Transcript Notes

Causes of Change
Energy and Change
Changes and Energy
 Both chemical and physical changes are accompanied by
either an increase or decrease in energy.
 Energy is the ability to do work or produce heat.
 There are two major types of energy: kinetic and potential.
 Kinetic energy is the energy of motion and depends on both
the mass and the velocity of the object.
 Potential energy is stored energy.
 Matter possesses both types of energy.
 Heat and temperature are used to describe the energy
content of matter.
What is temperature?
 Temperature is a measure of the average kinetic energy
(energy of motion) of particles in matter.
 As the kinetic energy of the particles increases, the
temperature increases.
 As the kinetic energy of the particles decreases, the
temperature decreases.
 The lowest possible temperature (the temperature at
which the particles stop moving) is called absolute zero.
 Temperature is an intensive property and does not
depend on the amount of matter present.
How is temperature measured?
 Temperature is measured with a thermometer.
 Almost all substances expand with an increase in
temperature (exception is water).
 Thermometers are designed so that the substances they
contain (mercury, etc.) expand and contract more than
the volume of the glass tube that contains them so that
the column height of the substance changes.
Temperature Scales
 Several temperature scales have been devised:
- Fahrenheit (oF): weather is measured using this
scale.
- Celsius (oC): the scale used in the metric system.
-Kelvin (K): based on absolute zero; used in the
International System (SI).
Comparison of Temperature Scales
•Celsius degrees are
larger in size than
Fahrenheit degrees.
•Celsius degrees are
the same size as
kelvins.
•The Kelvin scale is
the same as the
Celsius scale (only
273o higher).
•The only
temperature that is
the same on the
Celsius and
Fahrenheit scale is
-40o.
Converting Between Scales
 Celsius to Fahrenheit
oF

Fahrenheit to Celsius
oC

= 9/5oC + 32
= (oF – 32)5/9
Celsius to Kelvin
K = oC + 273

Kelvin to Celsius
oC
= K - 273
Practice Problems
 65o C = _______K
Answer: 65oC + 273 = 338 K
 25oC = _______K
Answer: 25oC + 273 = 298 K
 -50oC = ______K
Answer: -50oC + 273 = 223 K
 300 K = _______oC
Answer: 300 K -273 = 27oC
 273 K = _______oC
Answer: 273 K – 273 = 0oC
 150 K = _______oC
Answer: 150 K – 273 = -123oC
How Are Heat and Temperature
Related?
 Heat (q) is the total amount of energy that
transfers from one object to another.
 Heat is an extensive property and changes based on
the amount of matter present.
 Temperature measures the intensity of the energy
and heat measures the quantity of energy.
 The temperature of an object determines the
direction of heat transfer.
 When two objects of different temperatures are in
contact, heat moves from the object at the higher
temperature to the object at the lower
temperature until they are both at the same
temperature.
Methods of Heat Transfer
 Conduction
 Convection
 Radiation
Conduction
 Conduction occurs when heat is transferred by
particles that are in direct contact with each
other.
 Good conductors transfer heat easily. Metals
are good conductors.
 Insulators do not transfer heat easily. Glass,
rubber, wood, and plastic are good insulators.
 Examples of conduction: Food cooking on a
stove, burning your hand on a hot iron.
Walking on hot coals
Convection
 Convection occurs when heat is transferred through the
movement of gas or liquid particles.
 Warm air/water is less dense and rises.
 Cool air/water is more dense and sinks.
 These differences in density create a circulating
current.
 Examples: the heating of your home, deep ocean water
is cooler than at the surface.
Lava Lamps and Convection
Radiation
 Radiation of heat occurs when heat is transferred in
matter or space by means of electromagnetic waves.
 This type of heat travels outward from its source in all
directions.
 Examples: a fireplace heating a room, the sun heating
the earth.
A radiator
exemplifies all three
methods of heat
transfer.
1. Heat radiates outward from the radiator to heat the room through
radiation.
2. The man is warming his hands over the radiator through the
process of convection.
3. If the man accidentally touches the surface of the radiator, he will
burn his hands through the process of conduction.
Practice Questions
 Identify each of the following as conduction, convection, and
radiation.
1. A hot air balloon.
Answer: Convection
2. Warming yourself in front of a fireplace.
Answer: Radiation
3.Cooking food in a microwave oven.
Answer: Radiation
4. Frying bacon.
Answer: Conduction
5. Burning your hand on a curling iron.
Answer: Conduction
6. It is warmer in the attic than on the first floor of your house.
Answer: Convection
Transfer of Energy in Chemical and
Physical Changes
 All chemical and physical changes involve the
release or absorption of heat.
 The system includes the substances involved in
the change.
 The surroundings include everything else in the
universe.
 The system and the surroundings make up the
universe.
 Law of Conservation of Energy states that
energy cannot be created or destroyed-instead
it is transferred or changed from one form to
another.
Endothermic vs. Exothermic
 Endothermic processes are those in which heat
flows from the surroundings into a system (heat
is absorbed).
 The system gains heat and the surroundings
cool down.
 Exothermic processes are those in which heat
flows from the system to the surroundings
(heat is released).
 The system loses heat and the surroundings
heat up.
Energy of Phases and Phase
Changes
 Relative energy of particles making up the states (phases) of matter.
Solids < Liquids < Gases << Plasma
 Endothermic Phase Changes (Absorption of Energy)
Melting (SolidLiquid)
Vaporization (Liquid  Gas)
Sublimation (Solid  Gas)
 Exothermic Phase Changes (Release of Energy)
Freezing (Liquid  Solid)
Condensation (Gas  Liquid)
Deposition (Gas  Solid)
Phase Diagrams
•Triple point-the temperature and
pressure at which all three phases of
matter coexist.
•Critical point-the highest
temperature at which a liquid can
exist. (The substance exists only as
a vapor above this temperature no
matter what the pressure).
•Points on the line separating
phases of matter represent the
melting or boiling points.
•If the slope of the line separating
the solid and liquid phase is
positive, the solid is more dense
than the liquid.
Phase Diagram of Water
Note: The
line
separating
the solid and
liquid phases
has a
negative
slope.
Therefore,
the solid
phase is less
dense than
the liquid
phase and
the solid
floats in the
liquid.
Practice Interpreting Phase
Diagrams
 For additional interactive practice, visit the
following website.
 http://www.sciencegeek.net/Chemistry/taters/phasediagram.htm
Measuring Heat
 Heat is measured with a
calorimeter.
 The unit used when
measuring heat is the
calorie (cal) or joule (J).
 A calorie is the amount of
heat needed to raise the
temperature of 1 gram of
pure water 1oC.
 1 calorie = 4.18 joules.
 1000 calories (1 kcal) = 1
dietary calorie (Cal)
Heat Capacity
 The amounto of heat needed to raise the temperature of
an object 1 C is called the heat capacity of the object.
 The heat capacity depends on the mass and the
chemical make-up of the substance.
 Heat capacity is an extensive property.
 Specific heat capacity is the amount of heat it
takes to
o
raise the temperature of 1 g of a substance 1 C.
 Specific heat capacity is an intensive property.
 Water has a high specific heat capacity compared to
other substances-therefore, it requires more heat to
change the temperature of a sample of water.
Specific Heats of Various Substances
Substance
Specific Heat (J/goC)
Water
4.18
Ethanol (grain alcohol)
2.44
Ice
2.03
Steam
2.01
Aluminum
0.897
Concrete
0.84
Iron
0.449
Silver
0.235
Lead
0.129
Gold
0.129
Calculating Heat and Specific
Heat
 The equation for calculating heat:
heat = mass x change in temp x specific heat
or
q = m x ∆T x C
m is the mass and is measured in grams
∆T = final temp – initial temp (oC)
C is the specific heat
**when working with water, C = 1.00 cal/goC or 4.18 J/goC.
q is the heat measured in joules or calories . If q < 0, heat is
being released (exothermic) and if q>0, heat is being
absorbed (endothermic).
Practice Problem #1
If the temperature of 34.4 g of
ethanol increases from 25.5oC to
78.8oC, how much heat has been
absorbed by the ethanol?
Answer to Practice Problem #1
q=?
m= 34.4 g
∆ T = 78.8-25.5oC
C=2.44 J/g oC
q = mx∆TxC
q = 34.4 (78.8-25.5)2.44
q= 34.4(53.3)2.44
q=4,470 J or 4.47 kJ
Practice Problem #2
A 155 g sample of an unknown
substance was heated from 25.0oC
to 40.0oC. In the process, the
substance absorbed 5696 J of
energy. What is the specific heat
of the substance?
Answer to Practice Problem 2
q= 5696 J
m= 155 g
∆T = 40.0 – 25.0oC
C=?
q = mx∆TxC
5696 = 155 (40.0-25.0) x
5696 = 155 (15.0) x
5696 = 2325 x
x = 5696/2325 = 2.45 J/g oC
Practice Problem #3
A 4.50 g nugget of pure gold
absorbed 276 J of heat. The
initial temperature was 25.0oC.
What was the final temperature?
Answer to Practice Problem 3
q = 276 J
m = 4.50 g
∆T = Tf – 25.0 oC
C = 0.129 J/g oC
q = m x ∆T x C
276 = 4.50 (x – 25.0) 0.129
276 = .5805 (x – 25)
276/.5805 = .5805/.5805 ( x – 25)
475 = x – 25
X = 500oC
Practice Problem #4
 When a 58.8 g piece of hot alloy is placed in
125 g of cold water in a calorimeter, the
temperature of the alloy decreases by
106.1oC, while the temperature of the
water increases by 10.5oC. What is the
specific heat of the alloy?
Answer to Practice Problem #4
 In order to solve this problem, it is important to
recognize that the alloy lost heat (temperature
decreased) and the water gained heat (temperature
increased).
 Due to the law of conservation of energy,
heat lost = heat gained
 Heat gained by water: q = m x ∆T x C
q = 125 x 10.5 x 4.18 = 5486 J
 Heat lost by alloy: q = m x ∆T x C
5486 = 58.8 x (106.1) x C
C = 0.879 J/goC
Latent Heat and Changes in State
 Latent heat is the energy involved in matter
undergoing changes in state.
 Heat of fusion is the energy required to change
1 g of a substance from a solid to a liquid.
 The heat of fusion is unique to the substance.
 q = Hfusion x mass
Practice Problem
 How much energy is required to melt 10 g of
ice? (Heat of fusion of ice is 334 J/g)
 q = Hfusion x mass
 q = 10 x 334
 q = 3340 J
Latent Heat and Changes in State
 Heat of vaporization is the energy needed
to change 1 g of a substance from a liquid
to a gas.
 The heat of vaporization is also unique to
the substance.
 q = Hvaporization x mass
Practice Problem
 How much energy is required to boil 10 g of
water? (Hvap for water is 2260 J/g)
 q = Hvap x m
 q = 2260 x 10
 q = 22600 J
Transfer of Energy in Physical
Changes
Wherever there is a
change in temperature,
there is also a change in
kinetic energy of the
particles making up the
substance. (q=mx∆TxC)
Areas on the graph where
there is no change in
temperature represent a
change in potential
energy (or change in
state) of the particles
making up the substance.
(q = Hfus x m
or
q = Hvap x m)
Practice Problem (Honors
Chemistry only)
 Calculate the energy required to
change 10 g of ice at -10oC to steam
at 110oC.
 Hint: This problem must be solved in
five separate steps.
Answer to practice problem
 Step 1: Convert 10 g ice at -10oC to ice at 0oC.
Answer: q = mx∆TxC
q = 10 x (0 - -10oC) x 2.03 = 203 J
 Step 2: Melt the 10 g of ice.
Answer: q = mxHfus
q = 10 x 334 = 3340 J
 Step 3: Raise the temp of the water from 0oC to 100oC.
Answer: q = mx∆TxC
q = 10 x (100-0) x 4.18 = 4180 J
 Step 4: Boil the 10 g of water.
Answer: q= mxHvap
q = 10 x 2260 = 22600J
 Step 5: Raise the temp of the steam from 100oC to 110oC.
Answer: q = mx∆TxC
q = 10 x (110 – 100) x 2.01 = 201J
 Step 6: Calculate the total.
Answer: 203 + 3340 + 4180 + 22600 + 201 = 30524 = 31000 J
Practice Problem #2
 Calculate the energy required to change 1
mole of ice at 0oC to steam at 100oC.
Answer to Practice Problem #2.
 Step 1: Melt the ice.
Answer: q = mxHfus
q = 18 g x 334 = 6012 J
 Step 2: Raise the temp of the water from 0oC to 100oC.
Answer: q = m x ∆T x C
q = 18 x (100-0) x 4.18 = 7524 J
 Step 3: Convert the liquid water to steam
Answer: q = mxHvap
q = 18 x 2260 = 40680 J
 Step 4: Calculate the total.
Answer: 6012 + 7524 + 40680 = 54216 J = 54000J
Practice Interpreting Energy
Graphs Illustrating Phase Changes
 For additional interactive practice, visit the
following website.
 http://www.sciencegeek.net/Chemistry/taters/phasediagrams2.htm
Thermochemical Equations
 An equation that includes the heat change
involved in a chemical reaction is called a
thermochemical equation.
 Heat of reaction (∆H) is the heat change for
a particular equation.
 ∆H = Hproducts - Hreactants
Potential Energy Diagrams
 All reactions begin with the breaking of the
bonds of the reactants (endothermic step).
 In order for those bonds to break, collisions
between the reacting molecules must occur
and the reacting molecules must possess a
minimum amount of energy (activation
energy).
 During the collision, an unstable intermediate
molecule forms (activated complex).
 The molecules then rearrange and form new
bonds resulting in the formation of products
(exothermic step).
http://www.youtube.com/watch?v=VbIaK6PLrRM&feature=player_embedded
•If the initial endothermic step is greater than the
final exothermic step, the overall reaction is
endothermic.
•∆H = energy of products – energy of reactants
•If the reaction is endothermic, ∆H > 0.
•If the initial endothermic step is smaller than the
final exothermic step the reaction is exothermic.
•∆H = energy of products – energy of reactants
•If the reaction is exothermic, ∆H < 0.
Effect of a Catalyst
• A catalyst is a
substance that
speeds up a
reaction while
remaining
unchanged in the
process.
• One way in which
catalysts
accomplish this is
by lowering the
activation energy.
Practice Interpreting Potential
Energy Graphs
 For additional interactive practice, visit the
following website.
 http://www.sciencegeek.net/Chemistry/taters/energydiagram.htm
Writing Thermochemical Equations
 When 2 mol of solid magnesium combines with 1 mole of
oxygen gas, 2 mol of solid magnesium oxide is formed
and 1204 kJ of heat is released. Write the
thermochemical equation for this reaction.
 2Mg(s) + O2(g)  2MgO(s) + 1204 kJ
 Note: The energy is written on the reactant side of the
equation for endothermic reactions.
Practice Problem
 Use the following equation to answer the questions below:
129 kJ + 2 NaHCO3  Na2CO3 + H2O + CO2
1) Is the reaction exothermic or endothermic?
Answer: endothermic (heat appears on the left side of the
equation)
2) Sketch a potential energy curve that could represent this
reaction.
Answer:
3) How much heat is needed to decompose 2.24 moles of
NaHCO3?
Answer: x kJ = 2.24 moles NaHCO3
129 kJ
2 moles NaHCO3
x = 144 kJ
Review
 For additional interactive practice, visit the
following website.
 http://www.sciencegeek.net/Chemistry/taters/Unit7Thermochemistry.htm