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Chapter Six
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61.1% of students scored better on this test than
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Chapter Six
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 The
study of energetics of
chemical reactions
 Thermodynamics is the
branch of science which
studies the transformation of
energy from one form to
another
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Chapter Six
4
Because
energy!
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everything is
Chapter Six
5
The questions that chemical
thermodynamics ask are:
 How much heat is evolved
during a chemical reaction
("thermo chemistry")?
 What determines the direction of
spontaneous chemical change?

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Chapter Six
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

For example, if you put a flame to a 2:1
mixture of H2 and O2 there is an almighty
bang and the reaction proceeds
spontaneously to products, (water)
without any further supply of external
energy.
Why does it not happen that water
spontaneously goes to H2 and O2 when
we put a lit taper to it under the same
conditions?
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Chapter Six
7
 Energy
is the capacity to do
work (to displace or move
matter).
 Energy literally means “work
within”; however, an object
does not contain work.
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Chapter Six
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
Kinetic energy is the energy of
motion.
Ek = ½ mv2
m = mass
v = velocity
Energy has the units of joules (J or
kg . m2/s2) or calories
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Chapter Six
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


The energy STORED that is released
through motion
It is the energy within a system because
of its position or arrangement
Examples:
Ball in the air
 Battery
 Chemical energy

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Chapter Six
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At what point in each
bounce is the potential
energy of the ball at a
maximum?
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Chapter Six
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

Potential energy can be converted to
kinetic energy
Kinetic energy can be converted to
potential energy
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Chapter Six
12
1.
2.
3.
4.
5.
Light Energy – light bulb
Heat Energy – fire
Electrical Energy – putting your finger in
an electrical socket
Mechanical Energy – engine or a water
wheel
Chemical Energy – burning gasoline
(chemical energy is converted to heat
and light)
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Chapter Six
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

Chemical energy is stored inside a
substance until a substance
undergoes a chemical change. Then
it can be released or absorbed.
Chemical energy is stored in
chemical bonds between atoms.
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Chapter Six
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Thermodynamics – the study of
heat, energy, and the ability to
do work.
 Energy – the capacity to do work
or produce heat.

 Both
heat and work are energy
transfers
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Chapter Six
15
 The
First Law of
Thermodynamics – energy
is not created or destroyed
it is just transferred
 All
energy in the universe is
constant!
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Chapter Six
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 Work
– a force acts on
something causing it to
move (w)
 Heat – energy transfer from
thermal interaction (q)
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Chapter Six
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System – the part of the universe
we are interested in studying
(we choose the system)
 Surroundings – the rest of the
universe

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Chapter Six
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• Open: energy and
matter can be
exchanged with the
surroundings.
• Closed: energy can
be exchanged with
the surroundings,
matter cannot.
• Isolated: neither
energy nor matter
can be exchanged
with the
surroundings.
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A closed system;
energy (not matter)
can be exchanged.
After the lid of the jar
is unscrewed, which
kind of system is it?
Chapter Six
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
We are going to be talking about our
chemical system losing and gaining
energy, so it is logical to introduce a
quantity which represents the total
energy of the system. This is known
as the internal energy, E.

It’s important to note that your textbook
refers to internal energy as U sometimes.
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Chapter Six
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 Internal
energy is defined as
the total energy of our system
- chemical, nuclear, heat,
gravitational, any other type
of energy you can think of the sum of all the system's
energy (kinetic + potential).
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Chapter Six
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


A little thought will convince ourselves that it
would be impossible to actually measure the total
internal energy of our system.
So why define a quantity which we cannot
measure?
Well, although we cannot measure the absolute
value of the internal energy of a system, we can
measure changes in the internal energy. And since
thermodynamics is all about changes in energy
this makes the change in internal energy of a
system a very useful experimental quantity.
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Chapter Six
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
1.
2.
3.
The internal energy, E, of a system may
change in 3 different ways. The internal
energy, E, of a system will change if :
heat passes into or out of the
system;
work is done on or by the
system;
mass enters or leaves the
system.
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Chapter Six
23

The value of a state function is
independent of the history of
the system. All that matters
are the starting and ending
values, not how you got there.

Elevation change when climbing a
mountain is an example
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Chapter Six
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


The fact that internal energy is a state function
is extremely useful because we can measure the
energy change in the system by knowing the
initial energy and the final energy.
In other words, we don’t need all of the detail
of a process to measure the change in the value
of a state function.
In contrast, we do need all of the details to
measure the HEAT or the WORK of a system.
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Chapter Six
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

Internal energy (E) is the total energy
contained within a system
Part of E is kinetic energy (from
molecular motion)


Translational motion, rotational motion,
vibrational motion.
Collectively, these are sometimes called
thermal energy
• Part of E is potential energy
– Intermolecular and intramolecular forces of
attraction, locations of atoms and of bonds.
– Collectively these are sometimes called
chemical energy
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
There are two ways to classify
conversion of potential energy to kinetic
energy and vice versa:



Heat – q
Work – w
Heat and work are NOT state functions
(they are pathway dependent)

This makes it difficult to distinguish between
the energy from heat and the energy from
work. That’s again why ΔE being a state
function is so important.
 Car going down mountain example from
Crash Course.
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Chapter Six
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Energy entering a system
carries a positive sign (gains
energy)
 Energy leaving a system
carries a negative sign (loses
energy)

 Energy,
work, and heat are not
really positive or negative. The
sign just tells you the direction of
flow.
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Chapter Six
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
A negative quantity of work signifies
that the system loses energy.


A positive quantity of work signifies
that the system gains energy.


- w = work done by system (expansion)
+w = work done on system (compression)
In gases, work is related to changes in
volume.
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Chapter Six
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For now we will consider only
pressure-volume work.
work (w) = –PDV
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Chapter Six
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

Technically speaking,
heat is not “energy.”
Heat is energy
transfer between a
system and its
surroundings, caused
by a temperature
difference. Passes
from higher to lower
areas of heat.
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More energetic
molecules …
… transfer energy to
less energetic molecules.
Chapter Six
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
+ q = heat gained by system (heat is
transferred to the system)


Endothermic = the system gets colder
- q = heat lost by system (heat is
transferred from the system to the
surroundings)

Exothermic = the system gets warmer
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Surroundings are at 25 °C
25 °C
Typical situation:
some heat is released
to the surroundings,
some heat is absorbed
by the solution.
Hypothetical situation: all heat
is instantly released to the
surroundings. Heat = qrxn
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32.2 °C
35.4 °C
In an isolated system, all heat is
absorbed by the solution.
Maximum temperature rise.
Chapter Six
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Sign
Convention
+ΔE
-ΔE
+w
-w
+q
-q
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Meaning
System gains energy (produces E)
System loses energy (requires E)
Work done on system
Work done by system
Heat absorbed by system
(endothermic)
Heat given off by system
(exothermic)
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 calorie
- the quantity of heat
required to change the temp
of one gram of water one
degree Celsius
 SI Unit of basic energy is the
Joule (J)
1 cal = 4.184 J
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Chapter Six
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



“Energy cannot be created or destroyed.”
A change in the internal energy of a system is
due only to heat and work.
Inference: the internal energy change of a
system is simply the difference between its final
and initial states:
DE = Efinal – Einitial
Additional inference: if energy change occurs
only as heat (q) and/or work (w), then:
DE= q + w
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Chapter Six
36
Example 6.1
A gas does 135 J of work while expanding, and at the
same time it absorbs 156 J of heat. What is the change
in internal energy?
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Chapter Six
37
Enthalpy is the sum of
the internal energy and
the pressure-volume
product of a system:
H = E + PV
For a process carried out
at constant pressure,
so
The evolved H2 pushes
back the atmosphere;
work is done at
constant pressure.
Mg + 2 HCl --> MgCl2 + H2
q = DE + PDV
q = DH
Most reactions occur at constant
pressure, so for most reactions, the heat
evolved equals the enthalpy change.
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Chapter Six
38

Enthalpy is an extensive property.
It depends on how much of the substance is present.
 This makes sense because heat transfers into and
out of bonds, the more moles you have, the more
bonds you have, therefore more energy can be
released to or taken from the environment.

• Since E, P, and V are all state
functions, enthalpy (H) must be a
state function also.
• Enthalpy changes have unique
values. DH = q
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Two logs on a fire give
off twice as much heat
as does one log.
Chapter Six
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

Molar enthalpy is the amount of heat
lost or gained when one mole of a
compound is formed from its
constituent elements.
∆Hᵒf = Standard enthalpies of formation
(molar enthalpy at 25 degrees C and 1
atm) have been calculated for nearly all
compounds and are in the back of your
book. (Appendix C)
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Chapter Six
40

H2(g) + ½O2(g) → H2O(l)
ΔH = –285.8 kJ mol–1

2C(s) + H2(g) → C2H2(g)
ΔH = 227.0 kJ mol–1
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Chapter Six
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


Values of DH are measured experimentally.
Negative values indicate exothermic reactions.
Positive values indicate endothermic reactions.
A decrease in enthalpy
during the reaction; DH
is negative.
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An increase in enthalpy
during the reaction; DH
is positive.
Chapter Six
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

DH changes sign when a process is reversed.
Therefore, a cyclic process has the value DH = 0.
Same magnitude; different signs.
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Chapter Six
43
Example 6.3
Given the equation
(a) H2(g) + I2(s) --> 2 HI(g)
DH = +52.96 kJ
calculate DH for the reaction
(b) HI(g) --> ½ H2(g) + ½ I2(s).
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Chapter Six
44
The complete combustion of liquid octane, C8H18, to
produce gaseous carbon dioxide and liquid water at 25
°C and at a constant pressure gives off 47.9 kJ of heat
per gram of octane. Write a chemical equation to
represent this information.
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Chapter Six
45

What quantity of heat is associated with the
complete combustion of 1.00 kg of sucrose
(C12H22O11)?
KNOWN:
342.3 g/mol sucrose
-5.65 x103 kJ/mole
3 kJ
1
mol
-5.65
x
10
x
1.00 kg x 1000 g x
342.3 g
1 mol
1 kg
=
-1.65 x 104 kJ
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Chapter Six
46



For problem-solving, heat evolved (exothermic reaction)
can be thought of as a product. Heat absorbed
(endothermic reaction) can be thought of as a reactant.
We can generate conversion factors involving DH.
For example, the reaction:
H2(g) + Cl2(g) --> 2 HCl(g)
DH = –184.6 kJ
can be used to write:
–184.6 kJ
————
1 mol H2
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–184.6 kJ
————
1 mol Cl2
–184.6 kJ
————
2 mol HCl
Chapter Six
Example 6.5
What is the enthalpy change associated with the
formation of 5.67 mol HCl(g) in this reaction?
H2(g) + Cl2(g) --> 2 HCl(g)
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47
DH = –184.6 kJ
Chapter Six
48
6.5 A – What is the enthalpy change when 12.8
g H2 (g) reacts with excess Cl2 (g) to form HCl
(g)
H2 (g) + Cl2 (g) → 2 HCl (g)
∆H = -184.6 kJ

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Chapter Six
49



Turn in:
 Get out Notes
Our Plan:
 Clicker Review
 Notes
 Calorimetry POGIL (as practice)
 Find Someone Who
Homework (Write in Planner):
 Problems from the textbook (due 10/30)
 You can complete up through #60
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Chapter Six
50
 Heat
capacity – quantity of heat
required to change the
temperature of a system by one
degree
 Molar heat capacity – quantity
of heat required to change the
temperature of a mole of a
substance one degree
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Chapter Six
51


C = q/∆T
Example 6.6 – Calculate the heat capacity of an
aluminum block that must absorb 629 J of heat
from its surroundings in order for its
temperature to rise from 22ᵒC to 145ᵒC.
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Chapter Six
52



The science of measuring changes of
heat.
A calorimeter is a device used to
make this measurement.
General principle of calorimetery:
Heat Lost = - Heat Gained
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Chapter Six
53
For a reaction carried out in a calorimeter, the heat
evolved by a reaction is absorbed by the calorimeter and
its contents.
qrxn = – qcalorimeter
qcalorimeter = mass x specific heat x DT
By measuring the temperature change that occurs in a
calorimeter, and using the specific heat and mass of the
contents, the heat evolved (or absorbed) by a reaction can be
determined and the enthalpy change calculated.
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Chapter Six
54
Heat evolved in a reaction is
absorbed by the calorimeter and its
contents.
 The specific heat (c) is the heat
capacity of one gram of a pure
substance (or homogeneous
mixture).
q = c m DT

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Chapter Six
55



High specific heat means that the
substance is resistant to changes in
temperature.
Many metals have low specific heat
making them easy to heat up or cool
down.
Water has a high specific heat. If it did
not, life on Earth would hardly be
possible.
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Chapter Six
56






q = mass x specific heat x DT
If DT is positive (temperature increases), q is
positive and heat is gained by the system.
If DT is negative (temperature decreases), q is
negative and heat is lost by the system.
The calorie, while not an SI unit, is still used
to some extent.
Water has a specific heat of 1 cal/(g oC).
4.184 J = 1 cal
One food calorie (Cal or kcal) is actually equal
to 1000 cal.
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Chapter Six
57
Many metals have
low specific heats.
The specific heat of
water is higher than
that of almost any
other substance.
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Chapter Six
58
q = m x c x DT
 The mass and specific heat of an
object are often combined into a
single quantity (heat capacity –
C)
 Thus, q = C x DT

Prentice Hall © 2005
Chapter Six
59

Constant Pressure Calorimeter



“coffee cup” calorimeter
Open to atmosphere
Appropriate for solution chemistry

Constant Volume Calorimeter



Prentice Hall © 2005
Bomb calorimeter
Sealed and isolated
Appropriate for gas phase chemistry
Chapter Six
60
Example 6.7
How much heat, in joules and in kilojoules, does
it take to raise the temperature of 225 g of water
from 25.0 to 100.0 °C?
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Chapter Six
Example 6.8
61
What will be the final temperature if a 5.00-g silver
ring at 37.0 °C gives off 25.0 J of heat to its
surroundings? Use the specific heat of silver listed
in Table 6.1.
Prentice Hall © 2005
Chapter Six
62
Example 6.9
A 15.5-g sample of a metal alloy is heated to 98.9 °C and
then dropped into 25.0 g of water in a calorimeter. The
temperature of the water rises from 22.5 to 25.7 °C.
Calculate the specific heat of the alloy.
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Chapter Six
63
Example 6.11
A 50.0-mL sample of 0.250 M HCl at 19.50 °C is added to
50.0 mL of 0.250 M NaOH, also at 19.50 °C, in a
calorimeter. After mixing, the solution temperature rises to
21.21 °C. Calculate the heat of this reaction.
Prentice Hall © 2005
Chapter Six
64

A copper bottomed saucepan has a heat
capacity of 227.3 J/ºC. Calculate how much
heat the saucepan gains when its temperature
changes from 24.7 ºC to 110.2 ºC.
Prentice Hall © 2005
Chapter Six
65

6.8A – A 454 gram block of lead is at an initial
temperature of 22.5 degrees C. What will be the
temperature of the lead after it absorbs 4.22 kJ
of heat from its surroundings?
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Chapter Six
66

I have finished notes up through
homework problem #60. DO AT
LEAST THOSE PROBLEMS BY
MONDAY!
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Chapter Six
67



Turn in:
 Get out notes & textbook
Our Plan:
 Review Race
 Finish the Unit Notes
Homework (Write in Planner):
 Problems from the textbook (due 10/30)
Prentice Hall © 2005
Chapter Six
68



Some reactions cannot be carried out “as
written.”
Consider the reaction:
C(graphite) + ½ O2(g) --> CO(g).
If we burned 1 mol C in ½ mol O2, both CO and
CO2 would probably form. Some C might be
left over. However …
Prentice Hall © 2005
Chapter Six
69



… enthalpy change is a state function.
The enthalpy change of a reaction is the same
whether the reaction is carried out in one step
or through a number of steps.
Hess’s Law: If an equation can be expressed as
the sum of two or more other equations, the
enthalpy change for the desired equation is the
sum of the enthalpy changes of the other
equations.
Prentice Hall © 2005
Chapter Six
Example 6.14
70
Calculate the enthalpy change for reaction (a) given the data
in equations (b), (c), and (d).
(a) 2 C(graphite) + 2 H2(g) → C2H4(g)
DH = ?
(b) C(graphite) + O2(g) → CO2(g)
DH = –393.5 kJ
(c) C2H4(g) + 3 O2 → 2 CO2(g) + 2 H2O(l)
DH = –1410.9 kJ
(d) H2(g) + ½ O2 → H2O(l)
DH = –285.8 kJ
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Chapter Six
71
Example: Find the DHrxn of 2 NO (g) + O2 (g) 
N2O4 (g), given the DHrxn for the following
reactions:
N2O4 (g)  2 NO2 (g)
2 NO (g) + O2 (g)  2 NO2 (g)
Prentice Hall © 2005
DHrxn = 57.93 kJ/mol
DHrxn = -113.14 kJ/mol
Chapter Six
72
Calculate the enthalpy of reaction for the
following reaction: 2 Al(s) + 3 Cl2(g)  2 AlCl3(s),
given the reactions below.
2 Al(s) + 6 HCl(aq)  2 AlCl3(aq) + 3 H2(g) DH = -1049 kJ/mol
HCl(g)  HCl(aq)
DH = -74.8 kJ/mol
H2(g) + Cl2(g)  2 HCl(g)
DH = -185 kJ/mol
AlCl3(s)  AlCl3(aq)
DH = -323 kJ/mol
Prentice Hall © 2005
Chapter Six
73
H2 (g) + ½ O2 (g)  H2O (l)
2 Na (s) + ½ O2 (g)  Na2O (s)
Na (s) + ½ O2 (g) + ½ H2 (g)  NaOH (s)
DH = -286 kJ
DH = -414 kJ
DH = -425 kJ
Based on the information above, what is the standard
enthalpy change for the following reaction?
Na2O (s) + H2O  2 NaOH (s)
(A)
(B)
(C)
(D)
(E)
-1,125 kJ
-978 kJ
-722 kJ
-150 kJ
+275 kJ
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Chapter Six
74
• It would be convenient to be able to use the simple relationship
ΔH = Hproducts – Hreactants
to determine enthalpy changes.
• Although we don’t know absolute values of enthalpy, we don’t
need them; we can use a relative scale.
• We define the standard state of a substance as the state of the pure
substance at 1 atm pressure and the temperature of interest (usually
25 °C).
• The standard enthalpy change (ΔH°) for a reaction is the enthalpy
change in which reactants and products are in their standard states.
• The standard enthalpy of formation (ΔHf°) for a reaction is the
enthalpy change that occurs when 1 mol of a substance is formed
from its component elements in their standard states.
Prentice Hall © 2005
Chapter Six
75
When we say “The standard enthalpy of formation of
CH3OH(l) is –238.7 kJ”, we are saying that the reaction:
C(graphite) + 2 H2(g) + ½ O2(g) → CH3OH(l)
has a value of ΔH of –238.7 kJ.
We can treat ΔHf° values as though they were absolute
enthalpies, to determine enthalpy changes for reactions.
Question: What is ΔHf° for an element in its standard
state [such as O2(g)]? Hint: since the reactants are
the same as the products …
Prentice Hall © 2005
Chapter Six
76



DH°rxn = Snp x DHf°(products) – Snr x DHf°(reactants)
The symbol S signifies the summation of several terms.
The symbol n signifies the stoichiometric coefficient used
in front of a chemical symbol or formula.
In other words …
1. Add all of the values for DHf° of the products.
2. Add all of the values for DHf° of the reactants.
3. Subtract #2 from #1
(This is usually much easier than using Hess’s Law!)
Prentice Hall © 2005
Chapter Six
Example 6.15
77
Synthesis gas is a mixture of carbon monoxide and hydrogen that is
used to synthesize a variety of organic compounds. One reaction for
producing synthesis gas is
3 CH4(g) + 2 H2O(l) + CO2(g) --> 4 CO(g) + 8 H2(g) ΔH° = ?
Use standard enthalpies of formation from Table 6.2 to calculate the
standard enthalpy change for this reaction.
Prentice Hall © 2005
Chapter Six
78
Try it Out - Example 6.16
The combustion of isopropyl alcohol, common rubbing alcohol, is
represented by the equation
2 (CH3)2CHOH(l) + 9 O2(g) --> 6 CO2(g) + 8 H2O(l) ΔH° = –4011 kJ
Use this equation and data from Table 6.2 to establish the standard
enthalpy of formation for isopropyl alcohol.
Prentice Hall © 2005
Chapter Six
79
C2H4 (g) + 3 O2 (g)  2 CO2 (g) + 2 H2O (g)
For the reaction of ethylene represented above, DH is
-1,323 kJ. What is the value of DH if combustion
produced liquid water H2O (l), rather than water
vapor H2O (g)? (DH for the phase change H2O (g) 
H2O (l) is -44 kJ mol-1.)
(A)
(B)
(C)
(D)
(E)
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-1,235 kJ
-1,278 kJ
-1,323kJ
-1,367 kJ
-1,411 kJ
Chapter Six
80

We can apply
thermochemical
concepts to reactions
in ionic solution by
arbitrarily assigning
an enthalpy of
formation of zero to
H+(aq).
Prentice Hall © 2005
Chapter Six
81
Example 6.18
H+(aq) + OH–(aq) --> H2O(l)
ΔH° = –55.8 kJ
Use the net ionic equation just given, together with ΔHf°
= 0 for H+(aq), to obtain ΔHf° for OH–(aq).
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Chapter Six
82
Turn in:
 Homework
 Our Plan:
 Homework Questions
 Specific Heat Review Problem
 Hess’s Law Activity
 Begin Hand Warmer Lab
 Homework (Write in Planner):
 Work on Test Review
©Have
Pre-Lab and Procedure ready
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2005

Chapter Six
83

http://www.youtube.com/watch?v=JuWtBRrDQk
Prentice Hall © 2005
Chapter Six
84



Turn in:
 Nothing
Our Plan:
 Hand Warmer Lab
 Test Review
Homework (Write in Planner):
 TEST MONDAY
 Breakfast Club Monday morning

Lab Report Due Monday
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Chapter Six
85
 Turn
in:
 Nothing
 Our Plan:
 Test
 Homework (Write in Planner):
 Read Ch. 7 & 8
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Chapter Six