Energy, Enthalpy
Calorimetry & Thermochemistry
Energy is...
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The ability to do work.
Conserved.
made of heat and work.
a state function.
independent of the path, or how you get from point A
to B.
• Work is a force acting over a distance.
• Heat is energy transferred between objects because of
temperature difference.
The Universe
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is divided into two halves.
the system and the surroundings.
The system is the part you are concerned with.
The surroundings are the rest.
Exothermic reactions release energy to the
surroundings.
• Endothermic reactions absorb energy from the
surroundings.
Potential energy
CH 4 + 2O 2  CO 2 + 2H 2 O + Heat
CH 4 + 2O 2
Heat
CO 2 + 2 H 2 O
N 2 + O 2 + heat  2NO
Potential energy
2NO
Heat
N2 + O2
Units of Heat
• Calorie (cal)
– The quantity of heat required to change the
temperature of one gram of water by one
degree Celsius.
• Joule (J)
– SI unit for heat
1 cal = 4.184 J
Direction
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Every energy measurement has three parts.
A unit ( Joules or calories).
A number how many.
and a sign to tell direction.
negative - exothermic
positive- endothermic
Surroundings
System
Energy
DE <0
Surroundings
System
Energy
DE >0
First Law of Thermodynamics
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The energy of the universe is constant.
Law of conservation of energy.
q = heat
Take the systems point of view to decide signs.
q is negative when the system loses heat energy
q is positive when the system absorbs heat energy
Enthalpy
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Heat content of a substance
Depends on many things
Symbol is “H”
Can’t be directly measured
Changes in enthalpy can be calculated, DH =
Hfinal - Hinitial
• Changes in heat (q) can be used to find DH
Same rules for Enthalpy
• Heat energy given off by the system is
negative, DH < 0
• Heat energy absorbed by the system is
positive, DH > 0
q and DH
q is the “heat flow for a system”
It can be for any amount of mass or moles of substance
DH is “change in heat”
It is for a specific chemical or physical change and it
refers to that reaction or process.
Heat Capacity
• The quantity of heat required to change the temperature
of a system by one degree.
• Three “types” with different “systems:
– Specific Heat Capacity
– Molar Heat Capacity
– Heat Capacity
Heat Capacity Types
– Specific heat capacity
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System is one gram of substance
Units are J/goC
q = heat change
m = mass
C = Specific Heat Capacity
DT = final temperature – initial temperature
q = (m)(C)(DT)
Heat Capacity
Types
– Molar heat capacity.
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System is one mole of substance.
Units are J/mol oC
q = heat change
n = moles
C = molar heat capacity
DT = final temperature – initial temperature
q= nCDT
Heat Capacity Types
– Heat capacity of an object
• Specific for a given amount or unit
so it is not mass dependent
• Units are J/oC
• q = heat
• C = heat capacity
• DT = final temperature – initial temperature
q = CDT
Heat Capacity Types
• So how do you know which one to use when?
• Look at the known information
• Select the value with appropriate units to
cancel
Example
• The specific heat of graphite is 0.71 J/gºC.
Calculate the energy needed to raise the
temperature of 75 kg of graphite from 294 K
to 348 K.
Calorimetry
• Measuring temperature changes to calculate
heat changes.
• Use a calorimeter.
• Two kinds
– Constant pressure calorimeter (called a coffee cup
calorimeter)
– Constant volume calorimeter (called a bomb
calorimeter)
Calorimetry
• A coffee cup calorimeter measures temperature and
calculates q.
• An insulated cup, full of water (constant pressure, variable
volume)
• Water is the surroundings in which the system changes
• Calculate the heat change of water.
• The specific heat of water is 4.184 J /gºC
• Heat of water qH2O = CH2O x mH2O x DT
qsystem + qsurroundings = 0
• qH2O + qrxn = 0
• qH2O = - qrxn
In interactions between a system and its
surroundings the total energy remains constant—
energy is neither created nor destroyed.
Law of Conservation of Energy
Coffee Cup Calorimeter
• A simple calorimeter.
– Well insulated and therefore isolated.
– Measure temperature change.
qrxn = -qcal
Determination of Specific Heat
Example
Determining Specific Heat from Experimental Data.
Use the data presented on the last slide to calculate the specific
heat of lead.
qlead = -qwater
qwater = mcDT = (50.0 g)(4.184 J/g °C)(28.8 - 22.0)°C
qwater = 1.4x103 J
qlead = -1.4x103 J = mcDT = (150.0 g)(c)(28.8 - 100.0)°C
clead = 0.13 Jg-1°C-1
Example
• When a piece of copper (5.0 g) is heated for 2.0 seconds,
and 100 J of heat energy is transferred to the copper, the
temperature increases from 20.0 ˚C to 71.9 ˚C.
• What is the specific heat of the copper?
Examples
If 10.0 g of Cu is heated for 2.0 seconds from 20.0 ˚C and
200 J of heat are absorbed, what is the final temperature
of the block?
a. 71.9 ˚C
b. 100 ˚C
c. 46 ˚C
d. 123.9 ˚C
e. 719 ˚C
Example
Equal masses of liquid A, initially at 100 ˚C, and liquid B,
initially at 50 ˚C, are combined in an insulated container. The
final temperature of the mixture is 80 ˚C. Which has the
larger specific heat capacity, A or B?
• A
• B
• A and B have the same specific heat capacity.
Example
• When 86.7 grams of water at a temperature of 73.0 ˚C is mixed with an
unknown mass of water at a temperature of 22.3 ˚C the final
temperature of the resulting mixture is 61.7 ˚C. What was the mass of
the second sample of water?
a.
b.
24.9 g
302 g
c. 48.2 g
d.419 g
States of Matter
General Heating Curve
http://cwx.prenhall.com/petrucci/medialib/med
ia_portfolio/text_images/031_ChangesState.
MOV
Changes of State of Water
qphase change = DHphase change
(mass or moles)
Molar enthalpy of vaporization:
H2O (l) → H2O(g)
DHv = 40.65 kJ/mol @ 373.15 K
DHv = (40.65 kJ/mol)(1 mol/18.02 g) = 2256 J/g
Molar enthalpy of fusion (melting):
H2O (s) → H2O(l)
DHf = 6.01 kJ/mol @273.15 K
DHf = (6.01 kJ/mol)(1 mol/18.02 g) = 333 J/g
Heating Curve for Water
Total heat absorbed by the
water as it is warmed can
be determined by finding the
heat involved in each “step”
of the process.
q1+ q2 + q3 …. = qtotal
Note:
q=
(m)(C)(DT)
Cice= Csteam = 2.09 J/goC
qphase change = DHphase change (mass or moles)
Example
Calculate q for the process in which 50.0 g of water is converted
from liquid at 10.0°C to vapor at 125.0°C.
Example
• What is the heat of fusion of lead in J/g if 6.30
kilojoules of heat are required to convert 255
grams of solid lead at its melting point into a
liquid?
a. 0.0250 J/g
c.
1.61 J/g
b.24.7 J/g
d. 40.5 J/g
Example
• What quantity of heat is required to heat 1.00 g of lead from 25 ˚C to
the melting point (327 ˚C) and melt all of it? (The specific heat
capacity of lead is 0.159 J/g • K and it requires 24.7 J/g to convert lead
from the solid to the liquid state.)
a.
b.
2.47 J
48.0 J
c. 39.4 J
d. 72.7 J
Calorimetry
• Constant volume calorimeter is called a bomb
calorimeter.
• Material is put in a container with pure oxygen. Wires are
used to start the combustion. The container is put into a
container of water.
• The heat capacity of the calorimeter is known and tested.
Properties
• intensive properties not related to the amount
of substance.
• density, specific heat, temperature.
• Extensive property - does depend on the
amount of stuff.
• Heat capacity, mass, heat from a reaction.
Bomb Calorimeter
• thermometer
• stirrer
• full of water
• ignition wire
• Steel bomb
• sample
Bomb Calorimeter
qrxn = -qcal
qcal = qbomb + qwater + qwires +…
Define the heat capacity of the
calorimeter:
qcal = miciDT = CcalDT
heat
q and DH revisited
• If you know “q” for a process for a known
amount of sample
• And you have a balanced chemical equation
• “q” can be converted to DH
q to DH
If you know the heat per amount of
substance, find it per gram then use the molar
mass to find the heat per mole.
Thermochemical Equations
• Balanced chemical equation
• Includes a term which indicates the change in
enthalpy, DH
• Enthalpy term can be embedded in the
reaction or written after the reaction with DH
notation.
Examples
The combustion of 1.010 g sucrose, in a bomb calorimeter, causes the
temperature to rise from 24.92 to 28.33°C. The heat capacity of the
calorimeter assembly is 4.90 kJ/°C.
(a) What is the heat of combustion of sucrose, expressed in kJ/mol C12H22O11
(a) Verify the claim of sugar producers that one teaspoon of sugar (about 4.8 g)
contains only 19 calories.
Endothermic Thermochemical Equations
Al2O3(s)  2Al(s) + 3/2 O2(g)
DHrxn = 1,676kJ/mol rxn
• THE EXACT SAME AS WRITING:
• Al2O3(s) + 1,676 kJ  2Al(s) + 3/2 O2(g)
Exothermic Thermochemical Equations
H2(g) + ½ O2(g)  H2O(l) + 285.8kJ
Same as:
H2(g) + ½ O2(g)  H2O(l)
DHrxn = -285.8 kJ/mol rxn
Stoichiometry and Thermochemical Equations
• Just as the mole ratios can be used to
determine reacting and produced amounts
of chemicals
• The ratios can be used to relate amounts of
chemicals with energy
Example
Al2O3(s)  2Al(s) + 3/2 O2(g)
DHrxn = 1,676 kJ/mol rxn
• How many grams of aluminum would be produced if
the aluminum absorbed 3500 kJ of energy?
DHrxn
DHrxn is for ANY general chemical or physical
reaction
There are specific types of reactions which have
a special designation for their DH.
Some Important Types of Enthalpy Changes
Heat of combustion (DHcomb)
C4H10(l) + 13/2O2(g)
4CO2(g) + 5H2O(g)
Heat of formation (DHf)
K(s) + 1/2Br2(l)
KBr(s)
Heat of fusion (DHfus)
NaCl(s)
NaCl(l)
Heat of vaporization (DHvap)
C6H6(l)
C6H6(g)
Standard States and Standard
Enthalpy Changes
• Define a particular state as a standard state.
• Standard enthalpy of reaction, DH°
– The enthalpy change of a reaction in which all
reactants and products are in their standard states.
• Standard State
– The pure element or compound at a pressure of 1 bar
and at the temperature of interest.
Indirect Determination of DH:
Hess’s Law
• DH is an extensive property.
Enthalpy change is directly proportional to the
amount of substance in a system.
N2(g) + O2(g) → 2 NO(g)
DH = +180.50 kJ
½N2(g) + ½O2(g) → NO(g) DH = +90.25 kJ
 DH changes sign when a process is reversed
NO(g) → ½N2(g) + ½O2(g)
DH = -90.25 kJ
Hess’s Law
Hess’s law of constant heat summation
 If a process occurs in stages or steps (even
hypothetically), the enthalpy change for the overall
process is the sum of the enthalpy changes for the
individual steps.
½N2(g) + ½O2(g) → NO(g)
DH = +90.25 kJ
NO(g) + ½O2(g) → NO2(g)
DH = -57.07 kJ
½N2(g) + O2(g) → NO2(g)
DH = +33.18 kJ
Example
Find ΔH for 2H2(g) + O2(g)  2H2O(l)
Given:
CH3COOH(l) + 2O2(g) 2CO2(g) + 2H2O(l)
ΔH = -1,959.8 kJ
C(graphite) + O2(g)  CO2(g)
ΔH = -886.5 kJ
CH3COOH(l)  2C(graphite) + 2H2(g) + O2(g)
ΔH = 1,100.2 kJ
Example 11
• Find DH for 2NH3(g) N2H4 (l) + H2(g)
• Given:
• N2H4(l) + CH4O(l)  CH2O(g) + N2(g) + 3H2(g)
ΔH = -92.5KJ
N2(g) + 3H2(g)  2NH3(g)
ΔH = -115KJ
CH2O(g) + H2(g)  CH4O(l)
ΔH = 162.5KJ
Standard Enthalpies of Formation
D H°
f
• The enthalpy change that occurs in the formation of
one mole of a substance in the standard state from
the reference forms of the elements in their
standard states.
• The standard enthalpy of formation of a pure
element in its reference state is 0.
Standard Enthalpies of Formation
Standard Enthalpies of Formation
Writing Formation Equations
• Remember that DfHo refers to forming ONE
mole of a pure substance FROM ITS
ELEMENTS in their standard state
• Write thermochemical formation equations
for the formation of:
– Al2O3(s)
– NaHCO3(s)
Standard
Enthalpies of
Reaction
DH°rxn
DH°overall = -2DfH°NaHCO3+ DHf°Na2CO3+ DHf°CO2 + DHf°H2O
Enthalpy of
Reaction
DH°rxn = SDHf°products- SDHf°reactants
Enthalpies of Formation of Ions in
Aqueous Solutions