# Chapter 17 - Thermochemistry

```Quantitative
Thermochemistry
Chemistry Joke
Q: What weapon can you
make from the elements
potassium, nickel, and iron?
A: KNiFe!
Chemical Heat Changes
Stoichiometry can be used with
thermochemical equations.
 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.
3
Thermochemical Stoichiometry
CH4 + 2 O2  CO2 + 2 H2O + 802.2 kJ
 If
10.3 grams of CH4 are burned, how
much heat will be produced?
 Use 802.2 kJ in a stoichiometric ratio!
10. 3 g CH4
1 mol CH4
16.05 g CH4
- 802.2 kJ
1 mol CH4
= - 515 kJ
Thermochemical Stoichiometry
CH4 + 2 O2  CO2 + 2 H2O + 802.2 kJ
 If
you want to produce 500. kJ of heat, how
many grams of CH4 need to be burned?
500. kJ
1 mol CH4
- 802.2 kJ
16.05 g CH4
1 mol CH4
= 10.0 g
Physical Heat Changes
 Individual
substances respond differently
to heat changes.
 Some substances, like water, can absorb
a lot of heat before their temperature
rises.
 Other substances, like metals, only need
a little bit of heat to cause a temperature
change.
Physical Heat Change
 The
amount of heat it takes to raise the
temperature of 1g of a substance 1oC is
called the specific heat capacity of that
substance.
 “Specific” heat capacity is independent of
mass.
 Heat capacity is dependent on the mass
of the object.
Specific Heat Capacity
 Symbol
is “C”
 Units are J/g &middot;&ordm;C or cal/g &middot;&ordm;C
 When considering physical heat
changes, “q” is used to indicate heat.
q
C=
m x T
Specific =
heat
Heat
mass x temp change
T = Tfinal - Tinitial
Specific Heat Problem

When 435 J of heat are added to 3.4 g
of olive oil at 21oC, the temperature
increases to 85oC. What is the specific
heat of olive oil?
C = q / (m x T)
T = 85oC - 21oC = 64&ordm;C
435 J / (3.4 g x 64&ordm;C)
= 1.999  2.0 J/(g &middot;&ordm;C)
Specific Heat Problem
 It
takes 487.5 J to heat 25 grams of
copper from 25 &deg;C to 75 &deg;C. What is the
specific heat of copper?
 0.39 J/g&middot;&deg;C
Heat Problems
 The
equation for specific heat can be
rearranged to solve for heat.
 q = m C T
 Note that heat, q, and specific heat, C,
are two different things.
 q can be positive or negative depending
on the T.

(Remember, negative heat indicates an
exothermic process, and positive heat indicates
an endothermic process.)
Heat Problem

How much heat is required to raise the
temperature of 250 g of mercury 52oC? The
specific heat of Hg is 0.14 J/(g &middot;oC).
q = m x C x T
250 g x 0.14 J /(g &middot;&ordm;C) x 52&ordm;C
=1820  1800 J or 1.8 kJ
Heat Problem
 How
much energy must be absorbed by
20.0 g of water to increase its temperature
from 283.0 &ordm;C to 303.0 &ordm;C? The specific
heat of water is 4.184 J/g &middot;oC.
 1670 J
Phase Changes
 To
find the heat required to melt a
substance, q = m ∆Hfus.
 To find the heat required to vaporize a
substance, q = m ∆Hvap.
 The mass units have to match.
Phase Changes
Water: ∆Hfus = 333 J/g ∆HVap = 2250 J/g
 How much heat is required to change
25 g of ice at 0 &ordm;C to water at 0 &ordm;C?
 q = m ∆Hfus
 25 g x 333 J/g
 8300 J
Phase Changes
Water: ∆Hfus = 333 J/g
 How
∆HVap = 2250 J/g
much heat in kJ is required to
vaporize 2.0 moles of water?
 81 kJ
Using a Heating Curve
 How
much
energy would it
take to change
25 g of water at
30 &ordm;C to steam at
110 &ordm;C?
 Slanted
line: q = m C ∆T
 Flat lines: q = m ∆H
Using a Heating Curve
 Step1:
Draw a graph / label low and high
temp from problem
 Step 2: Determine any phase changes
 Step 3: Draw slanted lines to and from
phase changes and flat lines for phase
changes
 Step 4: Write one equation for each line
segment. Use the correct C, ∆T, or ∆H.
 Step 5: Add the energies together
Water: ∆Hfus = 333 J/g
Cwater = 4.184 J/g &ordm;C
∆HVap = 2250 J/g
Cice = 2.15 J/g &ordm;C
Csteam = 2.05 J/g &ordm;C
25 g of water at 30 &ordm;C to steam at 110 &ordm;C