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Calorimetry
Calorimetry
 Calorimetry
- the accurate and
precise measurement of heat
change for chemical and physical
processes.
 The device used to measure the
absorption or release of heat in
chemical or physical processes is
called a Calorimeter
Calorimetry
 Foam
cups are excellent heat
insulators, and are commonly used
as simple calorimeters
A Cheap
Calorimeter
 For
systems at constant pressure,
the heat content is the same as a
property called Enthalpy (H) of the
system
Calorimetry
in enthalpy = H
 q = H These terms will be used
interchangeably in this textbook
 Thus, q = H = m x C x T
 H is negative for an exothermic
reaction
 H is positive for an endothermic
reaction
 Changes
Calorimetry
 Calorimetry
experiments can be
performed at a constant volume
using a device called a “bomb
calorimeter” - a closed system
In terms of bonds
C
O
O
O
C
O
Breaking this bond will require energy.
O
C
O C O
O
Making these bonds gives you energy.
In this case making the bonds gives you
more energy than breaking them.
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Exothermic
 The
products are lower in energy
than the reactants
 Releases energy
2Al (s) + 3Cl2 (g) --> 2 AlCl3 (s) + 1408 kJ
∆H=1408 kJ
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Energy
C + O2  CO2+ 395 kJ
C + O2
395kJ
C O2
Reactants

Products
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Endothermic
 The
products are higher in energy
than the reactants
 Absorbs energy
2 H2O + 575 kJ ------> 2 H2 + 1 O2 (g)
∆H = + 572 kJ
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Energy
CaCO
 CaO
CaCO
CaO
+ CO+2 CO2
3 + 176
3 kJ
CaO + CO2
176 kJ
CaCO3
Reactants

Products
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Chemistry Happens in
MOLES
An equation that includes energy is
called a thermochemical equation
 CH4 + 2O2  CO2 + 2H2O + 802.2 kJ
 1 mole of CH4 releases 802.2 kJ of
energy.
 When you make 802.2 kJ you also
make 2 moles of water

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What is the molar enthalpy of CO2 (g) in the
reaction for the burning of butane below?
2 C4H10 +13 O2  8 CO2 +10 H2O
∆H=-5315 kJ
Answer:
Molar enthalpy is the enthalpy change in
equation divided by the balance of CO2
 Molar enthalpy, ∆H substance = 5315 kJ ÷ 8
mol

= 664 kJ / mol.

For each of the following rewrite the
equation in " H " notation, for one mole
of the underlined substance.
Fe2O3 (s)+3CO(g)→3CO2(g)+2Fe(s)+25kJ

Answer:
1/3 Fe2O3 (s)+CO(g)CO2(g)+2/3 Fe(s)
∆H = - 8.3 KJ

4 NH3(g)+5O2 (g)→4 NO(g)+6H2O(l)+1170kJ
2 HCl (g)+96 KJ → H2 (g)+Cl2 (g)
N2 (g)+3 H2 (g) → 2 NH3 (g)+92 KJ
2 CO2 (g)+566 KJ →2 CO (g)+ O2 (g)
4 Al (s) +3 O2 (g) →2 Al2O3 (s)+3360 KJ
Thermochemical Equations
 A heat
of reaction is the heat
change for the equation, exactly
as written
• The physical state of reactants
and products must also be given.
• Standard conditions for the
reaction is 101.3 kPa (1 atm.)
and 25 oC
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CH4 + 2 O2  CO2 + 2 H2O + 802.2 kJ

If 10. 3 grams of CH4 are burned
completely, how much heat will be
produced?
10. 3 g CH4
1 mol CH4
16.05 g CH4
802.2 kJ
1 mol CH4
= 514 kJ
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CH4 + 2 O2  CO2 + 2 H2O + 802.2 kJ
 How
many liters of O2 at STP would
be required to produce 23 kJ of
heat?
 How many grams of water would be
produced with 506 kJ of heat?
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
How much heat will be released if 65 grams of butane
is burned in a lighter according the equation:
2 C4H10 +13 O2  8 CO2 +10 H2O
∆H=-5315 kJ
 1molC4 H10  5315kJ 

65 gC4 H10 

 58.14 g  2molC4 H10 
= 2976.4 kJ
= 3.0 MJ
Calculate the heat released when 120 grams
of Iron (III) oxide is formed by the following
equation
2 Fe2O3 (s) → 4 Fe(s)+3 O2 (g)
∆H=1625 kJ

 1molFe2O3  1625kJ 
120 gFe2O3 


 159.70 g  2mol 
= 610.5 kJ
= 610 kJ

Q = n ∆H (substance)
Where n = # of moles
What mass of carbon dioxide must form
to create 1200 kJ of heat when the
following reaction occurs?
C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)
∆H=- 2808kJ
Answer: 110 grams

3) What mass of oxygen is needed to
completely react and release 550 kJ of
heat in the following reaction?
 4Fe (s)+3O2 (g) → 2 Fe2O3 (s)

∆H=- 1625 kJ
Answer: 32 grams
Summary, so far...
Enthalpy
 The
heat content a substance has at a
given temperature and pressure
 Can’t be measured directly because
there is no set starting point
 The reactants start with a heat content
 The products end up with a heat
content
 So we can measure how much
enthalpy changes
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Enthalpy
 Symbol
is H
 Change in enthalpy is H (delta H)
 If heat is released, the heat content of
the products is lower
H is negative (exothermic)
 If heat is absorbed, the heat content
of the products is higher
H is positive (endothermic)
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Energy
Change is down
H is <0
Reactants

Products
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Energy
Change is up
H is > 0
Reactants

Products
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Heat of Reaction
The heat that is released or absorbed in a
chemical reaction
 Equivalent to H
 C + O2(g)  CO2(g) + 393.5 kJ
 C + O2(g)  CO2(g)
H = -393.5 kJ
 In thermochemical equation, it is important
to indicate the physical state
 H2(g) + 1/2O2 (g) H2O(g) H = -241.8 kJ
 H2(g) + 1/2O2 (g) H2O(l) H = -285.8 kJ

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Heat of Combustion
 The
heat from the reaction that
completely burns 1 mole of a
substance
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 OBJECTIVES:
• Classify, by type, the heat changes
that occur during melting, freezing,
boiling, and condensing.
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 OBJECTIVES:
• Calculate heat changes that occur
during melting, freezing, boiling,
and condensing.
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Heats of Fusion and
Solidification
 Molar
Heat of Fusion (Hfus) - the
heat absorbed by one mole of a
substance in melting from a solid to
a liquid
 Molar Heat of Solidification (Hsolid)
- heat lost when one mole of liquid
solidifies
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Heats of Fusion and
Solidification
 Heat
absorbed by a melting solid is
equal to heat lost when a liquid
solidifies
• Thus, Hfus = -Hsolid
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Heats of Vaporization and
Condensation
 When
liquids absorb heat at their
boiling points, they become vapors.
 Molar Heat of Vaporization (Hvap) the amount of heat necessary to
vaporize one mole of a given liquid.
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Heats of Vaporization and
Condensation
 Condensation
is the opposite of
vaporization.
 Molar Heat of Condensation (Hcond)
- amount of heat released when one
mole of vapor condenses
 Hvap = - Hcond
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Heats of Vaporization and
Condensation
The large values for Hvap and Hcond
are the reason hot vapors such as
steam is very dangerous
• You can receive a scalding burn
from steam when the heat of
condensation is released!
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Heats of Vaporization and
Condensation
 H20(g)
 H20(l)
Hcond = - 40.7kJ/mol
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Heat of Solution
 Heat
changes can also occur when
a solute dissolves in a solvent.
 Molar Heat of Solution (Hsoln) heat change caused by dissolution
of one mole of substance
 Sodium hydroxide provides a good
example of an exothermic molar
heat of solution:
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Heat of Solution
NaOH(s)
H2O(l)

Na1+(aq) + OH1-(aq)
Hsoln = - 445.1 kJ/mol
 The heat is released as the ions
separate and interact with water,
releasing 445.1 kJ of heat as Hsoln
thus becoming so hot it steams
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