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Thermodynamics
the study of energy changes in reactions
Energy
Energy is the ability to do work or produce
heat.
Energy is divided into two basic types:
potential energy
kinetic energy.
Potential Energy
Potential energy is stored energy based on:
position - gravity-related, like a skateboarder at the
top of a half-pipe or stretched rubber band
chemical composition food or a battery
like the chemicals in
Kinetic energy
Kinetic energy is energy of motion
This includes the constant, random motion of atoms and
molecules
Is proportional to temperature
As temp increases, motion of particles increases
→Thermal Expansion
Chemical Energy
– Form of potential energy based upon
chemical composition
– Stored in bonds between atoms,
molecules, and ionic crystals
– Potential to
• form new substance
• Do work
• produce or require heat
– Chemical energy of food powers your body
– Chemical energy of gas powers your car
Thermal Energy:
Total internal energy
of system: depends on both Heat and Temperature
Heat: Form of
energy that flows
from warmer to
cooler object
Temperature:
measure of the
average KE of the
particles
Energy can be converted between the different forms
of energy, but no matter how many times energy is
converted the
Law of Conservation of Energy
States no energy can be created or destroyed
Heat (q): energy that moves from hot to cold
Will continue to flow until temps are equal
Heat is quantified by measuring change in temp,
20°C
10°C
Temp does not directly measure
heat
stored
What is final temp?
calorie: To measure heat directly, the calorie (cal)
was defined as the amount of heat needed to increase
the temp of 1 gram of water by 1 °C.
Note: The Calories reported on food wrappers are
measured as heat needed to increase the
temperature of 1000 grams of water by 1 °C,
1 Calorie (Cal) = 1000 calories (cal) = 1 kilocalorie (kcal)
Drinking a whole can of Pepsi
actually means drinking:
250 Cal = 250,000 calories!
Normal diet:
2,000 Cal = 2,000,000 cal
Calorimetry: measuring heat energy
Calorimeter: device used to measure heat lost or
absorbed during chemical reaction, including energy
from food
Joule: measure energy more accurately
Calories are still used in the nutrition field, but in chemistry
we use the Joule (J).
1 cal = 4.184 J
Ex. 1) If my breakfast contains 230 Calories, how much
energy in Joules will it supply?
Joule: measure energy more accurately
Calories are still used in the nutrition field, but in chemistry
we use the Joule (J).
1 cal = 4.184 J
Ex. 1) If my breakfast contains 230 Calories, how much
energy in Joules will it supply?
230 Cal
1000 cal
1 Cal
4.184 J
1 cal
Joule: measure energy more accurately
Calories are still used in the nutrition field, but in chemistry
we use the Joule (J).
1 cal = 4.18 J
Ex. 1) If my breakfast contains 230 Calories, how much
energy in Joules will it supply?
230 Cal
1000 cal
1 Cal
4.18 J
1 cal
= 961400 J
= 9.6 x 105 J
Both the calorie and Joule are
defined using water
But, other substances gain and
lose heat at different rates
Specific heat (c) – the amount
of energy required to raise 1
gram of a any substance by 1°C
Liquid water has a relatively high specific heat: loses &
gains heat slowly – moderates earth’s climate
The better the conductor, the lower the
specific heat
When a cake has been baked & removed from
the oven it would cause severe burns to touch the
metal pan, but the top of the cake can be touched
to test if the cake is ready…Why?
The metal of the pan has a low specific heat b/c it
can lose heat quickly
This heat would be absorbed by your finger!!!
The cake contains a lot of trapped air and other
materials with much higher specific heats. These
objects lose heat slowly, allowing more time before
your finger has absorbed enough heat to cause a
burn.
Density of a cake is much less than the density of the
metal pan, so for the same surface area touched by
your finger there are a lot more metal atoms that
transfer heat energy than cake molecules.
So… the mass also has an affect on the amount of
heat transferred
Plus… (old stuff)
Metals are good conductors of heat and
electricity
While air is a poor conductor of heat
and electricity (nonmetals)
To calculate the amount of heat gained or lost
by an object
q = m × c × ∆T
q = heat (J)
m = mass (g)
c = specific heat
J
g °C
∆T = Temp. final – Temp. initial, in °C
If the object is heating up:
the final temp is greater than the initial temp,
so ∆T is positive and
amount of heat gained is positive.
+q
If the object is cooling down:
the final temp is less than the initial temp,
so ∆T is negative and
the amount of heat lost is negative.
-q
If ∆T is positive the answer for heat is positive:
heat is gained.
If ∆T is negative the answer for heat is negative:
heat is lost.
-q
+q
Add the following to your notes
Specific heats for water:
for steam = 1.97 J/goC
for liquid = 4.18 J/goC
for ice = 2.09 J/goC
Specific heats (c) for various metals:
aluminum = 0.897 J/goC
copper = 0.385 J/goC
iron = 0.449 J/goC
lead = 0.129 J/goC
mercury = 0.140 J/goC
Ex. 2) What is the heat absorbed by 100. g of water
to raise it from 25 °C to 75 °C?
q = m × c × ∆T
q = ________ x ________ x (________ - _______)
q = _______________
Ex. 2) What is the heat absorbed by 100. g of water
to raise it from 25 °C to 75 °C?
q = m × c × ∆T
q = _100.g__ x _4.18 J/goC_ x (75oC_- 25oC)
q = __ ____ __
Ex. 2) What is the heat absorbed by 100. g of water
to raise it from 25 °C to 75 °C?
q = m × c × ∆T
q = _100. g__ x _4.18 J/goC _ x (75oC - 25oC)
q = __ 20900 J___ = 2.1 x 104 J or 21000 J
Ex. 3) What is the heat given off by 100. g of iron to
cool from 50. °C to 20. °C?
q = m × c × ∆T
q = ________ x ________ x (________ - _______)
q = _______________
Ex. 3) What is the heat given off by 100. g of iron to
cool from 50. °C to 20. °C?
q = m × c × ∆T
q = 100. x 0.449 J/goC x ( 20. oC – 50. oC)
q=
Ex. 3) What is the heat given off by 100. g of iron to
cool from 50. °C to 20. °C?
q = m × c × ∆T
q = 100. g x 0.449 J/goC x ( 20.oC – 50. oC)
q = 100.g x 0.449 J/goC x (-30. oC)
q=
Ex. 3) What is the heat given off by 100. g of iron to
cool from 50. °C to 20. °C?
q = m × c × ∆T
q = 100. x 0.449 x ( 20. – 50.)
q = 100. x 0.449 x (-30.)
q = -1347 J = -1300 J or -1.3 x 103 J
Heat Movement
The Law of Conservation of Energy states that heat
energy cannot just appear out of nowhere and
disappear to nowhere.
If an object warms up, that heat had to be absorbed
from somewhere and if an object cools down, that
heat had to be given off to something.
In the Earth’s ecosystem heat is gained by the sun’s
radiation and given off to space at night.
This is why a cloudy night is typically warmer than a
clear night – the heat gets trapped in by the clouds.
Heat can move from one object to another by
three ways:
1. Radiation
2. Convection
3. Conduction
The energy from the sun reaches earth by
radiation - electromagnetic waves:
The infrared part of the electromagnetic spectrum is the
“heat” form of radiation.
Ex: heat from fire, light bulb, coil on electric stove
Convection is the movement of heat by circulating
materials, so anything that is a fluid (gases and
liquids) can move heat by convection, often called
convection currents.
Ex. Most air-conditioning systems, the ocean
currents, and the warm and cold “fronts” the
weather-person talks about on the news.
Conduction is the movement of heat between objects
that are touching.
Most common form of heat movement when
cooking.
The food touches a pan that’s touching a burner. As
the pan absorbs heat from burner the food absorbs
heat from the pan.
When you touch a hot pan heat moves from the
pan into your hand which then burns you.
Often the lasting effects of a burn can be
minimized if you can quickly get the heat to move
out of your hand by running cold water over your
hand for a long time.
Heat energy will always naturally move from an
object with higher heat to an object with lower heat,
regardless of the method of heat transfer
The Law of Conservation of Energy also tells us
that the amount of heat lost by the hot object must
be equal to the amount of heat gained by the cold
object.
20°C
10°C
15°C
Hot becomes cooler, and cold becomes warmer.
-qlost = qgained
-- (mlost × clost × ∆Tlost ) = mgained × cgained × ∆Tgained
Phases or States of Matter: Solid, Liquid, Gas
Sometimes when an object gains or loses heat, it is
not just the temperature that changes but the phase
of the object can change.
The phase depends on how much heat energy is
in the object, often measured by the temperature.
The solid will have lower temperature than the
liquid form, which will have lower temperature
than the gas (vapor) form.
Energy in states of matter
Hi energy & velocity - Low attractive forces & density
Hi temp
Gas
Energy
Released
Exothermic
Liquid
Solid
Energy
In
Endothermic
Low energy & velocity - Hi attractive forces & density
Low temp
Phase changes occur when sufficient heat energy is
added or removed
+q: heat is added or absorbed – Endothermic
–q: heat is lost or released – Exothermic
Phase changes occur when sufficient heat
energy is added or removed to change the
object’s phase.
Melting is the solid phase changing to the
liquid phase
Freezing is the liquid phase turning to a
solid phase.
Melting and freezing both occur at the
melting point ~ the temp at which a solid
becomes a liquid.
If heat is being absorbed (+q), then melting
is occurring.
If heat is given off (-q), then freezing is
happening.
Boiling (vaporizing) is the liquid phase
changing to the gas phase, and
Condensation is the gas phase changing to
the liquid phase.
Boiling and condensing occur at the boiling
point ~ the temp at which a liquid becomes
a gas.
If heat is being absorbed (+q), then boiling
is occurring.
If heat is given off (-q), then condensing is
occurring.
Sometimes liquids evaporate instead of
boiling.
Evaporation is boiling, but occurs below the
boiling point. Evaporation can only happen at
the surface of the liquid (boiling happens
everywhere inside) and is a cooling process.
There are two other phase change possibilities,
but they are rare in everyday life.
Sublimation is the solid phase changing directly
to the gas phase (no liquid in-between)
Deposition is the gas phase changing directly to
the solid phase.
Ex. Dry ice, CO2
A +q would indicate sublimation
A –q would indicate deposition.
From this heat curve (for water) we can see how the
temperatures change and the phases change. Notice
that the temperature does not change during a phase
change (the two flat lines).
The heat equation q = m × c × ∆T will not work during
a phase change, so new equations are needed during a
phase change bc heat is still needed
For melting or freezing
q = m × Hf
q = heat (add “-“ sign for freezing), (J)
m = mass, (g)
Hf = heat of fusion,
J
g
For boiling/condensing
q = m × Hv
q = heat (add “-“ sign for condensing), (J)
m = mass, (g)
J
Hv = heat of vaporization,
g
Ex. 4) Calculate the heat needed for the heat curve
shown below for 150. g of water.
Hfwater = 334 J/g
Hvwater = 2260 J/g
c ice = 2.09 J/g °C
c water = 4.18 J/g °C
c steam = 1.97 J/g °C
Step 1: Ice from -40 °C to 0 °C.
q1 = m × c × ∆T
q1 = ________ x ________ x (________ - _______)
q1 = _______________
Step 1: Ice from -40.0 °C to 0 °C.
q1 = m × c × ∆T
q1 = 150. g x 2.09 J/goC x ( 0oC – (-40.0oC))
q1 = 12400 J
Step 2: Turning ice into liquid water
q2 = m × Hf
q2 = ___________ x ___________
q2 = _______________
Step 2: Turning ice into liquid water
q2 = m × Hf
q2 = 150. g x 334 J/g
q2 = 50100 J
Step 3: Water from 0 °C to 100 °C.
q3 = m × c × ∆T
q3 = ________ x ________ x (________ - _______)
q3 = _______________
Step 3: Water from 0 °C to 100 °C.
q3 = m × c × ∆T
q3 = 150. g x 4.18 J/goC x ( 100oC – 0oC)
q3 = 62700 J
Step 4: Turning liquid water into water vapor
(steam).
q4 = m × Hv
q4 = ___________ x ___________
q4 = _______________
Step 4: Turning liquid water into water vapor
(steam).
q4 = m × Hv
q4 = 150. g x 2260 J/g
q4 = 339000 J
Step 5: Water vapor from 100 °C to 140 °C
q5 = m × c × ∆T
q5 = ________ x ________ x (________ - _______)
q5 = _______________
Step 5: Water vapor from 100 °C to 140.0 °C
q5 = m × c × ∆T
q5 = 150. g x 1.97 J/goC x ( 140.0oC – 100oC)
q5 = 12200 J
Total heat involved = q1 + q2 + q3 + q4 + q5
Total heat involved = 12400 J + 50100 J + 62700 J +
339000 J + 12200 J
Total heat = __________
Total heat involved = 12400 J + 50100 J + 62700 J +
339000 J + 12200 J
Total heat = 476400 J
(with proper addition sig figs = 476000J)
Heat & Chemical Reactions
Thermochemical Equations: A balanced chemical
equation that includes all physical states (s, l, g) and
change in enthalpy (H) (heat content).
Heat released
4 Fe(s) + 3O2(g) → 2Fe2O3(s) + 1625 kJ
Heat can be written as either a reactant or product
Every chemical reaction involves a change in energy:
heat and/or light
.
If a chemical reaction feels cold:
absorbing heat.
endothermic reaction – energy in
heat energy is written on the reactant side
q is positive (+q)
27 kJ + NH4NO3(s) → NH4+(aq) + NO3-(aq)
Heat Added
If a chemical reaction feels warm,
giving off heat.
exothermic reaction - energy out
heat energy is written on the product side
q is negative (-q)
Heat released
4 Fe(s) + 3O2(g) → 2Fe2O3(s) + 1625 kJ
Ex. 5) If 515 kJ were needed in the dissolving
process, a) How many moles of ammonium
nitrate were originally used? b) How many
grams would that be?
27 kJ + NH4NO3(s) → NH4+(aq) + NO3-(aq)
Ex. 6) How many Joules of heat are released
when 77 grams of iron reacts with excess oxygen?
4 Fe(s) + 3O2(g) → 2Fe2O3(s) + 1625 kJ
Endothermic (a reactant)
solid + energy (heat) → liquid
Heat flows into the system from the surroundings.
This is why your skin gets
cold when you hold ice;
the heat is flowing out of
your skin to melt the ice.
System: ice
Surrounding: your skin
Exothermic (a product)
Iron + oxygen → iron(III) oxide + energy (heat)
heat is released from the system and is absorbed by
the surrounding environment
which is why it’s used
in instant hot pads.
System: hot pad
Surrounding: your skin
Enthalpy (H):
total heat energy of system
∆H = change in enthalpy
Exothermic
∆H is negative
Energy of system
decreased
Endothermic
∆H is positive
Energy of system
increased
To separate water into hydrogen and oxygen
(electrolysis) electricity is often used.
2H2O  2H2 + O2
Is electrolysis of water
endothermic
or exothermic?
Where does the energy
“go” in the equation?
To separate water into hydrogen and oxygen
(electrolysis) electricity is often used.
Is electrolysis of water
endothermic
or exothermic?
Where does the energy
“go” in the equation?
ΔH + 2H2O  2H2 + O2
The enthalpy of reaction (∆H) can be
calculated to determine if heat is
released or gained.
If ΔH is +, then heat is absorbed, so it
is an endothermic reaction
If ΔH is -, then heat is released, so it is
an exothermic reaction
ΔH = ∑∆nHfo(products) - ∑∆nHfo(reactants)
∆Hfo = standard enthalpy (heat) of formation
from the formation of 1 mole of compound
Ex. 7) Find the ∆Hrxn for liquid ethanol, C2H5OH,
undergoing combustion:
C2H5OH + 3O2 → 2CO2 + 3H2O
Is this an endothermic or exothermic reaction?
Thermodynamic values table
Compound
∆Hfo
C2H5OH
-277.69 kJ/mol
O2
0 kJ/mol
CO2
-393.509 kJ/mol
H2O
-285.830 kJ/mol
Ex. Find the ∆Hrxn for liquid ethanol, C2H5OH,
undergoing combustion:
C2H5OH + 3O2 → 2CO2 + 3H2O
Is this an endothermic or exothermic reaction?
Thermodynamic values table
Compound
∆Hfo
C2H5OH
-277.69 kJ/mol
O2
0 kJ/mol
CO2
-393.509 kJ/mol
H2O
-285.830 kJ/mol
∆Hfo(reactants):
∆Hfo(products):
ΔH = ∆Hfo(products) - ∆Hfo(reactants)
Ex. Find the ∆Hrxn for liquid ethanol, C2H5OH,
undergoing combustion:
C2H5OH + 3O2 → 2CO2 + 3H2O
Is this an endothermic or exothermic reaction?
Thermodynamic values table
Compound
∆Hfo
C2H5OH
-277.69 kJ/mol
O2
0 kJ/mol
CO2
-393.509 kJ/mol
H2O
-285.830 kJ/mol
Per mole
∆Hfo(reactants): (-277.69) + 3(0) = -277.69 kJ
∆Hfo(products): 2(-393.509) +3(-285.830) = -787.018 + (-857.49)
= -1644.508 kJ
ΔH = ∆Hfo(products) - ∆Hfo(reactants)
= (-1644.508) – (-277.69) = -1366.818 kJ
Ex. Find the ∆Hrxn for liquid ethanol, C2H5OH,
undergoing combustion:
C2H5OH + 3O2 → 2CO2 + 3H2O
Is this an endothermic or exothermic reaction?
Thermodynamic values table
Compound
∆Hfo
C2H5OH
-277.69 kJ/mol
O2
0 kJ/mol
CO2
-393.509 kJ/mol
H2O
-285.830 kJ/mol
Per mole
∆Hfo(reactants): (-277.69) + 3(0) = -277.69 kJ
∆Hfo(products): 2(-393.509) +3(-285.830) = -787.018 + (-857.49)
= -1644.508 kJ
ΔH = ∆Hfo(products) - ∆Hfo(reactants)
= (-1644.508) – (-277.69) = -1366.818 kJ Neg = Exothermic
An additional way to show a change in enthalpy
without a calculation is with an enthalpy diagram.
What is the key to knowing if the diagram shows
exothermic or endothermic? Energy in or out
An additional way to show a change in enthalpy
without a calculation is with an enthalpy diagram.
Energy out
Energy in
What is the key to knowing if the diagram shows
exothermic or endothermic? Energy in or out