# Thermochemistry 2022 ```Warm- up
•
in the holder
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Log on to eClass and check the agenda
•
Fill in the notes from the back as you go
through the Nearpod
Thermochemistry
notes
The chemistry of Heat &amp;
Energy
Part 1 What is Energy?
Energy is the
ability to do work
or produce heat.
What is Energy?
Energy is the ability to do work or
produce heat.
Energy exists in 3 forms
• potential energy
• kinetic energy
• in the form of heat
Potential Energy
“stored” energy that has the
potential to do work; this is also
energy due to the position of an
object
Kinetic Energy
• the energy of motion
*Most
energetic!
• Which particles have the most
kinetic energy?
Law of Conservation of Energy
• Energy cannot be created nor destroyed
only rearranged (transferred).
• Energy can be converted from one form to
another
Energy
Heat (q) is energy that is in the process of
flowing from a warmer object to a colder
object
Heat can flow in one of two
directions:
Exothermic
To give off heat; energy
is lost (-q)
Endothermic
To absorb heat; energy is
Examples of Exothermic
Reactions
Burning
Fireworks
Examples of Endothermic
Processes
melting
Evaporating
Temperature &amp; Heat
Temperature is related to the average kinetic
energy of the particles in a substance.
Cold
Hot
Measuring Temperature
Temperature may be measured using
Celsius or Kelvin.
Celsius
Kelvin
Part 2
Measuring Heat
• A unit for measuring heat is the calorie
(cal)
• calorie -the amount of heat needed
to raise 1 g of water 1&deg;Celsius
1 Calorie = 1000
cal
A fruit and oatmeal bar contains 142 Calories.
Convert this energy to calories.
142 Cal 1000 cal
1 Cal
1 Cal = 1000 cal
=142,000 cal
Joules
The SI unit for measuring heat and
energy is the Joule (J)
1 cal = 4.184 J
Burning a peanut releases 30 Calories
a) how many calories are released?
30 Cal
1000 cal
= 30,000cal
1 Cal
b) how many Joules are released?
Burning a peanut releases 30 Calories
a) how many calories are released?
30 Cal
1000 cal
= 30,000cal
1 Cal
b) how many Joules are released?
30000 cal
4.184 J
1 cal
=125,520 J
Part 3
Storing Heat
The following slide shows a graph of
different substances being heated
for up to 60 minutes..
After 30 minutes, the metal temp is ~45oC.
Storing Heat
Different substances have different
capacities for storing heat energy
(Specific Heat)
0.6 J/g&deg;C
4.184
J/g&deg;C
1.0 J/g&deg;C
Specific Heat (c)
The amount of heat
needed to raise the
temp of 1g of a
substance by 1&deg;C is
called specific heat.
Why does water have such a
high specific heat?
water
metal
• Water molecules form strong bonds with
each other; therefore it takes more heat
energy to break them.
• Metals have weak bonds and do not need as
much energy to break them.
How to calculate changes in
heat energy
Heat calculations
Use the formula:
q = mc△T
q
m
c
△T
How to calculate changes in heat
energy
q = mc△T
Heat lost or gained (J)
q =_____________________________
mass of the substance (g)
m =_____________________________
change in temperature (Tf – Ti)
△T = ___________________________
specific heat of substance (j/g ◦C)
C = ____________________________
Warm up-Practice Problem 1 on page 7
in the notes
How much heat is transferred to a 5.00 g piece
of iron, initially at 20.0 oC, when it is placed in a
beaker of boiling water at 1 atm?
q = m x c x △T
specific heat of iron is
0.449 J/goC.
Heat calculations
Use the formula:
q = mc△T
q
m
c
△T
Practice Problem 1
How much heat is transferred to a 5.00 g piece
of iron, initially at 20.0 oC, when it is placed in a
beaker of boiling water at 1 atm?
q = m x c x △T
q = 5.00 x 0.449 x (100-20)
q = 179.6 J
specific heat of iron is
0.449 J/goC.
Practice Problem 2
How many joules are required to lower
the temperature of 100.0 g of iron from
75.0 oC to 25.0 oC?
q = m x c x △T
q = 100 x 0.449 x (25-75)
q = - 2245 J
specific heat of iron is
0.449 J/goC.
Practice Problem 3
How many calories of energy are given off to
lower the temperature of 100.0 g of iron from
150.0 oC to 35.0 oC?
q = m c△T
q = 100 x 0.449 x (35-150)
q = -5163.5 J
specific heat of iron is
0.449 J/goC.
Solution: Problem 4
How much heat is transferred to a 10.00 g piece of
iron, initially at 20.0 oC, when it is placed in a beaker
of boiling water at 1 atm?
q = m x c x △T
specific heat of
iron is 0.449
J/goC.
q = 359.2 J
Calorimetry
and Specific
Heat
The amount of calories contained in a food sample
can be determined using a calorimeter.
What is a calorimeter?
• Device used to measure the heat given
off during chemical reactions
The specific heat of liquid water is
4.184
0
J/g C
We can use this known information
to calculate the specific heat of
other substances
Assuming that all of the heat from a burning peanut is absorbed by
50ml (50g) of water in a calorimeter, calculate the amount of
calories in the peanut if the water temp increased from 24˚C to
64˚C. (specific heat of water = 4.184 J/g˚C)
q = m c△T
1 cal = 4.184J
q = ? (this is the unknown in this
question)
m=
c=
ΔT=
1 cal
4.184 J
Assuming that all of the heat from a burning peanut is absorbed by
50ml (50g) of water in a calorimeter, calculate the amount of
calories in the peanut if the water temp increased from 24˚C to
64˚C. (specific heat of water = 4.184 J/g˚C)
q = m c△T
1 cal = 4.184J
q = ? (this is the unknown in this
question)
m = 50 g
c= 4.184 J/g0C
ΔT= (64 - 24) = 400C
8368 J
1 cal
4.184 J
= 2000 cal
THIS
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How much heat is in your snack
•Purpose
• To calculate how many calories a Cheeto has
•Safety
• Goggles
• It’s fire, be careful
Clean up, clean up, everybody
clean up
•Empty the soda cans
•Throw away any leftover Cheeto
•Clean up any crumbs
•Clean up any spills
Calculations
•Flaming Hot Cheeto Puffs
•0.44 Cal/g
•Regular Cheeto Puffs
•0.44 Cal/g
A fruit and oatmeal bar contains 142 Calories.
Convert this energy to calories.
142 Cal 1000 cal
1 Cal
1 Cal = 1000 cal
=142,000 cal
Burning a peanut releases 30 Calories
a) how many calories are released?
30 Cal
1000 cal
= 30,000cal
1 Cal
b) how many Joules are released?
Burning a peanut releases 30 Calories
a) how many calories are released?
30 Cal
1000 cal
= 30,000cal
1 Cal
b) how many Joules are released?
30000 cal
4.184 J
1 cal
=125,520 J
Practice Problem 3
How many joules of energy are given off to
lower the temperature of 100.0 g of iron from
150.0 oC to 35.0 oC?
q = m c△T
specific heat of iron is
0.449 J/goC.
Practice Problem 3
How many calories of energy are given off to
lower the temperature of 100.0 g of iron from
150.0 oC to 35.0 oC?
q = m c△T
q = 100 x 0.449 x (35-150)
q = -5163.5 J
specific heat of iron is
0.449 J/goC.
Problem 4
How much heat is transferred to a 10.00 g piece of
iron, initially at 20.0 oC, when it is placed in a beaker
of boiling water at 1 atm?
q = m x c x △T
specific heat of
iron is 0.449
J/goC.
q=
Solution: Problem 4
How much heat is transferred to a 10.00 g piece of
iron, initially at 20.0 oC, when it is placed in a beaker
of boiling water at 1 atm?
q = m x c x △T
specific heat of
iron is 0.449
J/goC.
q = 359.2 J
Warm up
•
•
•
Finish/ turn in How much heat
Finish/ turn in WORKSHEET:
Energy Units Conversions from
Tuesday
Today
•
•
Notes on Specific Heat
Calculations
Specific Heat Calculations
● You can determine the
specific heat of an unknown
metal by placing a heated
piece of the metal in water
in a calorimeter.
● Once you have measured
the final temperature of
both the metal and the
water, you can work out the
specific heat of the metal.
q metal = 30.5 x c x -61.9
q water = 50.0 x 4.184 x 9.1
q metal = q water
1) A 100. g sample of water at 25.3 oC was placed in
a calorimeter. 45.0 g of lead shots (at 100 oC) was
added to the calorimeter and the final temperature
of the mixture was 34.4 oC. How much heat was
given off by the lead? What is the specific heat of
First, find the q for water
q = m c△T
Hot
1) A 100. g sample of water at 25.3 oC was placed in
a calorimeter. 45.0 g of lead shots (at 100 oC) was
added to the calorimeter and the final temperature
of the mixture was 34.4 oC. How much heat was
given off by the lead? What is the specific heat of
First, find the q for water
q = m c△T
Hot
q = 100 x 4.184 x (34.4- 25.3)
=
1) Then, use that (q) to calculate the (c) of Lead.A
100. g sample of water at 25.3 oC was placed in a
calorimeter. 45.0 g of lead shots (at 100 oC) was
added to the calorimeter and the final temperature
of the mixture was 34.4 oC.
*Law of conservation of energy: heat given
off by lead = heat transferred to water
q = m c△T
q = 100 x 4.184 x (34.4- 25.3)
= 3807.4 J
Hot
-3807.4 = 45.0 x c x (34.4 - 100)
=c
1) Then, use that (q) to calculate the (c) of Lead.A
100. g sample of water at 25.3 oC was placed in a
calorimeter. 45.0 g of lead shots (at 100 oC) was
added to the calorimeter and the final temperature
of the mixture was 34.4 oC.
*Law of conservation of energy: heat given
off by lead = heat transferred to water
q = m c△T
q = 100 x 4.184 x (34.4- 25.3)
= 3807.4 J
Hot
-3807.4 = 45.0 x c x (34.4 - 100)
1.29 J/g0C
=c
1) Then, use that (q) to calculate the (c) of Lead.A
100. g sample of water at 25.3 oC was placed in a
calorimeter. 45.0 g of lead shots (at 100 oC) was
added to the calorimeter and the final temperature
of the mixture was 34.4 oC.
*Law of conservation of energy: heat given
off by lead = heat transferred to water
q = m c△T
-3807.4 = 45.0 x c x (34.4 - 100)
-3807.4 = -2952 x c
Hot
-2952
-2952
1.29 J/g0C
=c
A 17.9 g sample of unknown metal was heated to
48.31 oC. It was then added to 28.05 g of water in an
insulated cup. The water temperature rose from
21.04 oC to 23.98oC. What is the specific heat of the
metal in J/goC?
q = m c△T
First find the heat given off from the metal to the
water
q = 28.05 x 4.184 x (23.98- 21.04)
= 345 J
Second, find the specific heat (c) of the metal.
-345 = 17.9 x c x (23.98- 48.31)
0.792 J/g&deg;C = c
Practice
The temperature of a piece of copper with a mass
of 95.4g increases from 25oC to 48 oC when the
metal absorbs 849J of heat. What is the specific
heat of copper in J/goC?
q = m c△T
849 = 95.4 x c x (48-25)
849
= c = 849
(95.4 x 23)
2194.2
0.387 J/g&deg;C = c
Remember:
(-q) = Exothermic = releases energy
(+q) = Endothermic = absorbs energy
Today
•
Grab a device
•
form
•
Warm up
•
particles that could
be in the solid,
liquid or gas phase.
descriptions and
identify the phases
of matter (what
phase is 1, 2, and
3)?
Today
•
•
Heat and Phase Changes
Notes
Phase Changes and Heating
Heat &amp; Phase
Changes
Remember:
(-q) = Exothermic = releases energy
(+q) = Endothermic = absorbs energy
Kinetic Theory of Matter
An explanation of how particles in matter behave.
There are three basic assumptions to the kinetic theory:
1. All matter is composed of small particles
(atoms, molecules, and ions).
2. These particles are in constant, random
motion.
3. These particles are colliding with
each other and the walls
of their container.
Kinetic Theory-Solids
• Particles of a solid are tightly packed
vibrating in a fixed position.
• Solids have a definite shape and
volume.
• Solids do not change shape (volume)
without outside forces
Kinetic Theory-Liquids
•
Particles of liquids are not as tightly packed
as solids, they are able to to slide over one
another.
• Liquids have indefinite shape and definite
volume
• Liquids will spread out, but will not fill any
container
Kinetic Theory-Gases
• Particles of gases are very far apart and move
freely.
• Gases do not have a definite volume or
shape.
• Gases will completely fill any container
Warm up
particles that could
be in the solid,
liquid or gas phase.
descriptions and
identify the phases
of matter (what
phase is 1, 2, and
2- Liquid;3)?3- Gas
•
1- Solid;
Plasma
Endothermic PHASE CHANGES
Description of
Phase Change
Solid to
liquid
Liquid to
gas
Solid to gas
Term for Phase
Change
Heat Movement During
Phase Change
Melting
Heat goes into the
solid as it melts.
(endothermic)
Evaporation
Heat goes into the
liquid as it vaporizes
(endo)
Sublimation
Heat goes into the solid
as it sublimates (endo)
Exothermic PHASE CHANGES
Description of
Phase Change
Term for Phase
Change
Condensation
Gas to liquid
Liquid to
solid
Freezing
Heat Movement During
Phase Change
Heat leaves the
gas as it
condenses (exo)
Heat leaves the
liquid as it freezes
(exothermic)
What is happening
-As temperature increases/ heat is added,
intermolecular forces are being broken
Heating Curves
A heating curve shows the temperature change as
thermal energy, or heat, is added to
a substance
Heating curve video
Which region(s) represent kinetic energy (temperature) of the
substance increasing?1, 3 and 5
Which region(s) represent kinetic energy of the substance staying
the same?___________
Does the graph represent an endothermic or exothermic process?
2 and 4 represent
potential energy
increasing. There is no
increase in temperature
Which region(s) represent kinetic energy (temperature) of the
substance increasing?1, 3 and 5
Which region(s) represent kinetic energy of the substance staying
the same?2 and 4
Does the graph represent an endothermic or exothermic process?
Which region(s) represent kinetic energy (temperature) of the
substance increasing?1, 3 and 5
Which region(s) represent kinetic energy of the substance staying
the same?2 and 4
Does the graph represent an endothermic or exothermic process?
Heating Curve for Water
1. What is happening to water at each letter?
A. ____________
B. ____________
C. ____________
D. ____________
E. ____________
Heating Curve for Water
1. What is happening to water at each letter?
(ice)
A. Solid
____________
B. Melting
____________
(water)
C.Liquid
____________
D. ____________
Boiling
E. Gas
____________
(steam)
Assignment
•
•
Phase Changes and Heating Curves
Heating Curves
To calculate the energy changes as a substance warms
or cools the equation q=mCΔT is used.
Ex: Calculate the heat required to pass through
region A-B if 20g of ice was being heated.
(specific heat ice = 2.09J/g&deg;C)
q = mc∆T =
20 x 2.09 x ( 0 - -25) = 1045J
Calculate the heat required to pass through
region C-D if 20 g of water was being heated.
(specific heat liquid water = 4.184 J/g&deg;C)
q = mc∆T =
20 x 4.184 x ( 100 - 0) = 8368J
Calculate the heat required to pass through
region E-F if 20 g of steam was being heated.
(specific heat water vapor= 2.02 J/g&deg;C)
q = mc∆T = 20 x 2.02 x ( 125 - 100) = 1263J
Enthalpy and Hess’ Law
System &amp; Surroundings
• Scientists study heat changes in systems and
surroundings.
• A system is the part of the universe we are
studying.
• The surroundings are everything else outside
of the system
• System + surroundings = the universe
System and Surroundings
Beaker &amp; solution
Table &amp; air
Heat is Exchanged Between the
System &amp; the Surroundings
Energy out
Energy in
Enthalpy (H)
• Heat transferred from a system to its
surroundings during a chemical reaction at
constant pressure is called Enthalpy (H) .
• Can’t measure H, only ΔH
• At constant pressure, ΔH = q = mCΔT
What is Enthalpy?
Enthalpy- a measure of heat-flow within a
close system
-can only be measured as a change,
not a state of being.
-noted as ΔH
ΔH = q
ΔH = mcΔT
**this only works in a
closed system
Measuring Enthalpy (ΔH)
If heat leaves system and goes into
surroundings, and it feels hot to the touch
ΔH = - (exothermic)
Measuring Enthalpy (ΔH)
If heat goes from surroundings into system
and it feels cool to the touch
ΔH = + (endothermic)
Heat of Formation Values (ΔHf )
• Every substance has an energy associated with it
called ΔHf (ENTHALPY OF FORMATION).
• REMEMBER to make sure the substance you
look up also have the correct states of matter.
Enthalpies of Reaction
• All reactions have some ΔH associated with it
2H2(g) + O2(g) → 2H2O(l)
ΔH = - 572 kJ
• How did we calculate that ΔH?
ΔHrxn = (sum of ΔHproducts) - (sum of the ΔHreactants)
Enthalpies of Reaction
• How do we know the ΔH’s of the products and
reactants?
ΔHrxn = (sum of ΔHproducts) - (sum of the ΔHreactants)
First: elements have a ΔH of zero (0).
Second: we use a table of known values for
compounds
Enthalpies of Reaction
• All reactions have some ΔH associated with it
2H2(g) + O2(g) → 2H2O(l)
• ΔH H2 = 0 kJ/mol
• ΔH O2 = 0 kJ/mol
• ΔH H2O = -285.83 kJ/mol
ΔH = - 572 kJ
ΔHrxn = (sum of ΔHproducts) - (sum of the ΔHreactants)
BUT!! It is 2H2O, so:
•
•
ΔH = (2*-285.83) - (2*0 + 0)
ΔH = -572 kJ
Hess’ Law
Hess’s Law: If you add two or
more chemical equations to
get an overall equation, then
you can also add the heat
changes (ΔHs) to get the
overall heat change.
Hess Law Example
• 3A + 3B 🡪 4C + D
• 2A + 2B 🡪 2C
• A + B 🡪 2C + D
• 3A + 3B 🡪 4C + D
ΔH = ????
ΔH = -594kJ
ΔH = 198kJ
ΔH = -396kJ
• To calculate the heat of reaction we add all the
Enthalpy changes (ΔH) for all of the reactions.
equations:
1. If two identical substances are on
opposite sides of the arrow, they will
cancel (reduce).
2. If two identical substances are on
same side of the arrow, add the
coefficients together.
3. Keep substances on the same side of
the arrow in the final equation.
equations.
C + O2 🡪 CO2
CO2 🡪 CO + &frac12; O2
equations.
C + O2 🡪 CO2
CO2 🡪 CO + &frac12; O2
equations.
C + O2 🡪 CO2
CO2 🡪 CO + &frac12; O2
C + &frac12;O2 🡪 CO
equations.
2 Cu + O2 🡪 2 CuO
4 Cu + O2 🡪 2 Cu2O
6Cu + 2O2 🡪 2CuO + 2Cu2O
Hess’ Law
• Basically: if we can add the reactions
together, then we can add the ΔH’s
• A + 2B → C + D
•
B → 2C + D
• A + 3B → 3C + 2D
•
ΔH= -20 kJ
ΔH= -45 kJ
ΔH = ?
Practice
• A + 2B → C + D
•
B → 2C + D
• A + 3B → 3C + 2D
H= -20 kJ
H= -45 kJ
ΔH = -65 kJ
Practice
• 2A + 5B → C + D
•
A→C+B
• 3A + 4B → 2C + D
H= -237 kJ
H= 200 kJ
ΔH = -37 kJ
C2H4 🡪 2 C + 2 H2
2 C + 3 H2 🡪 C2H6
ΔH= 152kJ
ΔH= -85kJ
C2H4 + H2 → C2H6
ΔH = 67
kJ
End of material
-ticket-out-of-the-door
-Asynchronous assignment tomorrow
-Test for Unit 5- Thermo/Kinetics is May
6th
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