1. Chapter 6: Thermochemistry
AP Chemistry Exam Part 2
Chapter 6: Thermochemistry
final exam chapter 6

2. 6.1 Nature of energy Concept 1: What is energy?
Energy is the capacity to do work (or to produce heat)
Work is a force acting over a distance
Heat is actually a form of energy
Kinetic energy: energy due to the motion of an object
KE = ½mv2
KE measured in joules
final exam chapter 6

3. 6.1 Nature of energy Concept 1: What is energy?
What is the kinetic energy of a 2.25 kg baseball moving at 113 m/s?
Try below
final exam chapter 6

4. 6.1 Nature of energy Concept 1: What is energy?
KE=1/2mv2
(1/2)225*1132
1,436,512 joules
final exam chapter 6

5. 6.1 Nature of energy Concept 1: What is energy?
Energy is the flow of heat
Endothermic reaction = reaction gain heat
Exothermic reaction = reaction releases heat
Energy is E, change in energy equation below
∆E =q +w
q= heat
q is positive in endothermic reactions
q is negative in exothermic reactions b. w = work
w is negative if the system does work
w is positive if work is done on the system
final exam chapter 6

6. 6.1 Nature of energy Concept 1: What is energy?
final exam chapter 6

7. 6.1 Nature of energy Concept 1: What is energy?
final exam chapter 6

8. 6.1 Nature of energy Concept 1: What is energy?
w = -P∆V
a. by a gas (through expansion) = ∆V is positive, w is negative
b. to a gas (by compression) = ∆V is negative
w is positive
final exam chapter 6

9. 6.1 Nature of energy Concept 1: What is energy?
final exam chapter 6

10. 6.1 Nature of energy Concept 1: What is energy?
final exam chapter 6

11. 6.2 Enthalpy and Calorimetry Concept 2: Calculating Enthalpy and Calorimetry
Calorimetry - science of measuring heat energy
equations = q = mcΔT  where  q = heat energy  m = mass  c = specific heat 
ΔT = change in temperature 
Specific Heat Capacity = 1calorie
- Energy required to raise the temp of 1 gram of a substance by 1OC
final exam chapter 6

12. 6.2 Enthalpy and Calorimetry Concept 2: Calculating Enthalpy and Calorimetry
What is the heat in Joules required to raise the temperature of 25 grams of water from 0 °C to 100 °C? What is the heat in calories?  specific heat of water = 4.18 J/g·°C 
Try below
final exam chapter 6

13. 6.2 Enthalpy and Calorimetry Concept 2: Calculating Enthalpy and Calorimetry
q = (25 g)x(4.18 J/g·°C)[(100 °C - 0 °C)]  q = (25 g)x(4.18 J/g·°C)x(100 °C)  q = J  J = 1 calorie  x calories = J x (1 cal/4.18 J)  x calories = 10450/4.18 calories  x calories = 2500 calories  Answer:  J or 2500 calories of heat energy are required to raise the temperature of 25 grams of water from 0 °C to 100 °C.
final exam chapter 6

14. Change in enthalpy (∆H) or change in heat of system
Chapter 6.3 Hess’s Law Concept 3: Hess’s Law and calculating change in change in enthalpy
Change in enthalpy (∆H) or change in heat of system
N2 (g) + 2O2 (g) 2NO2 (g) ∆H = +68
Endothermic
2NO2 (g)  N2 (g) + 2O2 (g) ∆H = - 68kJ
Exothermic
final exam chapter 6

15. Calculate ∆H for N2 (g) + 2O2 (g) 2NO2 (g)
Chapter 6.3 Hess’s Law Concept 3: Hess’s Law and calculating change in change in enthalpy
Calculate ∆H for N2 (g) + 2O2 (g) 2NO2 (g)
N2 (g) + O2 (g)  2NO (g) ∆H = 180kJ
2NO2  2NO (g) + O2 (g) ∆H = 112kJ
See how you need to reverse the second equation and then change from endothermic to exothermic
Add the two together to get final answer
∆H = +68
final exam chapter 6

16. Chapter 6.3 Hess’s Law Concept 3: Hess’s Law and calculating change in change in enthalpy
final exam chapter 6

17. Chapter 6.3 Hess’s Law Concept 3: Hess’s Law and calculating change in change in enthalpy
final exam chapter 6

18. Chapter 6.3 Hess’s Law Concept 3: Hess’s Law and calculating change in change in enthalpy
final exam chapter 6

19. Chapter 6.3 Hess’s Law Concept 3: Hess’s Law and calculating change in change in enthalpy
final exam chapter 6

20. Chapter 6.3 Hess’s Law Concept 3: Hess’s Law and calculating change in change in enthalpy
final exam chapter 6