Thermochemistry- Heat and
Chemical Change
Chapter 11
Energy & Chemistry
ENERGY is the capacity to
do work or transfer heat.
HEAT (q) is the form of
energy that flows between
2 objects because of their
difference in temperature.
Chemical Potential Energy
is the energy stored in the
structural units of
chemical substances.
Energy
Every reaction has an energy change
associated with it
 System- Part of universe focused on
 Surroundings- Everything else in universe
 Law of conservation of energy – In any
chemical or physical process, energy is
neither created nor destroyed
 Energy is stored in bonds between atoms

3
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
4
Exothermic



The products are
lower in energy
than the reactants
Releases energy
EXOthermic: heat
transfers from
SYSTEM to
SURROUNDINGS.
5
Energy
C + O2  CO2+ 395 kJ
C + O2
-395kJ
CO2
Reactants

Products
6
Endothermic
The products are
higher in energy
than the reactants
 Absorbs energy
 ENDOthermic:
heat transfers from
SURROUNDINGS
to the SYSTEM.

7
Energy
CaCO
 CaO
CaCO
CaO
+ CO+2 CO2
3 + 176
3 kJ
CaO + CO2
+176 kJ
CaCO3
Reactants

Products
8
UNITS OF ENERGY
*1 calorie = heat required to
raise temp. of 1.00 g of
H2O by 1.0 oC.
*1000 cal = 1 kilocalorie = 1
kcal
*1 kcal = 1 Calorie (a food
“calorie”)
*In SI, use the unit called
the JOULE
*1 cal = 4.184 joules
James Joule
1818-1889
Finding “How Much” Heat
q= mass x specific heat x change in
Temperature
 q = m x C x ∆T
 q is heat or is referred to as enthalpy
 Enthalpy = ∆H
 ∆H is the heat content in a chemical
reaction that happens under constant
pressure

Heat Capacity
Heat Capacity- the amount of heat
required to increase the temperature of
something by 1ْC.
 Specific Heat Capacity or Specific Heatamount of heat needed to increase the T
of 1g of something by 1ْ C.
 Specific Heat= C (J/g•◦ C) or (cal/g•◦ C)
 ∆T = Tfinal – Tinitial
 Use table 11.2 p.296 for values of C

Examples

How much heat is required to raise the
temperature of 400.0g of silver 45
ْC?

If 25.0 g of Al cool from 310 oC to 37
oC, how many joules of heat energy
are lost by the Al?
Calorimetry
Measuring heat.
 Use a calorimeter.
 Two kinds
 Constant pressure calorimeter
(called a coffee cup calorimeter)
 An insulated cup, full of water.
 The specific heat of water is
4.184j/gºC
 ∆H= (q) = specific heat x m x DT

Example

A chemical reaction is carried out in a
coffee cup calorimeter. There are 75.8 g
of water in the cup, and the temperature
rises from 16.8 ºC to 34.3 ºC. How
much heat was released?
Calorimetry
Second type 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/or tested. (cal/ ºC)
 Multiply temperature change by the heat
capacity to find heat

Bomb Calorimeter

thermometer

stirrer

full of water

ignition wire

Steel bomb

sample
Example

A lead mass is heated and placed in a
coffee cup calorimeter containing 40.0
mL of water at 17.0 ْC. The water
reaches a temperature of 20.0 ْC. How
many joules were released by the lead?
Heats of Reaction
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

20
Enthalpy
Symbol is H
 Change in enthalpy is DH
 delta H
 If heat is released the heat content of
the products is lower
 DH is negative (exothermic)
 If heat is absorbed the heat content
of the products is higher
 DH is positive (endothermic)

21
Energy
Change is down
DH is <0
Reactants

Products
22
Energy
Change is up
DH is > 0
Reactants

Products
23
Heat of Reaction
The heat that is released or absorbed in a
chemical reaction
 Equivalent to DH
 C + O2(g)  CO2(g) +393.5 kJ
 C + O2(g)  CO2(g)
DH = -393.5 kJ
 In thermochemical equations, it is
important to say what state
 H2(g) + ½ O2 (g) H2O(g) DH = -241.8 kJ
 H2(g) + ½ O2 (g) H2O(l) DH = -285.8 kJ

24
Example

The burning of magnesium is a highly
exothermic reaction. How many
kilojoules of heat are released when
0.75 mol of Mg burn in an excess of O2?
2Mg (s) + O2 (g)
→ 2MgO (s) + 1204kJ
Heat of Combustion
The heat from the reaction that
completely burns 1 mole of a
substance
 Remember to Balance and Use Table
11.4 p. 305





?C2H2 + ?O2  ? CO2 + ? H2O
?CH4 + ? O2  ?CO2 + ?H2O
?C6H6 + ? O2  ?CO2 + ? H2O
?C8H18 + ? O2  ?CO2 + ? H2O
26
Hess’s Law of Heat Summation
Most reactions happen in a series of
steps
 Many times a direct measurement
CANNOT be made
 Hess’s Law allows the calculation of the
heat of a reaction without DOING the
reaction
 Thermochemical equations can be
added to get final heat of reaction

Why Does It Work?
If H2(g) + 1/2 O2(g) H2O(g) DH=-285.5 kJ
 then
H2O(g) H2(g) + 1/2 O2(g) DH =+285.5 kJ
 If you turn an equation around, you
change the sign
 2 H2O(g) 2H2(g) + O2(g) DH =+571.0 kJ
 If you multiply the equation by a number,
you multiply the heat by that number.

28
Example

Given
5
C2H2 (g) + O2 (g)  2CO2 (g) + H2O(l)
2
DHº= -1300. kJ
C(s) + O2 (g)  CO2 (g)
DHº= -394 kJ


H 2 (g) +
1
O 2 (g)  H 2 O(l)
2
DHº= -286 kJ

calculate DHº for this reaction
2C(s) + H2 (g)  C2H2 (g)
Example
Given
DHº= +77.9kJ
DHº= +495 kJ
DHº= +435.9kJ
Calculate DHº for this reaction
O2 (g) + H2 (g)  2OH(g)
O2 (g)  2O(g)

H2 (g)  2H(g)


O(g) + H(g)  OH(g)


How do you get good at this
31
Standard Heat of Formation
The DH for a reaction that produces 1
mol of a compound from its elements at
standard conditions
 Standard conditions 25°C and 1 atm.
0
 Symbol is DH f

The
standard heat of formation of an
element is 0
This

includes the diatomics
32
What good are they?
There is a table (pg. 316) of heats of
formations
 The heat of a reaction can be calculated
by subtracting the heats of formation of
the reactants from the products

DH = DH (products) - DH (reactants)
0
f
0
f
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Examples

CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
0
DH f CH4 (g) = -74.86 kJ
0
DH f O2(g) = 0 kJ
0
DH f CO2(g) = -393.5 kJ
0
DH f H2O(g) = -241.8 kJ
DH= [-393.5 + 2(-241.8)]-[-74.86 +2 (0)]
 DH= -802.2 kJ

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Example

2 SO3(g)  2SO2(g) + O2(g)
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