Thermochemistry
Chapter 17
Thermochemistry
• Thermochemistry is the study of energy
changes that occur during chemical reactions
and changes in state of matter.
Thermochemistry
The energy is stored in bonds between atoms
and is called chemical energy. It is a form of
potential energy.
Heat
• Heat (q) is energy transferred from one object
to another because of the temperature
difference between them.
• Heat always flows from a warmer object to a
cooler object.
Types of Reactions
• In an endothermic process a reaction
absorbs/requires/gains heat.
– Melting, photosynthesis, cooking
• In an exothermic process a reaction
releases/loses heat.
– Freezing, cellular respiration, fire
Measuring Heat
• Heat flow is measured in two common units, the
calorie (c) and the joule (J).
• A calorie is the amount of heat needed to raise
the temperature of one gram of water 1°C.
• 1 J = 0.2390 cal
• 4.184 J = 1 cal
Measuring Heat
• The specific heat of a substance is the amount of heat
it takes to raise the temperature of 1 g of a substance
1°C
• Specific heat (C) is calculated with the following
equation:
C =
q
m x ΔT
Where q is heat, m is mass, and ΔT is change in
temperature
Practice specific heat problem
• The temperature of a 98.4 g piece of copper
increases from 25°C to 48°C when copper
absorbs 849J of heat. What is the specific heat
of copper?
Practice specific heat problem
• The temperature of a 98.4 g piece of copper
increases from 25°C to 48°C when copper
absorbs 849J of heat. What is the specific heat
of copper?
C =
q
m x ΔT
Practice specific heat problem
• The temperature of a 98.4 g piece of copper
increase from 25°C to 48°C when copper
absorbs 849J of heat. What is the specific heat
of copper?
C = 849J
(98.4g) x (23°C)
Practice specific heat problem
• The temperature of a 98.4 g piece of copper
increase from 25°C to 48°C when copper
absorbs 849J of heat. What is the specific heat
of copper?
C = 849J
(98.4g) x (23°C)
C = 0.375132556
Practice specific heat problem
• The temperature of a 98.4 g piece of copper
increase from 25°C to 48°C when copper
absorbs 849J of heat. What is the specific heat
of copper?
C = 849J
(98.4g) x (23°C)
C = 0.375132556 = 0.38 J/g°C
One more practice problem
• When 435 J of heat is added to 3.4 g of olive
oil at 21°C, the temperature increases to
85°C. What is the specific heat of the olive
oil?
C = q
m x ΔT
One more practice problem
• When 435 J of heat is added to 3.4 g of olive
oil at 21°C, the temperature increases to
85°C. What is the specific heat of the olive
oil?
C = 435
3.4 g x 64°C
One more practice problem
• When 435 J of heat is added to 3.4 g of olive
oil at 21°C, the temperature increases to
85°C. What is the specific heat of the olive
oil?
C = 435
3.4 g x 64°C
= 1.999080882 = 2.0 J/g°C
Enthalpy
• The heat constant of a system is called
enthalpy (H), measured in Joules (J)
• The heat absorbed or released by a system or
reaction is called the change in enthalpy or ΔH
• You can calculate enthalpy with this equation:
ΔH=-m x C x ΔT
m= mass, C = specific heat, ΔT = change in temp
Practice Enthalpy Problem
• 50.0g of water is heated from 25°C to 32°C.
The specific heat of water is 4.18 J/g°C.
Calculate the enthalpy change in this reaction.
Practice Enthalpy Problem
• 50.0g of water is heated from 25°C to 32°C.
The specific heat of water is 4.18 J/g°C.
Calculate the enthalpy change in this reaction.
• ΔH=-m x C x ΔT
Practice Enthalpy Problem
• 50.0g of water is heated from 25°C to 32°C.
The specific heat of water is 4.18 J/g°C.
Calculate the enthalpy change in this reaction.
• ΔH=-(50.0g) x (4.18) x (32-25)
Practice Enthalpy Problem
• 50.0g of water is heated from 25°C to 32°C.
The specific heat of water is 4.18 J/g°C.
Calculate the enthalpy change in this reaction.
• ΔH=-(50.0g) x (4.18) x (32-25)
=-1463 J = -1500 J
Another Practice
• Calculate the change in enthalpy when you
increase the temperature of 250g of water
from 20 to 46C. The specific heat of water is
4.18 J/g°C.
Another Practice
• Calculate the change in enthalpy when you
increase the temperature of 250g of water
from 20 to 46C. The specific heat of water is
4.18 J/g°C.
• ΔH=-m x C x ΔT
Another Practice
• Calculate the change in enthalpy when you
increase the temperature of 250g of water
from 20 to 46C. The specific heat of water is
4.18 J/g°C.
• ΔH=-(250) x (4.18) x (46-20)
Another Practice
• Calculate the change in enthalpy when you
increase the temperature of 250g of water
from 20 to 46C. The specific heat of water is
4.18 J/g°C.
• ΔH=-(250) x (4.18) x (46-20)= -27170
Another Practice
• Calculate the change in enthalpy when you
increase the temperature of 250g of water
from 20 to 46C. The specific heat of water is
4.18 J/g°C.
• ΔH=-(250) x (4.18) x (46-20)= -27170
=-30000J
Math Practice
• Complete the practice problems for specific heat and
enthalpy.
Phase Changes:
Freezing and Melting
• When a substance melts, it absorbs heat, called the
heat of fusion.
Phase Changes:
Freezing and Melting
When a substance freezes, it loses heat, called the heat
of solidification.
• The quantity of heat absorbed by one mole of a
melting solid is exactly the same as the quantity of
heat released when one mole of the liquid freezes.
Phase changes:
Vaporization and Condensation
• When liquid evaporates, it absorbs heat,
called heat of vaporization.
Phase changes:
Vaporization and Condensation
When gas condenses to liquid, it releases heat,
called heat of condensation.
• The quantity of heat absorbed by a vaporizing
liquid is exactly the same as the quantity of
heat released when the vapor condenses.
Specific Heat of Metals Lab
• 1. Read and annotate the lab procedure.
• 2. Complete the pre-lab.
• 3. This is a stand-alone lab report! You will
either handwrite or type and print it – it does
not go in your notebooks!
• 4. Be ready to conduct the lab next class
period.