AP Chemistry Exam Part 2
Chapter 6: Thermochemistry
final exam chapter 6
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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
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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
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6.1 Nature of energy
Concept 1: What is energy?
• KE=1/2mv2
• (1/2)225*1132
• 1,436,512 joules
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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
•
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6.1 Nature of energy
Concept 1: What is energy?
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6.1 Nature of energy
Concept 1: What is energy?
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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
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6.1 Nature of energy
Concept 1: What is energy?
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6.1 Nature of energy
Concept 1: What is energy?
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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
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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
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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 = 10450 J
4.18 J = 1 calorie
x calories = 10450 J x (1 cal/4.18 J)
x calories = 10450/4.18 calories
x calories = 2500 calories
Answer:
10450 J or 2500 calories of heat energy are required to raise the
temperature of 25 grams of water from 0 °C to 100 °C.
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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
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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)
– 2NO2  2NO (g) + O2 (g)
∆H = 180kJ
∆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
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Chapter 6.3 Hess’s Law
Concept 3: Hess’s Law and calculating change in change in enthalpy
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Chapter 6.3 Hess’s Law
Concept 3: Hess’s Law and calculating change in change in enthalpy
final exam chapter 6
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Chapter 6.3 Hess’s Law
Concept 3: Hess’s Law and calculating change in change in enthalpy
final exam chapter 6
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Chapter 6.3 Hess’s Law
Concept 3: Hess’s Law and calculating change in change in enthalpy
final exam chapter 6
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Chapter 6.3 Hess’s Law
Concept 3: Hess’s Law and calculating change in change in enthalpy
final exam chapter 6
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