Transcript Chemistry You Need to Know

Chapter 7: Hot & Cold Packs
Introductory Activity
How many things can you think of in
everday life that either give off heat or
absorb heat?
Which of these things are physical
processes?
Which are chemical processes?
Hot & Cold Packs
This chapter will introduce the chemistry
needed to understand how Hot & Cold
Packs work
Section 7.1: Endothermic & Exothermic
Section 7.2: Calorimetry and heat capacity
Section 7.3: Changes in State
Section 7.4: Heat of a Chemical Reaction
Section 7.5: Hess’s Law
Hot/Cold Packs
Use
Transfer of energy
between
System &
Surroundings
Is determined with
Calorimetry
Physical
change
Can be done in
Effect on
temperature
depends on
Chemical
change
Materials ability to
absorb energy
without noticeable
temperature
change
Section 7.1—Endothermic and
Exothermic
Why are hot packs “hot” and cold packs feel “cold”
Endothermic & Exothermic
When the system absorbs energy from the
surroundings, it’s an endothermic process
When the system releases energy to the
surroundings, it’s an exothermic process
System & Surroundings
It’s very important to define the system &
surroundings correctly to use the endoand exothermic definitions!
Most people define the system too broadly
They include everything in the beaker or
container as the system
However, the system is only the molecules
undergoing the change
System & Surroundings
The system is only made of the
molecules undergoing the change
The water molecules & the container…
Your hand and the air…
Even the thermometer…
They are all the surroundings
Note that water is made up of water
molecules—not a solid chunk of water…but
for this picture, it’s best to represent water as
one thing since it’s the surroundings and focus
on the molecules reacting as the system.
Exothermic & You
You touch the beaker and it feels
hot
Energy is being transferred TO YOU
You are the surroundings
When energy moves from system to
surroundings, it’s exothermic
Exothermic & the Thermometer
The temperature (measured by the
thermometer) is related to the
average kinetic energy of the
molecules in the container
The majority of the molecules in a
solution are water
If the temperature is increasing, the
energy of the water molecules is
increasing
Since water is the surrounding (it’s
not actually reacting), energy is being
transferred to the surroundings
Exothermic shows an increase in the temperature within the container
Endothermic
The opposite is also true
If the container feels cold to you, energy is
being transferred FROM YOU (the
surroundings) into the system—
endothermic
If the thermometer goes down, energy is
being transferred FROM the water
molecules (surroundings) into the system-endothermic
Let’s Practice
Example:
Identify the system
and surroundings
when you hold an
ice cube while it
melts. Is this endoor exothermic?
Let’s Practice
Example:
Identify the system
and surroundings
when you hold an
ice cube while it
melts. Is this endoor exothermic?
System: Water molecules in the form of ice
Surroundings: You and the air
It feels cold to you…so energy is leaving you (surroundings)
When energy goes from surroundings to system it’s endothermic
Section 7.2—Calorimetry &
Heat Capacity
Why do some things get hot more quickly than others?
Temperature
Temperature – proportional to the
average kinetic energy of the molecules
Energy due to motion
(Related to how fast the
molecules are moving)
As temperature
increases
Molecules move
faster
Heat & Enthalpy
Heat (q)– The flow of energy from
higher temperature particles to lower
temperature particles
Enthalpy (H)– Takes into account the
internal energy of the sample along with
pressure and volume
Under constant pressure (lab-top conditions), heat and enthalpy
are the same…we’ll use the term “enthalpy”
Energy Units
The most common energy units are Joules (J) and
calories (cal)
Energy Equivalents
4.18 J
=
1.00 cal
1000 J
=
=
1 kJ
1000 cal
1 Cal (food calorie)
These equivalents can be used in dimensional analysis to convert units
Heat Capacity
Specific Heat Capacity (Cp) – The
amount of energy that can be absorbed
before 1 g of a substance’s temperature
has increased by 1°C
Cp for liquid water = 1.00 cal/g°C or 4.18 J/g°C
Heat Capacity
High Heat Capacity
Low Heat Capacity
Takes a large amount of
energy to noticeably
change temp
Small amount of energy can
noticeably change
temperature
Heats up slowly
Cools down slowly
Heats up quickly
Cools down quickly
Maintains temp better with
small condition changes
Quickly readjusts to new
conditions
A pool takes a long time to warm up and remains fairly warm over night.
The air warms quickly on a sunny day, but cools quickly at night
A cast-iron pan stays hot for a long time after removing from oven.
Aluminum foil can be grabbed by your hand from a hot oven because it
cools so quickly
What things affect temperature change?
 Heat Capacity of substance
The higher the heat capacity, the slower the
temperature change
 Mass of sample
The larger the mass, the more molecules there are to
absorb energy, so the slower the temperature change
Specific heat capacity of
substance
H  m  C p  T
Energy added
or removed
Mass of sample
Change in temperature
Positive & Negative T
 Change in temperature (T) is always T2 – T1
(final temperature – initial temperature)
If temperature increases, T will be positive
 A substance goes from 15°C to 25°C.
 25°C - 15°C = 10°C
 This is an increase of 10°C
If temperature decreases, T will be negative
 A substance goes from 50°C to 35°C
 35°C – 50°C = -15°C
 This is a decrease of 15°C
Positive & Negative H
Energy must be put in for temperature to
increase
A “+” T will have a “+” H
Energy must be removed for temperature
to decrease
A “-” T will have a “-” H
Example
Example:
If 285 J is added to 45 g of
water at 25°C, what is the
final temperature? Cp water
= 4.18 J/g°C
Let’s Practice #1
Example:
How many joules must be
removed from 25 g of water
at 75°C to drop the
temperature to 30°? Cp
water = 4.18 J/g°C
Let’s Practice #2
Example:
If the specific heat capacity
of aluminum is 0.900 J/g°C,
what is the final temperature
if 437 J is added to a 30.0 g
sample at 15°C
Calorimetry
Conservation of Energy
1st Law of Thermodynamics – Energy
cannot be created nor destroyed in
physical or chemical changes
This is also referred to as the Law of Conservation of Energy
If energy cannot be created nor destroyed, then energy lost by the
system must be gained by the surroundings and vice versa
Calorimetry
Calorimetry – Uses the energy change
measured in the surroundings to find
energy change of the system
Because of the Law of Conservation of Energy,
The energy lost/gained by the surroundings is equal to but opposite
of the energy lost/gained by the system.
Hsurroundings = - Hsystem
(m×Cp×T)surroundings = - (m×Cp×T)system
Don’t forget the “-” sign on one side
Make sure to keep all information about surroundings together and all
information about system together—you can’t mix and match!
Two objects at different temperatures
Thermal Equilibrium – Two objects at
different temperatures placed together
will come to the same temperature
So you know that T2 for the system is the same as T2 for the surroundings!
An example of Calorimetry
Example:
A 23.8 g piece of unknown metal is heated to
100.0°C and is placed in 50.0 g of water at
24°C water. If the final temperature of the
water is 32.5°,what is the heat capacity of
the metal?
An example of Calorimetry
Example:
A 23.8 g piece of unknown metal is heated to
100.0°C and is placed in 50.0 g of water at
24°C water. If the final temperature of the
water is 32.5°,what is the heat capacity of
the metal?
Metal:
H metal  H water
m = 23.8 g
T1 = 100.0°C
m Cp  T metal  m Cp  T water
T2 = 32.5°C


23.8 g  C p  32.5 C  100.0 C    50.0 g  4.18 J   32.5 C  24.5 C 
Cp = ?
gC


Water:


m = 50.0 g
  50.0 g  4.18 J   32.5 C  24.5 C 
gC

T1 = 24°C
Cp  


23.8 g  32.5 C  100.0 C 
T2 = 32.5°C
Cp = 4.18 J/g°C
C = 1.04 J/g°C
p
Let’s Practice #3
Example:
A 10.0 g of aluminum (specific heat capacity
is 0.900 J/g°C) at 95.0°C is placed in a
container of 100.0 g of water (specific heat
capacity is 4.18 J/g°C) at 25.0°. What’s the
final temperature?
Section 7.3—Changes in
State
What’s happening when a frozen ice pack melts?
Change in State
To melt or boil,
intermolecular
forces must
be broken
Breaking
intermolecular
forces
requires
energy
The energy being
put into the
system is used
for breaking
IMF’s, not
increasing
motion
(temperature)
A sample with solid & liquid will not rise above the melting point until all
the solid is gone.
The same is true for a sample of liquid & gas
Melting
 When going from a solid to a liquid, some of the
intermolecular forces are broken
 The Enthalpy of Fusion (Hfus) is the amount of
energy needed to melt 1 gram of a substance
The enthalpy of fusion of water is 80.87 cal/g or 334 J/g
 All samples of a substance melt at the same
temperature, but the more you have the longer it
takes to melt (requires more energy).
H  m  H fus
Energy needed
to melt
Mass of the
sample
Energy needed
to melt 1 g
Example
Example:
Find the enthalpy of
fusion of a substance
if it takes 5175 J to
melt 10.5 g of the
substance.
Vaporization
 When going from a liquid to a gas, all of the rest
of the intermolecular forces are broken
 The Enthalpy of Vaporization (Hvap) is the amount
of energy needed to boil 1 gram of a substance
The Hvap of water is 547.2 cal/g or 2287 J/g
 All samples of a substance boil at the same
temperature, but the more you have the longer it
takes to boil (requires more energy).
H  m  Hvap
Energy needed
to boil
Mass of the
sample
Energy needed
to boil 1 g
Example
Example:
If the enthalpy of
vaporization of water
is 547.2 cal/g, how
many calories are
needed to boil 25.0 g
of water?
Increasing molecular motion (temperature)
Changes in State go in Both Directions
Vaporizing or
Evaporating
Liquid
Melting
Freezing
Solid
Gas
Condensing
Going the other way
 The energy needed to melt 1 gram (Hfus) is the
same as the energy released when 1 gram
freezes.
If it takes 547 J to melt a sample, then 547 J would be
released when the sample freezes. H will = -547 J
 The energy needed to boil 1 gram (Hvap) is the
same as the energy released when 1 gram is
condensed.
If it takes 2798 J to boil a sample, then 2798 J will be
released when a sample is condensed.
H will = -2798 J
Example
Example:
How much energy is
released with 157.5 g of
water is condensed?
Hvap water = 547.2 cal/g
Heating Curves
Heating curves show how the temperature changes as energy is added to the
sample
Melting &
Freezing
Point
Temperature
Boiling &
Condensing
Point
Heating curve of w ater
150
100
50
0
-50
Energy input
Going Up & Down
Moving up the curve requires energy, while moving down releases energy
Temperature
Heating curve of w ater
150
100
50
0
-50
Energy input
States of Matter on the Curve
Liquid & gas
Energy added breaks remaining IMF’s
Liquid Only
Energy added
increases temp
Temperature
Heating curve of w ater
150
100
50
0
-50
Energy input
Solid Only
Energy added
increases temp
Solid & Liquid
Energy added
breaks IMF’s
Gas Only
Energy added
increases temp
Different Heat Capacities
Liquid Only
Cp = 1.00 cal/g°C
The solid, liquid and gas states absorb
water differently—use the correct Cp!
Temperature
Heating curve of w ater
150
100
50
0
-50
Energy input
Gas Only
Cp = 0.48 cal/g°C
Solid Only
Cp = 0.51 cal/g°C
Changing States
Liquid & gas
Hvap = 547.2 cal/g
Temperature
Heating curve of w ater
150
100
50
0
-50
Energy input
Solid & Liquid
Hfus = 80.87 cal/g
Adding steps together
Temperature
Heating curve of w ater
150
100
50
0
-50
Energy input
If you want to heat ice at -25°C to water at 75°C, you’d have to first warm the ice
to zero before it could melt.
Then you’d melt the ice
Then you’d warm that water from 0°C to your final 75°
You can calculate the enthalpy needed for each step and then add them together
Example
Example:
How many calories are
needed to change 15.0 g of
ice at -12.0°C to steam at
137.0°C?
Useful information:
Cp ice = 0.51 cal/g°C H  m  Cp  T
Cp water = 1.00 cal/g°C
H  m  H fus
Cp steam = 0.48 cal/g°C
Hfus = 80.87 cal/g
H  m  Hvap
Hvap = 547.2 cal/g
Let’s Practice
Example:
How many needed to
change 40.5 g of water at
25°C to steam at 142°C?
Useful information:
Cp ice = 0.51 cal/g°C H  m  Cp  T
Cp water = 1.00 cal/g°C
H  m  H fus
Cp steam = 0.48 cal/g°C
Hfus = 80.87 cal/g
H  m  Hvap
Hvap = 547.2 cal/g
Section 7.4—Energy of a
Chemical Reaction
What’s happening in those hot/cold packs that contain chemical reactions?
Enthalpy of Reaction
Enthalpy of Reaction (Hrxn) – Net
energy change during a chemical
reaction
+Hrxn means energy is being added to the system—endothermic
-Hrxn means energy is being released from the system—exothermic
Enthalpy of Formation
Enthalpy of Formation (Hf) – Energy
change when 1 mole of a compound is
formed from elemental states
Heat of formation equations:
H2 (g) + ½ O2 (g)  H2O (g)
C (s) + O2 (g)  CO2 (g)
A table with Enthalpy of Formation values can be found in the Appendix of your text
Be sure to look up the correct state of matter:
H2O (g) and H2O (l) have different Hf values!
Enthalpy of Formation & Enthalpy of Reaction
Reactants are
broken apart
and Products
are formed.
Breaking apart
reactants is
the opposite
of Enthalpy of
Formation.
Forming
products is
the Enthalpy
of Formation.
The overall
enthalpy of
reaction is the
opposite of Hf
for the reactants
and the Hf for
the products
This is not the way a reaction occurs—reactants break apart and then
rearrange…remember Collision Theory from Chpt 2! But for when discussing
overall energy changes, this manner of thinking is acceptable.
Hrxn   H f prod   H f react
Hrxn = sum of Hf of all products – the sum of Hf reactants
Example
Example:
Find the Hrxn for:
CH4 (g) + 2 O2 (g)  2 H2O (g) + CO2 (g)
Hf (kJ/mole)
CH4 (g)
-74.81
O2 (g)
0
H2O (g)
-241.8
CO2 (g)
-393.5
Let’s Practice #1
Example:
Find the Hrxn for:
2 CH3OH (l) + 3 O2 (g)  2 CO2 (g) + 4 H2O (l)
Hf (kJ/mole)
CH3OH (l)
-238.7
O2 (g)
0
H2O (l)
-285.8
CO2 (g)
-393.5
Enthalpy & Stoichiometry
The Enthalpy of Reaction can be used
along with the molar ratio in the balanced
chemical equation
This allows Enthalpy of Reaction to be
used in stoichiometry equalities
Example
Example:
If 1275 kJ is released, how many grams of B2O3 is
produced?
B2H6 (g) + 3 O2 (g)  B2O3 (s) + 2 H2O (g)
H = -2035 kJ
Let’s Practice #2
If you need to produce 47.8 g B2O3, how many
kilojoules will be released?
B2H6 (g) + 3 O2 (g)  B2O3 (s) + 2 H2O (g)
H = -2035 kJ
Section 7.5—Hess’s Law
How can we find the enthalpy of a reaction using step-wise reactions?
Hess’s Law
Hess’ Law – The sum of the energy
changes during a series of reactions is
equal to the sum of the reaction.
In other words…if you go from Reactant A to Product Z all in one step, you
will have the same total energy change as someone that went from A to Z in
7 step—the energy from each of their 7 steps would add up to your 1 step
energy change.
Example
1
Label each step-wise equations with letters (“a”, “b”, “c”) if not
already done for you.
Calculate the enthalpy of the reaction
2N2 (g) + 5O2 (g)  2N2O5 (g)
Hrxn = ?
Use:
a 2 H2(g) + O2 (g)  2 H2O (l)
b N2O5 (g) + H2O (l)  2HNO3 (l)
c N2 (g) + 3 O2 (g) + H2 (g)  2 HNO3 (l)
Hrxn = -571.6 kJ
Hrxn = -76.6 kJ
Hrxn = -74.1 kJ
Example
2
For the first reactant in the overall reaction, find the step-wise reaction that has
the same chemical and the same state of matter. It doesn’t have to be on the
reactants side of the step-wise equation
If it is on the correct “side” write it as is. Write it’s label beside it, too (“a”, “b”)
If it’s on the wrong “side”, flip the equation. If you flip it, write it’s label as “-a” or “-b”.
Calculate the enthalpy of the reaction
2N2 (g) + 5O2 (g)  2N2O5 (g)
Hrxn = ?
Use:
a 2 H2(g) + O2 (g)  2 H2O (l)
b N2O5 (g) + H2O (l)  2HNO3 (l)
c N2 (g) + 3 O2 (g) + H2 (g)  2 HNO3 (l)
c
Hrxn = -571.6 kJ
Hrxn = -76.6 kJ
Hrxn = -74.1 kJ
N2 (g) + 3 O2 (g) + H2 (g)  2 HNO3 (l)
Example
3
Repeat Step 2 for each reactant & product in the overall
equation. If a reactant appears in more than one step-wise
reaction, skip that reactant or product and move onto the next
one.
Calculate the enthalpy of the reaction
2N2 (g) + 5O2 (g)  2N2O5 (g)
Hrxn = ?
Use:
a 2 H2(g) + O2 (g)  2 H2O (l)
b N2O5 (g) + H2O (l)  2HNO3 (l)
c N2 (g) + 3 O2 (g) + H2 (g)  2 HNO3 (l)
c
Hrxn = -571.6 kJ
Hrxn = -76.6 kJ
Hrxn = -74.1 kJ
N2 (g) + 3 O2 (g) + H2 (g)  2 HNO3 (l)
Example
3
Repeat Step 2 for each reactant & product in the overall
equation. If a reactant appears in more than one step-wise
reaction, skip that reactant or product and move onto the next
one.
Calculate the enthalpy of the reaction
2N2 (g) + 5O2 (g)  2N2O5 (g)
Hrxn = ?
Use:
a 2 H2(g) + O2 (g)  2 H2O (l)
b N2O5 (g) + H2O (l)  2HNO3 (l)
c N2 (g) + 3 O2 (g) + H2 (g)  2 HNO3 (l)
c
-b
Hrxn = -571.6 kJ
Hrxn = -76.6 kJ
Hrxn = -74.1 kJ
N2 (g) + 3 O2 (g) + H2 (g)  2 HNO3 (l)
2HNO3 (l)  N2O5 (g) + H2O (l)
Example
4
Use any un-used step-wise equations to get rid of unwanted
things. Putting them on opposite sides will allow them to
cancel.
Calculate the enthalpy of the reaction
2N2 (g) + 5O2 (g)  2N2O5 (g)
Hrxn = ?
Use:
a 2 H2 (g) + O2 (g)  2 H2O (l)
b N2O5 (g) + H2O (l)  2 HNO3 (l)
c N2 (g) + 3 O2 (g) + H2 (g)  2 HNO3 (l)
c
-b
-a
Hrxn = -571.6 kJ
Hrxn = -76.6 kJ
Hrxn = -74.1 kJ
N2 (g) + 3 O2 (g) + H2 (g)  2 HNO3 (l)
2 HNO3 (l)  N2O5 (g) + H2O (l)
2 H2O (l)  2 H2 (g) + O2 (g)
Example
Begin to cancel things out that appear on both the reactants and products side.
Your goal is to add up all the step-wise equations to equal the overall equation.
5
Multiply equations by a whole number if you need more of something to match
the overall reaction or to fully cancel something out that you don’t want in the
overall equation.
Calculate the enthalpy of the reaction
2N2 (g) + 5O2 (g)  2N2O5 (g)
Hrxn = ?
Use:
a 2 H2 (g) + O2 (g)  2 H2O (l)
b N2O5 (g) + H2O (l)  2 HNO3 (l)
c N2 (g) + 3 O2 (g) + H2 (g)  2 HNO3 (l)
Hrxn = -571.6 kJ
Hrxn = -76.6 kJ
Hrxn = -74.1 kJ
2 c 2 N2 (g) + 63 O2 (g) +2H2 (g)  42 HNO3 (l)
-b 2 HNO3 (l)  N2O5 (g) + H2O (l)
-a
2 H2O (l)  2 H2 (g) + O2 (g)
Example
Begin to cancel things out that appear on both the reactants and products side.
Your goal is to add up all the step-wise equations to equal the overall equation.
5
Multiply equations by a whole number if you need more of something to match
the overall reaction or to fully cancel something out that you don’t want in the
overall equation.
Calculate the enthalpy of the reaction
2N2 (g) + 5O2 (g)  2N2O5 (g)
Hrxn = ?
Use:
a 2 H2 (g) + O2 (g)  2 H2O (l)
b N2O5 (g) + H2O (l)  2 HNO3 (l)
c N2 (g) + 3 O2 (g) + H2 (g)  2 HNO3 (l)
Hrxn = -571.6 kJ
Hrxn = -76.6 kJ
Hrxn = -74.1 kJ
2 c 2 N2 (g) + 63 O2 (g) +2H2 (g)  42 HNO3 (l)
2 -b 42 HNO3 (l) 2N2O5 (g) +2H2O (l)
-a
2 H2O (l)  2 H2 (g) + O2 (g)
Example
Begin to cancel things out that appear on both the reactants and products side.
Your goal is to add up all the step-wise equations to equal the overall equation.
5
Multiply equations by a whole number if you need more of something to match
the overall reaction or to fully cancel something out that you don’t want in the
overall equation.
Calculate the enthalpy of the reaction
2N2 (g) + 5O2 (g)  2N2O5 (g)
Hrxn = ?
Use:
a 2 H2 (g) + O2 (g)  2 H2O (l)
b N2O5 (g) + H2O (l)  2 HNO3 (l)
c N2 (g) + 3 O2 (g) + H2 (g)  2 HNO3 (l)
Hrxn = -571.6 kJ
Hrxn = -76.6 kJ
Hrxn = -74.1 kJ
5
2 c 2 N2 (g) + 63 O2 (g) +2H2 (g)  42 HNO3 (l)
2 -b 42 HNO3 (l) 2N2O5 (g) +2H2O (l)
-a
2 H2O (l)  2 H2 (g) + O2 (g)
2 N2 (g) + 5 O2 (g)  2 N2O5 (g)
Example
6
Use the step-wise “labels” as a math expression for solving for
Hrxn.
Calculate the enthalpy of the reaction
2N2 (g) + 5O2 (g)  2N2O5 (g)
Hrxn = ?
Use:
a 2 H2 (g) + O2 (g)  2 H2O (l)
b N2O5 (g) + H2O (l)  2 HNO3 (l)
c N2 (g) + 3 O2 (g) + H2 (g)  2 HNO3 (l)
Hrxn = -571.6 kJ
Hrxn = -76.6 kJ
Hrxn = -74.1 kJ
5
2 c 2 N2 (g) + 63 O2 (g) +2H2 (g)  42 HNO3 (l)
2 -b 42 HNO3 (l) 2N2O5 (g) +2H2O (l)
-a
2 H2O (l)  2 H2 (g) + O2 (g)
2 N2 (g) + 5 O2 (g)  2 N2O5 (g)
2 × (-74.1 kJ)
2 × -(76.6 kJ)
1 × -(-571.6 kJ)
270.2 kJ
What did you learn about
hot/cold packs?
Hot/Cold Packs
Use
Transfer of energy
between
System &
Surroundings
Is determined with
Calorimetry
Physical
change
Can be done in
Effect on
temperature
depends on
Chemical
change
Materials ability to
absorb energy
without noticeable
temperature
change