Constant Pressure Calorimetry:
Another Example
A piece of chromium metal weighing 24.26 g is
heated in boiling water to a temperature of 98.3°C
and then dropped into a coffee cup calorimeter
containing 82.3g of water at 23.3°C. When
thermal equilibrium is reached, the final
temperature is 25.6°C. Calculate the Cp of
chromium.
© 2009 Brooks/Cole - Cengage
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Constant Volume Calorimetry: It’s The Bomb!
• Technique can be used to obtain the heat content
of combustion of compounds
• Used in the food, fuel and pharmaceutical
industries to know how much energy would be
released by completely consuming the compound
• Uses a BOMB Calorimeter
© 2009 Brooks/Cole - Cengage
CALORIMETRY
Measuring Heats of Reaction
•Place sample of known
mass inside the bomb
•Place oxygen in the sample
chamber and immerse
bomb into water
•Ignite the bomb and
measure temperature of
water
•Since the volume doesn’t
change, no P-V work is
done, so the qr is a
measurement of the ΔU
© 2009 Brooks/Cole - Cengage
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Calorimetry
Some heat from reaction warms
water
qwater = (sp. ht.)(water mass)(∆T)
Some heat from reaction warms
“bomb”
qbomb = (heat capacity, J/K)(∆T)
Total heat evolved = qtotal = qwater + qbomb
© 2009 Brooks/Cole - Cengage
Measuring Heats of Reaction
CALORIMETRY
Calculate energy of combustion (∆U) of
octane.
2C8H18 + 25O2 --> 16CO2 + 18H2O
• Burn 1.00 g of octane
• Temp rises from 25.00 to 33.20 oC
• Calorimeter contains 1200. g water
• Heat capacity of bomb = 837 J/K
© 2009 Brooks/Cole - Cengage
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Measuring Heats of Reaction
CALORIMETRY
Step 1 Calc. energy transferred from reaction to water.
q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J
Step 2 Calc. energy transferred from reaction to bomb.
q = (bomb heat capacity)(∆T)
= (837 J/K)(8.20 K) = 6860 J
Step 3 Total energy evolved
41,200 J + 6860 J = 48,060 J
Energy of combustion (∆U) of 1.00 g of octane
= - 48.1 kJ
© 2009 Brooks/Cole - Cengage
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The First Law of Thermodynamics
• Up until now, we have only considered the changes in
the internal energy of a system as functions of a
single change: either work or heat
• However, these changes rarely occur singly, so we
can describe the change in internal energy as:
U = q + w
(The 1st Law)
• The change in internal energy is dependent upon the
work done by the system and the heat gained or lost
by the system
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The First Law: Put Another Way
1st Law of Thermodynamics
A system can store energy. A change in the
energy of a system means that there must be a
change in the heat or the work done BY or TO
the system.
OR
The Total Energy of the Universe is Constant
© 2009 Brooks/Cole - Cengage
The First Law
• Internal energy is an example
of a state function
• A state function is a property
that only depends on the
current state of the system
and is independent of how
that state was reached
• Pressure, Volume,
Temperature and density are
all examples of state
functions
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State Functions and Things that Aren’t
• Work and heat ARE NOT state
functions
• The amount of work done
depends on how the change
was brought about
© 2009 Brooks/Cole - Cengage
Change in Internal Energy
(Implications of a State Function)
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• It doesn’t matter what path we take to get to the final point,
the change in internal energy is only dependent on where we
started and where we finished
• Let’s think about this on a molecular level…
– If we expand an ideal gas isothermally, the molecules will have the same
kinetic energy and will move at the same speed
– Despite the fact that the volume has increased, the potential energy of the
system remains the same because there are no forces between molecules
(KMT)
– Since neither the kinetic nor potential energy has changed, the change in
internal energy is…
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Zero!
U = 0 for the isothermal expansion of an ideal gas
© 2009 Brooks/Cole - Cengage
Enthalpy
• In a constant volume system in which no work is done (neither
expansion nor non-expansion), we can rearrange the first law to:
U = q + w
(but w=0)
U = q
• Most systems are constant pressure systems which can expand and
contract
• When a chemical reaction takes place in such a system, if gas is
evolved, it has to push against the atmosphere in order to leave the
liquid or solid phase
– Just because there’s no piston, it doesn’t mean that no work is done!
© 2009 Brooks/Cole - Cengage
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Enthalpy
• Let’s look at an example:
• If we supply 100J of heat to a system at constant
pressure and it does 20J of work during expansion, the
U of the system is +80J (w=-20J)
– We can’t lose energy like this
• Enthalpy, H, is a state function that we use to track
energy changes at constant pressure
H=U + PV
• The change in enthalpy of a system (H) is equal to the
heat released or absorbed at constant pressure
© 2009 Brooks/Cole - Cengage
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Enthalpy
• Another way to define enthalpy is at constant pressure:
H = q
© 2009 Brooks/Cole - Cengage
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Enthalpy and Chemical Reactions
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• Enthalpy is a tricky thing to grasp, but we can look at it this
way:
Enthalpy is the macroscopic energy change (in the form of
heat) that accompanies changes at the atomic level (bond
formation or breaking)
• Enthalpy has the same sign convention as work, q and U
– If energy is released as heat during a chemical reaction the enthalpy has a ‘-’
sign
– If energy is absorbed as heat from the surrounding during a reaction, the
enthalpy has a ‘+’ sign
© 2009 Brooks/Cole - Cengage
Heat Transfers at Constant Pressure
•
H = U + PV
The change in the enthalpy of a system is equal to the heat
released or absorbed at constant pressure
WTF?
1. In a coffee cup calorimeter (constant pressure calorimeter), the
heat, q, that is released or absorbed is equal to the change in
enthalpy, H
2. When we add heat to a constant pressure system, the enthalpy
increases by that amount
3. H<0 for exothermic reactions
H>0 for endothermic reactions
© 2009 Brooks/Cole - Cengage
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USING ENTHALPY
Consider the formation of water
H2(g) + 1/2 O2(g) --> H2O(g) + 241.8 kJ
Exothermic reaction — energy is a “product” and
∆H = – 241.8 kJ
© 2009 Brooks/Cole - Cengage
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Energy Transfer
with Change of State
Changes of state involve energy (at constant T)
Ice + 333 J/g (heat of fusion) f Liquid water
q = (heat of fusion)(mass)
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Energy Transfer and
Changes of State
Liquid --> Vapor
Requires energy (heat).
This is the reason
a)you cool down after
swimming
b)
you use water to put
out a fire.
© 2009 Brooks/Cole - Cengage
+ energy
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Heating/Cooling Curve for Water
Note that T is
constant as ice melts
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Heat & Changes of State
What quantity of heat is required to melt
500. g of ice and heat the water to
steam at 100 oC?
Heat of fusion of ice = 333 J/g
Specific heat of water = 4.2 J/g•K
Heat of vaporization = 2260 J/g
+333 J/g
© 2009 Brooks/Cole - Cengage
+2260 J/g
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Heat & Changes of State
What quantity of energy as heat is required to melt 500. g
of ice and heat the water to steam at 100 oC?
1.
To melt ice
q = (500. g)(333 J/g) = 1.67 x 105 J
2.
To raise water from 0 oC to 100 oC
q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 105 J
3.
To evaporate water at 100 oC
q = (500. g)(2260 J/g) = 1.13 x 106 J
4. Total energy = 1.51 x 106 J = 1510 kJ
© 2009 Brooks/Cole - Cengage
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USING ENTHALPY
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Making liquid H2O from H2 +
O2 involves two exothermic
steps.
H2 + O2 gas
H2O vapor
© 2009 Brooks/Cole - Cengage
Liquid H2O
Hess’s Law
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The ∆rHo for a reaction that is the sum of one or
more reactions is the sum of the ∆rHo values for
all of the reactions
Remember: Since ∆H is a state function, it
doesn’t matter how we get to where we end up,
all that matters is where we started and where
we finished.
© 2009 Brooks/Cole - Cengage
USING ENTHALPY
Making H2O from H2 involves two steps.
H2(g) + 1/2 O2(g) f H2O(g) + 242 kJ
H2O(g) f H2O(liq) + 44 kJ
-----------------------------------------------------------------------
H2(g) + 1/2 O2(g) f H2O(liq) + 286 kJ
Example of HESS’S LAW—
If a rxn. is the sum of 2 or more others,
the net ∆H is the sum of the ∆H’s of
the other rxns.
© 2009 Brooks/Cole - Cengage
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Hess’s Law
& Energy Level
Diagrams
Forming H2O can occur
in a single step or in a
two steps. ∆Htotal is the
same no matter which
path is followed.
© 2009 Brooks/Cole - Cengage
USING ENTHALPY
Making CO from C (graphite) and O2 is a single step, but
the CO reacts with oxygen to form CO2.
1) C(graphite) + ½O2(g) --> CO (g) ? kJ/mole-rxn
2) CO(g) + O2(g) --> CO2(g) + -283 kJ/mole-rxn
-----------------------------------------------------------------------
3) C (graphite) + ½O2(g) --> CO2(g) -393.5 kJ/mole-rxn
Example of HESS’S LAW—
If a rxn. is the sum of 2 or more others, the net ∆H
is the sum of the ∆H’s of the other rxns.
∆Ho3= ∆Ho1 + ∆Ho2
© 2009 Brooks/Cole - Cengage
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Using Enthalpy
∆Ho3= ∆Ho1 + ∆Ho2
-393.5 kJ/mole-rxn = ∆Ho1 + (-283.0 kJ/mole-rxn)
∆Ho1 = -110.5 kJ/mole-rxn
© 2009 Brooks/Cole - Cengage
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Hess’s Law
& Energy Level
Diagrams
Forming CO2 can occur
in a single step or in a
two steps. ∆Htotal is the
same no matter which
path is followed.
Active Figure 5.16
© 2009 Brooks/Cole - Cengage
Hess’s Law: Another Example
Use Hess’s Law to calculate the enthalpy change for the formation of
CS2(l) from C(s) and S(s).
The overall equation is: C(s) + 2S(s) --> CS2(l)
C(s) + O2(g) --> CO2(g)
∆Ho = -393.5 kJ/mole-rxn
S(s) + O2(g) --> SO2(g)
∆Ho = -295.8 kJ/mole-rxn
CS2(l) + 3O2(g) --> CO2(g) +2SO2(g)
∆Ho = -1103.9 kJ/mole-rxn
© 2009 Brooks/Cole - Cengage
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Hess’s Law: Another Example
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How do we start?
-Let’s set our overall reaction up:
C(s) + O2(g) --> CO2(g)
∆Ho = -393.5 kJ/mole-rxn
S(s) + O2(g) --> SO2(g)
∆Ho = -295.8 kJ/mole-rxn
CO2(g) +2SO2(g) --> CS2(l) + 3O2(g)
∆Ho = 1103.9 kJ/mole-rxn
----------------------------------------------------------------------------------C(s) + 2S(s) --> CS2(l)
∆Ho = ?
Note: We reversed reaction 3 and changed the sign
Now: We need to make sure that all of the molecules
balance and take care to multiply their ∆rHo values if
necessary
© 2009 Brooks/Cole - Cengage
Hess’s Law: Another Example
C(s) + O2(g) --> CO2(g)
∆Ho = -393.5 kJ/mole-rxn
S(s) + O2(g) --> SO2(g)
∆Ho = -295.8 kJ/mole-rxn
CO2(g) +2SO2(g) --> CS2(l) + 3O2(g) ∆Ho = 1103.9 kJ/mole-rxn
----------------------------------------------------------------------------------C(s) + 2S(s) --> CS2(l)
∆Ho = ?
We need 2 S(s) atoms, so let’s multiply reaction2
by 2:
2S(s) + 2O2(g) --> 2SO2(g)
© 2009 Brooks/Cole - Cengage
∆Ho = -591.6 kJ/mole-rxn
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Hess’s Law: Another Example
Now let’s look at what we’ve got:
C(s) + O2(g) --> CO2(g)
∆Ho1 = -393.5 kJ/mole-rxn
2S(s) + 2O2(g) --> 2SO2(g)
∆Ho2 = -591.6 kJ/mole-rxn
CO2(g) +2SO2(g) --> CS2(l) + 3O2(g) ∆Ho3 = 1103.9 kJ/mole-rxn
----------------------------------------------------------------------------------C(s) + 2S(s) --> CS2(l)
∆Ho = ∆Ho1 + ∆Ho2 + ∆Ho3
Everything looks good, so let’s add it up:
∆Ho = (-393.5 kJ/mole-rxn)+(-591.6 kJ/mole-rxn)+(1103.95 kJ/mole-rxn)
∆Ho = 118.9 kJ/mole-rxn
© 2009 Brooks/Cole - Cengage
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Standard Enthalpy Values
Most ∆H values are labeled ∆Ho
Measured under standard conditions
P = 1 bar
Concentration = 1 mol/L
T = usually 25 oC
with all species in standard states
e.g., C = graphite and O2 = gas
© 2009 Brooks/Cole - Cengage
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Standard Enthalpy Values
NIST (Nat’l Institute for Standards and
Technology) gives values of
∆Hfo = standard molar enthalpy of
formation
— the enthalpy change when 1 mol of
compound is formed from elements under
standard conditions.
See Appendix
© 2009 Brooks/Cole - Cengage
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∆Hfo, standard molar
enthalpy of formation
Equations for Standard Molar Enthalpy of Formation:
KNO3
K(s) + ½ N2(g) + 3/2O2(g) --> KNO3
FeCl3
Fe(s) + 3/2 Cl2(g) --> FeCl3(s)
C12H22O11
12C(s) + 11H2(g) + 11/2 O2(g) --> C12H22O11(s)
© 2009 Brooks/Cole - Cengage
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Using Standard Enthalpy Values
In general, when ALL enthalpies
of formation are known,
∆Ho =  ∆fHo (products) -  ∆fHo (reactants)
Remember that ∆ always = final – initial
© 2009 Brooks/Cole - Cengage
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Using Standard Enthalpy Values
Calculate the heat of combustion of
methanol, i.e., ∆Ho for
CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g)
∆Ho =  ∆Hfo (prod) -  ∆Hfo (react)
© 2009 Brooks/Cole - Cengage
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Using Standard Enthalpy Values
CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g)
∆rHo =  ∆Hfo (prod) -  ∆Hfo (react)
∆Hro = ∆Hfo (CO2) + 2 ∆Hfo (H2O)
- {3/2 ∆Hfo (O2) + ∆Hfo (CH3OH)}
= (-393.5 kJ) + 2 (-241.8 kJ)
- {0 + (-201.5 kJ)}
∆Hro = -675.6 kJ per mol of methanol
© 2009 Brooks/Cole - Cengage
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Using Standard Enthalpy Values
Calculate the heat of reaction of nitric oxide, i.e., ∆Hro for
4NH3 (g) + 5 O2(g) --> 4NO(g) + 6 H2O(g)
∆Ho =  ∆Hfo (prod) -  ∆Hfo (react)
∆Ho (prod) = 4(∆Hfo NO) + 6(∆Hfo H2O)
= 4(90.29) + 6(-241.830)
= -1089 kJ/mole-rxn
∆Hfo (react) = 4(∆Hfo NH3) + 5(∆Hfo O2)
= 4(-45.90) + 5(0)
= -183.6 kJ/mole-rxn
∆Hro =  ∆Hfo (prod) -  ∆Hfo (react)
= -1089 kJ/mole-rxn + 183.6 kJ/mole-rxn = -906.2 kJ/mole-rxn
© 2009 Brooks/Cole - Cengage
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