 = endothermic
H < 0 = exothermic

17. Examples
Determine the sign of the enthalpy change in each of the following scenarios:
a. An ice cube melts
b. 1g of butane is combusted in sufficient oxygen to give complete combustion to CO2 and H2O

18. Enthalpy of Reaction
The change in enthalpy, H, also called the heat of reaction
Enthalpy is an extensive property.
Ex: CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ΔH = -890kJ
The energy given off by reacting 1mol CH4 with 2mol O2 is -890kJ
The energy given off by reacting 2mol CH4 with 4mol O2 is 2(-890kJ) = -1780kJ
© 2009, Prentice-Hall, Inc. 18

19. Enthalpy
H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction.
Ex. CH4 +2O2 CO2 + 2H2O H =-890kJ
CO2 + 2H2O  CH4 +2O2 H = +890kJ

20. Examples
How much heat is released when 4.50g of methane gas is burned in a constant pressure system?
CH4(g) + 2O2(g)  H2O(l) + O2(g) H = -890kJ
4.50 g CH4
× 𝟏 𝒎𝒐𝒍 𝑪𝑯 𝟒 𝟏𝟔.𝟎𝟓 𝒈 𝑪𝑯 𝟒
× −𝟖𝟗𝟎 𝒌𝑱 𝟏 𝒎𝒐𝒍 𝑪𝑯 𝟒
= -250 kJ

21. Hess’s Law:
If a reaction is carried out in a series of steps, the ΔH for the reaction is the sum of the individual ΔH for each step.
We can estimate H using published H values and the properties of enthalpy.

22. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Example
Calculate ΔH for the reaction:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Given:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔH = -802 kJ
H2O(l) → H2O(g) ΔH = +44 kJ
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔH = -802 kJ
2 H2O(g) → 2 H2O(l) ΔH = 2 x - 44 = -88 kJ
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔH = -802 kJ kJ = -890 kJ

23. Standard enthalpy of formation, ΔHfo
enthalpy change for the formation of 1 mol of compound with all substances in their standard states (1 atm and 25°C)
Units of kJ/mol; Values will be given
The ΔHfo for the most stable form of any element is 0 (ex. ΔHfo = 0 for H2, O2 etc)
Hrxn = Hf°products – Hf° reactants

24. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H = [3( kJ) + 4( kJ)] – [1( kJ) + 5(0 kJ)]
= [( kJ) + ( kJ)] – [( kJ) + (0 kJ)]
= ( kJ) – ( kJ) = kJ 24