Transcript Module 2 - chem534

Module 3
Heat and Energy in Chemical Reactions.
Introduction: Thermochemistry
Chemical reactions often involve changes in
temperature, and in the heat energy that causes
temperature to change. This branch of chemistry that
studies heat changes is called THERMOCHEMISTRY.
Some Quick Definitions
 Energy: The ability to (a) do work or (b)
to supply heat.
 Kinetic Energy: energy associated with
movement of objects or particles.
 Potential Energy: energy that has not
produced heat or motion yet. Potential
energy may be associated with
gravitation and with chemical bonds.
Background Information
(notes optional)
 Most energy on earth comes from the
sun, but a small amount comes from the
radioactive decay of certain atoms, and
an even smaller amount from the tidal
effects of the moon.
 Most of the energy that we experience is
thermal energy, that is, one of several
forms of energy related to heat
 (Background Info: Notes optional)
 Thermal energy can be transferred in
many ways:
 Conduction: heat transferred between touching
objects. This form of heat transfer is the most
important in chemistry.
 Convection: heat transferred to air molecules,
which then may expand and may rise due to their
reduced density. This is important in meteorology
 Radiant heat: A hot object, like a heat lamp, may
directly radiate heat in the form of infrared
radiation. This is important in physics
Heat and Temperature
 Heat is a form of thermal energy that is
transferred when two systems with
different temperatures come into contact
with each other.
 Heat “flows” from one object to another.
 Temperature is a measure of the
agitation of atoms or molecules in a
system.
 Temperature is directly related to the average
kinetic energy of the molecules.
Heat vs. Work
 When energy transfer is in the form of
“work” the transfer is orderly. A whole
object or group of molecules move in an
orderly fashion in one direction.
 When energy transfer is in the form of
“heat” the transfer is more random. The
molecules of the heated object move
more quickly in random directions.
 P. 128-130 Questions p. 130
Law of Conservation of Energy
Energy can be neither created nor
destroyed in any chemical process.
 However:


Energy can be transferred from one object to
another.


Energy can be transformed from one type of
energy to another.


Eg. Collisions transfer energy between objects.
Eg. Potential energy can be transformed into
kinetic energy, and vice versa.
Nuclear reactions can change matter into
energy or vice versa
Systems and Their Surroundings



System: the part of the universe on
which we are focusing our attention
(usually a beaker, flask or test tube)
Surroundings: Everything else in the
universe.
A system may be:
Open: matter & energy easily exchanged between
the system and its surroundings
 Closed: matter cannot be exchanged, but some
energy can escape or enter the system
 Isolated: neither matter nor energy can be
exchanged between system and surroundings.

Calorimeters




A calorimeter is an instrument used to measure
the amount of heat released or absorbed during
a change.
A calorimeter is a closed system containing a
fixed amount of water that can absorb heat
The ΔH can be calculated by measuring the
water temperature before and after the reaction
Calorimeters can be as complicated as the
“bomb” calorimeter illustrated on p. 132 or as
simple as a Styrofoam cup
The Calorimeter Formula
Δ
Q=mc T
Where
Q=Heat Energy
m = mass of the water
c = specific heat capacity of the water
ΔT
= the change in temperature
As You Remember!


Specific Heat capacity (c) is the amount
of energy required to raise the
temperature of one gram of a substance
by one degree Celsius.
For water, c is 4.19 J / g°C
Simple examples: copy and complete*
Mass of
Water
Initial
Temp
Final
Temp
Change
ΔT
250 g
22°C
47 °C
+25°C
26187.5 J
200 g
20 °C
18 °C
-2°C
-1676 J
30 °C
+12 °C
7542 J
-6704 J
150 g
100 g
95.5g
18°C
25 °C
9°C
-16 °C
22 °C
52°C
+30 °C
Heat
Q
12000 J
*Assume measurements are precise, c = 4.19J/g° for
water and that no heat is lost in the calorimeter.
Molar heat of Reaction



The molar heat of a reaction is the
amount of heat released / absorbed by
the reaction of one mole of the material
ΔH
(Molar) = Q / n
A.K.A. molar heat of combustion, molar
heat of solution, etc.
Example: copy and try this

The following data was recorded for the partial
combustion of wax (C25H52) in a calorimeter. Use this
data to calculate the molar heat of combustion of wax.





Initial mass of wax
Final mass of wax
Volume of water in calorimeter
Initial temperature of water
Final temperature of water
22.35 g
12.08 g
352.5 mL
12.6 °C
43.5 °C
Solution: part 1





heat captured
Mass of water:
m=352.5 g (water)
Change of temp: ΔT= 30.9 °C
Specific heat:
c= 4.19 J/g °C
Q=mc ΔT
= 352.5×4.19 × 30.9
Q
= 45638.5275 Joules
But… that is not for a whole mole.
Solution: part 2 –






number of moles
Mass (wax)
m=10.27 g
Molar mass (wax) M=352.61 g/mol
Moles of wax:
n= m ÷M
n=
10.27 ÷ 352.61
n=
0.0291 mol
The molar heat of combustion is the heat
produced per mole of wax
Solution: part 3 - molar heat


Q ÷n = 45638.5275 ÷ 0.0291
Solution= 1568334.278
Round to 3 significant digits
ANSWER:


The molar heat of combustion of wax is
about 1570000 Joules/mole


or 1.57x106 J/mol
or 1570 kJ/mol
Heat Transfer


When two bodies (or substances) of different
temperatures come into contact, heat is
transferred from the warmer body (or
substance) to the cooler body, until equilibrium
is reached, when the temperature is the same.
At equilibrium, the heat lost (Q-loss) by the
substance that was warmer, must equal the
heat gained (Q-gain) by the substance which
was cooler.
Heat Transfer
Heat moves from Hot to Cold
-Q1
Q
= Q2
loss
=
Hot Body Equilibrium
Q
gain
Cool Body
Until they are the same temperature
If
Δ
Q=mc T
and
-Q1 = Q2
Then
Δ
-m1c1 T1
=
Δ
m2c2 T2
Example: Copy and try this!

A calorimeter contains 230.0g of water at
25.0°C. A 200.0g sample of copper at 47.0
°C is placed inside the calorimeter.
Calculate the final temperature of this
system. (the specific heat capacity of
copper is 0.390 J/g °C)
Solution









Q=mcΔT
For water: Qgain=230 × 4.19 × ΔTwater
For copper: Qloss=200 × 0.39 × ΔTCu
Since -Q1 = Q2 this means that:
-(200 × 0.39 × ΔTCu) = 230x4.19x ΔTwater
Let x represent the final temperature
ΔT
So
water = (x–25 ) °C
ΔT
And
= (x– 47) °C
Cu
We can substitute:
m(copper) c
ΔT
(copper)
(copper)
m
(water)
c
(water)
ΔT
(water)

-200 g × 0.39 j/g°C × (x-47)°C = 230 g × 4.19 j/g°C× (x-25) °C

-78 g•j/g°C (x - 47) °C = 963.7 g•j/g°C(x-25)

-78x + 3666 = 963.7x

-963.7x - 78x = -3666 - 24092.5

1041.7

26.647

Answer:
- 24092.5
x = 27758.5
(round to 3 sig. dig.)
26.6°C
This is true when liquids are mixed
too!


When liquids are mixed, the heat lost by
the warm liquid must be the same as the
heat gained by the cooler one.
In this case, not only do the substances
transfer heat to each other, but they will
mix as well.
Example: copy and try

A mixture is made of 100.0 mL of
water at 90.0°C and 100.0 mL of
water at 25.0°C. What will the final
temperature be?
Data: M(hot)=100g
m(cold)=100g
C(hot)=4.19 j/g
c(cold)=4.19 j/g
Ti(hot)=90
Ti(cold)=25
Tf=x
The same
-ΔT(hot) =Tf –Ti(hot)
Tf=x
ΔT
cold
=Tf –Ti(cold)
Solution: part 1

Qgain=mcold x ccold x ΔTcold
-Qloss = mhot x chot x ΔThot
For cold water, Qgain=100 × 4.19 × ΔTcold
For hot water
Qloss=100 × 4.19 × ΔThot
Since –Q(lost) = Q(gained) this means that
-100 × 4.19 × ΔThot = 100 × 4.19 × ΔTcold

ΔTcold = (x-25)

-ΔThot= (x-90)





Q=mcΔT so…
let x= final temp
Note: since ΔTcold = ΔThot there is a short cut
you can use, we’ll look at that later.
Solution: part 2

-100 × 4.19 (x-90)

–(x – 90) = x -25

x-25=-x+90


x + x = 90 + 25
2 x = 115
x = 57.5

Answer: the final temperature is 57.5 °C


=100
×
4.19 (x-25)
Of course, in this case you could have done the problem
more simply, since you knew you had equal amounts of the
same substance (ie. same mass & specific heat capacity) ,
you could have just averaged the temperature!
Homework Questions:



What will the final temperature be if you mix 100.0 g. of
water at 20.0°C and 40.0 g. of water at 80.0 °C?
What is the molar heat of combustion of methane (CH4)if
burning 1.00 gram of it in a calorimeter raises the
temperature of 800.0g of water from 27.0 °C to 42.0 °C?
What will the final temperature be if 50.0g of copper
(c=0.39 J/g °) at 80.0 °C is dropped into 200.0 mL of
20.0°C water?
Review Concepts

There are three types of Change


Physical change: does not change the
composition for example a change of state
Chemical change: Changes the composition,
examples: effervescence, decomposition, change of
colour, precipitation, combustion.

Nuclear change: Changes in the atom nucleii,
examples: formation of isotopes, radioactive decay,
nuclear fission, nuclear fusion
Remember:
Kinetic Energy of particles:

Particles (ie. Molecules) can have 3 types
of motion, giving them kinetic energy



Vibrational kinetic energy
Rotational kinetic energy
Translational kinetic energy
State & Kinetic Energy

Solids exhibit only vibrational energy


Liquids exhibit mostly rotational energy


Virtually none of their kinetic energy comes from
translation or rotation.
A small portion of their kinetic energy can come
from vibration or translation.
Gases exhibit mostly translational energy

a tiny portion of their kinetic energy can come from
vibration or rotation.
Melting a Pure Solid and
Boiling a Pure Liquid



As you slowly add thermal energy to a pure
solid, its temperature will rise as the molecular
vibration increases.
There will come a point, however, where heating
the solid does not increase the temperature.
Instead, the increased energy is used to change
the type of motion, making the molecules
tumble or rotate. This is called the melting
point.
The solid becomes a liquid, and the temperature
again rises with increased thermal energy, until
the liquid begins to boil or evaporate.
Temperature (°C or K ) 
Phase Diagram
Boiling
Melting
Rotation
Translation
Vibration
&
Rotation
Thermal Energy Added (joules or kilojoules) 
Info from a phase diagram
(diagram shown in endothermic direction, increasing absorption of heat)
100
80
60
Temperature °C 
40
20
Specific Heat capacity
(c) of the solid is the
inversePoint
of the(56°C)
slope, ie:
Boiling
Run over Rise
30÷60 = 0.5 J/g°
Melting Point (18°C)
0
-20
-40
10
20
30
40
Heat of Fusion (ΔHfus=10 J/g)
Heat of vaporization (ΔH vap=15 J/g)
50
Energy Added (Joules / gram) 
60
70
80
90
Info from a phase diagram
(diagram shown in Exothermic direction, increasing release of heat)
100
80
Condensation Point (56°C)
Temperature °C
60
40
20
Freezing Point(18°C)
0
-20
-40
Heat of condensation (ΔHcond=15 J/g)
10
20
30
40
Heat
50 of solidification
60
70
80(ΔH=10
90 J/g)
Energy Released (Joules / gram) 
Heat Diagrams for Mixtures
Melting Range
Boiling Range
So far we have only shown heat diagrams for pure
substances. For mixtures the “plateaus” are less clear, and
the melting and boiling points less clearly defined.
Third fractional distillation point
Second fractional distillation point
First fractional distillation point
Sometimes the difference in
boiling points for different
substances in a liquid mixture
can be used to separate them
by fractional distillation.
Assignments


Textbook Reading pp. 125-136
Textbook Questions pp. 145, #1-19
Module 3, Lesson 2
Endothermic and Exothermic
Reactions
Exothermic Reactions




Exo=outer, Therm=heat
Exothermic reactions are chemical
reactions which release heat energy.
You can recognize exothermic reactions
because the products are hotter than the
reactants.
Example: burning wood is an exothermic
reaction.
Endothermic Reactions




Endo = inner; Therm = heat
Endothermic reactions absorb heat from
their surroundings.
You can recognize endothermic reactions
because the products become colder than
the reactants were.
Examples: instant “ice packs”,
More About Energy



Potential Energy (EP) is stored energy. In
chemistry it is usually stored as chemical bonds.
Kinetic Energy (EK) is energy of motion. In
chemistry it usually revealed by temperature,
caused by moving molecules.
Heat Energy (Q) is energy transferred from one
body to another due to a difference in
temperature between the bodies
Types of Heat Change in Chemistry



Enthalpy (H): The amount of total energy in a
substance, most of it is in the form of potential
or “hidden” heat energy. (Detailed discussion will follow)
Heat of Reaction (ΔH): amount of energy
absorbed or released during a reaction. It
represents an amount that the enthalpy has
changed.
In addition to chemical reactions, energy can be
absorbed or released in other changes. The
symbol ΔH can also be used for these.
Variations on ΔH
(enthalpy can change in many types of reaction)






Heat of Formation (ΔHF): the amount of energy
absorbed/released when compound is made from its
elements.
Heat of Dissolution (ΔHd): The amount of energy
absorbed/released when a solute dissolves.
Heat of Neutralization (ΔHn): The amount of energy
absorbed/released when a solute dissolves.
Heat of Combustion (ΔHcombustion): amount of energy
released when a material burns.
Heat of Fusion (melting) or Solidification (freezing)
(ΔHf)=-(ΔHs) : amount of energy absorbed when a solid
melts or released when a liquid freezes
Heat of Vaporization or Condensation (ΔHv)=-(ΔHcondensation)
: amount of energy absorbed when a liquid evaporates
or released when a gas condenses
Recapping Enthalpy




Enthalpy is the total heat content of a
substance, including the energy that was
stored in the bonds of the substance during
its formation.
Enthalpy is mostly potential energy.
Enthalpy cannot be measured directly, but
it can be calculated by the amount of
energy released/absorbed during reactions.
The Heat of Reaction (ΔH) or enthalpy
change is the difference between the Heats
of Formation (ΔHF) of the products and the
reactants
Next
Potential Energy (Enthalpy)
Enthalpy Diagram: Exothermic
Reactants
ΔH is Negative
(ΔH <0)
Products
Progress of Reaction (Time)
Potential Energy (Enthalpy)
Enthalpy Diagram: Endothermic
Products
ΔH is Positive
(ΔH >0)
Reactants
Progress of Reaction (Time)
Some books show Enthalpy Graphs like
this:
Exothermic
ΔH
ΔH
Endothermic
It means the same as the ones
before!
Another Type of Enthalpy Graph
This is the type shows what happens during a reaction.
We will examine this type in more detail in the next module.
Enthalpy
Activation Energy
ΔH
Reaction progress (time)
Trick Questions
• Some physical processes are tricky to classify
as exothermic or endothermic. The following
are guides:
EXOTHERMIC
Condensing (gasliquid)
Freezing (liquidsolid)
Solidifying (gassolid)
ΔH is negative
ENDOTHERMIC
Evaporating (liquidgas)
Sublimating (solidgas)
Boiling (liquidgas)
Melting ΔH positive
Dissolving can be EXOTHERMIC or
ENDOTHERMIC depending on the solute.
Example 1:

Which of the followings Statements about
enthalpy is true?




A) enthalpy always increases during a chemical
change
B) enthalpy always decreases during a chemical
change
C) enthalpy remains unchanged when chemical
bonds form
D) enthalpy always decreases during a
exothermic reaction.
Example 2:

Which of the following processes releases more
energy than it absorbs:
 A) Water cools down when NH4Cl is dissolved in
it.
 B) Solid margarine melts on a hot plate
 C) Gasoline is burned in an automobile engine
 D) River water evaporates
Example 3:

The molar heat of formation of a compound
is:




A) The heat required to atomize one mole of the
compound.
B) The heat of reaction for making one mole of
that compound from its elements.
C) The heat of reaction for formation of one mole
of the compound from the reaction of two other
compounds.
D) The heat liberated during bond formation for
one mole of the compound.
Example 4:
ΔH
What type of reaction is this?
Exothermic
References and practice



Textbook Reading: p. 653-671
Study guide reading: 3-5 to 3-6
Extra Practice: Student study guide
pages 3-6 to 3-10 # 1-14

Correct these yourself using the answer
key.
Module 3, Lesson 3
Calorimeters and Heat Transfer
Extra Practice

Student Study Guide


pp. 3-14 to 3-16 #1-12
Do these on your own and correct
yourself. Keep your answers in your
folder.
Lab Activity
Heat Exchange
Lab Activity: Heat Exchange
• Purpose:
– (a) to investigate heat exchange when mixing liquids
of different temperatures and verify the formula:
Qloss = Qgain, or mcΔT(hot)=mcΔT(cold).
– (b) to determine the specific heat capacity of an
unknown metal, and to use that information to identify
the metal.
• Materials:
– Styrofoam cups, Graduated cylinder, Thermometer,
hot water (from a water bath), cold water (from tap).
Never leave a thermometer unattended!
Procedure (part 1-4)
• Place a measured amount of cold water in a
Styrofoam cup (measure to 1/10 mL).
• Place a measured amount of hot water into
water a Styrofoam cup.
• Take the temperature of both cups (to 1/10 °C).
• Pour the cold water into the hot water.
• Stir and take the temperature of the mixture.
• Repeat this with four different volume
combinations (see suggested volumes)
Rewrite these instructions in procedure format
Procedure (part 5)
• Put about 50 ml (measure an exact amount) of
cold water into a Styrofoam cup and measure its
temperature.
• Take a piece of metal out of boiling water and
add it to the cold water (boiling water should be
100°C, but use a thermometer to be sure).
• Use the beaker tag to find the mass of metal.
• Find the temperature of the water after the metal
has been added.
Sample Beaker Tag
found at the corner stations for step 5
Metal #
Mass of metal
Metal #5
Mass = 23.58g ± 0.03g
H2O use ≈ 60mL
Note: use both pieces!
Suggested amount of water to use
Other instructions
Suggested volumes
(do not try to get these exactly. Come close but don’t pour
back and forth. Measure to 3 significant digits)
Run
1
Mass*water Temp
Mass*
Temp
Final
(cold)
(hot)
(hot)
Temp
(cold)
≈100 g
≈ 8 °C
≈100 g
≈ 55 °C
? °C
2
≈ 50 g
°C
≈100 g
°C
? °C
3
≈100 g
≈ 8 °C
≈50 g
≈ 55 °C
? °C
4
≈100 g
°C
≈75 g
°C
If time allows
5
50-70 g
≈ 8 °C
metal g
≈100 °C
? °C
(see beaker tag
for suggestion)
(see beaker tag
for mass of metal)
*for water, mass in grams = volume in mL
Specific heat capacities of common
metals.
Aluminum
0.91
Magnesium
1.05
Antimony
0.21**
Mercury
0.14*
Copper
0.39
Nickel
0.59
Gold
0.13*
Platinum
0.13*
Iron
0.46
Tin
0.21
Lead
0.13*
Zinc
0.39
•*Gold, lead, platinum and mercury have similar specific
heats, but are easy to tell apart by other properties
** Antimony and Tin are similar, but we don’t use antimony.
Module 3, Lesson 4
Thermo-chemical Equations
Thermo-chemical Equations
 An
equation that includes the heat change
is a thermo-chemical equation.
 Thermo-chemical
equations are just
ordinary chemical equations, but they also
show how much heat (or energy) was
given off or absorbed during a reaction.
 Unless stated otherwise, the amount of
energy is in kilojoules per mole.
2 Types of Thermo-chemical Equation
 Thermo-chemical
equations can be
expressed using two methods:
 First method:

By adding the heat to the equation directly, or
 Second

method:
By writing the ΔH value at the end of the
equation.
*Preferred Method
Method 1: Examples




Add the heat factor to either side:
For exothermic reactions:
+
 NaOH(s)  Na
+
OH
(aq)
(aq) + 44.51 kJ/mol
For endothermic reactions:
+ + NO  NH4NO3(s) + 25.7 kJ/mol  NH4
3
This method is simple, but it is not widely used.

Remember, in exothermic reactions the heat is listed
after the arrow, in endothermic reactions the heat is
listed before the arrow.
Method 2: Examples
 Add
the ΔH value to the end of the
equation. In this case…


Negative means exothermic
Positive means endothermic
 Exothermic


NaOH(s)  Na+ (aq) + OH-
(aq)
ΔH
=-44.51 kJ/mol
ΔH
=+ 25.7 kJ/mol
Endothermic

NH4NO3(s)  NH4+ + NO3-
Examples:

Are the following equations endothermic or
exothermic?
Exothermic




C2H5OH + 3O2  2CO2 + 3H2O ΔH=-1235 kJ
2NaHCO3+129kJ Na2CO3+H2O+CO2 Endothermic
Exothermic
C+O2CO2 + 349 kJ
NH4NO3 + H2O NH4++NO3-+H2O ΔH=25.7kJ
Endothermic
Convert the following “type 1”
thermochemical equations into “type 2”
thermochemical equations.





NaOH(s)  Na+(aq) + OH-(aq)+ 44.5kJ/mol
C2H5OH + 3O2  2CO2 + 3H2O + 1234kJ/mol
C(s) + S2(s) + 89kJ/mol  CS2(l)
C2H5Cl(l) + 26.4kJ/mol  C2H5Cl(g)
H2O(g)  H2O(l) + 40.7kJ/mol
NaOH(s)  Na+(aq) + OH-(aq)
ΔH=–44.5kJ/mol
C2H5OH + 3O2  2CO2 + 3H2O
ΔH=–1234kJ/mol
C + S2(s)  CS2(l)
C2H5Cl(l)  C2H5Cl(g)
H2O(g)  H2O(l)
ΔH=+89kJ/mol
ΔH=+26.4kJ/mol
ΔH=–40.7kJ/mol
Module 3, Lesson 5
Hess’ Law
Hess’s Law

If two or more thermo-chemical equations
are added together to give a final
equation, then the enthalpy changes can
be added to give the enthalpy change for
the final equation.
Example of Adding Equations

Find the equation for:


N2 + 2O2  2 NO2
ΔH=?
By combining the equations:


N2+O2  2 NO
2 NO+O2  2 NO2
ΔH=+180.6 kJ/mol
ΔH=-144.4 kJ/mol
Adding Equations
Reactants
Products
Enthalpy (ΔH)
N2 +O2  2 NO
+180.6kJ
2 NO+O2  2 NO2
- 144.4 kJ
N2+2NO+2O2  2 NO + 2NO2
N2 + 2O2  2 NO2
+ 36.2 kJ
+ 36.2 kJ

In the example, we cancelled the like
substances at the end of the problem.


You may also cancel them earlier, as long as
you are careful to only cancel a substance in the
reactant column with an identical substance in
the product column.
You may multiply an entire equation, including
the ΔH value by a simple number to make the
coefficients match.


You can only cancel if coefficents they match exactly!
You may switch the substances in the reactant
column with the substances in the product
column, but you then have to change the sign of
the ΔH value.
Example 2:
(see study guide page 47 question 2)

Find the heat of combustion of methanol
(CH3OH) from the following reactions:




C(s) + O2(g)  CO2(g) + 394 kJ
H2(g) + ½ O2(g)  H2O(l) + 242 kJ
C(s) + 2H2(g) + ½ O2(g)  CH3OH(l)+239 kJ
Note: for simplicity, the “phase markers” ie.(s) (l)
(g) (aq) will be left out of the solution this time. Be
aware that they are important if a material’s
changes state before or after it reacts.

The Reactions we know:
C(s) + O2(g)  CO2(g) + 394 kJ
 H2(g) + ½ O2(g)  H2O(l) + 242 kJ
 C(s) + 2H2(g) + ½ O2(g)  CH3OH(l)+239 kJ


The reaction we are trying to get will contain
the following:
CH3OH(l)+
O2(g)  CO2 +
H2 O
We can balance it to find the coefficients, but I’m not
going to do that yet, since all I need to know for now is
what side of the arrow each substance is on.


Solution
Reverse
Step
Step
Step
the
3:
1:
CH
double
copy
3rd3OH
equation
the
is
everything
on
reactions
wrong
and
in
side
change
equation
the
2 sign
Step
4:2:
simplify
by
cancellation
Reactants
Products
C + O2  CO2
x2
flip
2H
+ O
O22 
 2H
H22+½
H22OO
C+2H2+½ O2 
CH3OH
ΔH
-394 kJ
-484
kJ
-242 kJ
+-239 kJ
CH3OH+ 2O2CO2+2H2O+ ½ O2 -639 kJ
CH3OH+ 1½ O2 CO2+2H2O
-639 kJ
Assignments
• Read Textbook pages 672 to 673
• Practice problems: p. 674 #9-10
– (correct yourself from answer key in back, keep
answers in your assignment folder)
• Assigned problems: p. 681 # 31-32
– (put the answers in your assignment folder, I will
correct later)
• Student Study Guide: Read pp. 3-17 to 3-18
• Study Guide Extra practice: 3-18 to 3-20 #1-6
– (correct yourself but keep answers in folder)
Module 3, Lesson 6
Heat of Formation
Heat of Formation Tables
Standard Heat of Formation


The standard heat of formation of a
compound is the change in enthalpy that
accompanies the formation of one mole of
the compound from its elements, with all
substances in their standard states at
25°C
The symbol for heat of formation is ΔHf
In other words:


It’s how much heat was given off (or
absorbed) when you combined the
elements into a compound
For example:

When you burn hydrogen and oxygen
together you make water (steam actually)
and a lot of heat is given off (usually a
flaming explosion!)
The Example of Water


Hydrogen + Oxygen  water
Balanced Equation:


But we want it per mole, so ÷2


2 H2 + O2  2 H2O, ΔH = -570 kJ (for 2 mol)
H2 + ½ O2  H2O, ΔH = -285 kJ/mol
So the standard heat of formation for
water from hydrogen & oxygen is -285
kJ/mol
Some background info:

When calculating the standard heat of
formation we arbitrarily set the enthalpy
of the most common form of the elements
to zero. This gives us a negative heat of
formation if creating the compound was
an exothermic reaction, or positive if
making the compound was endothermic
Heat of Formation
Exothermic
0
compounds
elements
ΔH
0
elements
ΔH
compounds
Endothermic
Heat of Formation Tables


Table 8.4 on page 418 of your text book is
an example of a Heat of Formation Table.
It gives the heat of formation of several
common organic compounds.
Using Heat of Formation


The heat of reaction is equal to the
difference between the heat of formation
of the reactants and products.
ΔHReaction = ΔHF(Products) – ΔHF(Reactants)
Example: Find the heat of reaction for the
formation of hydrogen peroxide:
2 H2O(l) + O2(g)  2 H2O2(l)

Step 1. look up heats of formation in tables:






2 (-285.8) + 1(0) = -571.6 kJ
2 (-187.8) = -375.6 kJ
Step 4. Subtract: ΔHf


(table 8.4 p. 418)
(element)
(table 8.4 p. 418)
(just kJ, no longer kJ/mol)
Step 3. Find the total of the product:


= -285.8 kJ/mol
= 0.00 kJ/mol
= -187.8 kJ/mol
Step 2. Find the total ΔHf of the reactants:


H2O(l)
O2(g)
H2O2
(products)
– ΔHf
(reactants)
-375.6 – (-571.6) = -375.6+571.6 = +196 kJ
Answer: The heat of reaction is: ΔH =+196 kJ
(for 2 moles of peroxide).
The molar heat” of reaction is
ΔH°=
98 kJ/mol)
Assignments
• Read pp. 674 -676 in textbook
• Practice: question 11 and 12 on p. 677
(check your own answers in answer key)
• Assigned: #33 on page 681
Heat of Reaction/Hess’ Law
Laboratory Activity
• In this lab you will be calculating the heat
energy given off by dissolving and
neutralizing sodium hydroxide
• You should also be able to use these
calculations to demonstrate the accuracy
of Hess’ Law.
NaOH(s)
NaOH(aq) + HCl(aq)
 NaOH(aq)
 NaCl(aq) +H2O(l)
NaOH(s) + HCl(aq)
 NaCl(aq) +H2O(l)
Heat of Reaction/Hess’ Law
Laboratory Activity

Pointers for this lab:




Be careful with the thermometers! Keep them
in the case when not using them!!!!
DON’T put contaminated NaOH back into the
vial. Throw excess down the sink.
Measure close to, but not above the required
measurements, for example don’t try to get
exacty 2.00 g, just very close.
For reaction#3 use the colour coded
graduated cylinders from the drawers
Yellow cylinder for NaOH solution
 Red cylinder for HCl solution

Heat of Reaction/Hess’ Law
Laboratory Activity


These slides will show some of the
calculations required to complete the Lab.
Note: The computer program will calculate
the class results, and will use adjustments
to improve the accuracy of individual
results, but you should do your own
calculations as well.
Data for Reaction #1
NaOH (s)  NaOH (aq)
Symbol






Temperature of Water
Volume of the Water
Final temperature
Mass of Cup
Mass of Cup+NaOH
Mass of NaOH
=_____°C Ti
=_____ mL mw
=_____ °C Tf
=__ (used to calculate mNaOH)
=__ (used to calculate mNaOH)
=_____g
mNaOH
Calculations for Reaction #1





Find
Find
Find
Find
Find




Δ T:
MNaOH
Q (calorimeter)
n (#mols NaOH)
ΔH
ΔT=
Tf – Ti
(use periodic table)
Q=mwatercwater ΔT
n=mNaOH ÷MNaOH
ΔH= Q ÷ n
Where ΔT=temperature change,
MNaOH =molar mass, n= moles of NaOH
Q = overall heat of reaction,
ΔH =molar heat of reaction
Data for Reaction #2
NaOH(s) + HCl(aq) NaCl(aq)+ H2O






Temperature of HCl
Volume of the HCl
Final temperature
Mass of Cup
Mass of Cup+NaOH
Mass of NaOH
=_____°C Ti
=_____ mL mS
=_____ °C Tf
(used to calculate mass of NaOH)
(used to calculate mass of NaOH)
=_____g
mNaOH
Calculations for Reaction #2





Find
Find
Find
Find
Find

Δ T:
MNaOH
Q (calorimeter)
n (#mols NaOH)
ΔH
ΔT=
Tf – Ti
(use periodic table)
Q=mSc ΔT
n=mNaOH ÷MNaOH
ΔH= Q ÷ n
Assume that HCl solution has same specific heat
capacity and density as water. This is not exactly
true, but it is close enough for your own calculation.
The computer analysis will be more accurate.
Data for Reaction #3
NaOH(aq) + HCl(aq) NaCl(aq)+ H2O





Temperature of HCl
Volume of the HCl
Temperature of NaOH
Volume NaOH
Final Temperature
=_____°C T1
=_____ mL m1
=_____ °C T2
=_____ mL  m2
=_____ °C  Tf
Calculations for Reaction#3






Find combined solution mass: ms = m1 + m2
Average Initial Temp:
Ti = T1+ T2 )
(note: we are using a simple average here)
2
ΔT= T – T
Find ΔT:
m2=v2
f
i
Find Q (calorimeter)
Q=mSc ΔT
Find n (#mols NaOH)
(v2 / 1000)L * 1.00 mol/L



Divide volume of NaOH Solution by 1000 to get Litres
Multiply by the concentration (mol/L)… in this case just 1
Find ΔH

ΔH=
Q÷n
Notes: To more accurately compare with other sources,
you may wish to change your H values to kilojoules/mol
instead of joules/mol (divide them by 1000)
Submit Your Data Sheet



One partner from each pair should submit
a data collection sheet to the teacher at
the end of the lab class.
The other partner should keep a copy of
the data collection sheet in a safe place
(ie. Their lab folder)
The submitted sheet will be used for
computer analysis of your results.
Lab Write Up


Wait until the class results are back from
computer analysis before finishing your
good copy.
Do and record all your calculations in your
rough copy, so that you will be ready to
finish your good copy and compare your
calculations to the computer analysis
Individual & Class Results
Our Results
(calculated)
Our Results
(computer)
Our Results
(corrected)
Class Results
(computer)
Class Results
(corrected)
Expected
(from references)
Reaction
Reaction
Reaction
Reaction
#1
#3
#1+ #3
#2
44
56
100
100
Results in kJ/mol, with decimal values rounded.
Discussion Points





Were your results as good as the class average?
Were the adjusted or unadjusted results closer
to the expected results?
What were the percentage differences between
the calculated and observed results?
What were the three reactions? Why can we say
reaction #2 is the sum of reaction#1 and
reaction#3?
How can this experiment be used to justify Hess’
Law?
Assignment


All questions on page 680-681 that were
not previously assigned,
Module 3 (chapter 27) Test first class next
week:
Answers to page 680
• 13. Kinetic energy is active energy of motion,
such as heat energy. Potential energy is hidden
or stored energy, such as enthalpy
• 14. Not all of gasoline’s energy is used to give
the car motion. Some of its energy is converted
into heat that is “wasted” in the radiator. Other
energy is used to overcome mechanical friction
or air resistance or to brake the car.
• 15. Two components of internal energy change
are work (w) and heat flow (q)
• 16. The processes are classified as:
–
–
–
–
A) exothermic (the battery gives off energy)
B) endothermic (plants absorb sun energy)
C) exothermic (burning releases heat)
D) endothermic (the potato absorbs heat)
• 17. In this case q is positive and w is negative,
so…
–
-24.6kJ + 14.6kJ = 10.0 kJ
• 18. A calorimeter measures the amount of heat
absorbed or released in a physical or chemical
process.
• 19. A Styrofoam cup has several problems
as a calorimeter:
• heat escapes through the top,
• vapour or gases formed during reactions escape
• . It is difficult to calculate how much heat passes
into the Styrofoam.
• 20. There are two ways to do this:
• 1st method: Q=mcΔT where m=100g. (mass of
both liquids), c=4.19 (assume liquids are mostly
water) and ΔT= 26.3-24.4 = 1.9°C, so Q=100 x
4.19 x 1.9 = 791.6J
• 2nd method uses the data about the calorimeter in
the problem335 J/°C x 1.9°C= 636.5J
• The two methods give different results. In this
case the 2nd method is better, since the first
method used several assumptions.
• 21. q=ΔH if the pressure is constant
• 22. The work must be negligible (≈0)
• 23. if the reaction of 1 mol Fe2O3 produces
26.3 kJ then the reaction of 3.4 mol would
produce 3.4 x 26.3 = 89.42 kJ
• 24.
– The formation of 1 mol of Al2O3 would have
an enthalpy change of 3352 kJ / 2 or 1676 kJ
ie ΔH=+1676 kJ
– The reaction would be endothermic.
• 25. (a) heat of vaporization, (b) heat of
solution, (c ) heat of fusion, (d) heat of
condensation, (e) heat of solidification
• 26. (a) endo, (b) endo, (c ) endo, (d) exo,
(e) exo
• 27. The heat that flows into the system is
used to melt the ice (change vibrations
into rotation) instead of raising the
temperature
#1: hot solid and cold water
1.
Qloss
Data
= Qgain
mcΔT (hot)
= mcΔT(cold)
mlet(hot)
=2.05gthe final temperature…
t represent
2.05g • 0.519 j/g˚ (74.21˚C - t )
c(hot)
=0.519 J/g˚C
1.064 (74.21 – t)
Ti78.95-1.064t
(hot)=74.21˚C
3048
so..
3048 / 110.2
mt=(cold)
=26.05g
t=27.66˚C
c(cold)=4.19 J/g ˚C
= 26.05 g• 4.19 j/g ( t - 27.2˚C)
= 109.1 ( t – 27.2)
= 109.1t - 2969
= 110.2t
27.66˚C is the final temperature.
ΔT (water) = 0.46˚C
Ti(cold)=27.2 ˚C
ΔT (solid) = 46.55˚C
b) yes, this makes sense because: i) the final temperature is
between the hot and cold temperature, ii)the specific heat
capacity of water is much greater than that of the solid
a)
2. Calorimeter Problem (Q = mcΔT )
Formula:
Data
Q= mc
mc=1.56 kJ/K
Q=1.56 • 3.2
ΔT= 3.2 K
Q=4.992 kJ
m(quinine) =0.1964g
Heat per gram=Q/m
M(quinine) = 108 g/mol
= 4.992 kJ / 0.1964 g
ΔT
= 25.4 kJ/g
n(quinine) = m / M
=0.1964/108
= 0.001818 mol
(heat of combustion per gram)
molar heat formula:
ΔH=Q / n
ΔH= 4.992 kJ / 0.001818 mol
ΔH= 2745 kJ/mol
(heat of combustion per mole)
#3. (Hot metal in cold ethanol)
Data
Halfnium=hot
Ethanol = cold
mHot=15.6g
cHot = 0.146
Ti
Hot=160˚C
Qloss
mcΔThot
= Qgain
=mcΔTcold
15.6 •0.146 (160-t) = 125 • 2.45 (t-20)
2.27(160-t)
= 306.25 (t-20)
364.4 – 2.27 t
=306.3 t – 6125
6489.4
= 308.6 t
So…
t = 21.03
mcold=125g
Ccold=2.45
Ti
cold
= 20˚C
The final temperature is 21.03˚C
#4
Data:
Q=1430 J
ΔT=1.93˚C
Heat capacity =mc
Formula: Q=mcΔT
1430 = mc (1.93)
1430 / 1.93 =mc
So:
mc =740.9 J/K
The heat capacity is 741 J/K
#5
Data:
Formula: Q=mcΔT
mc=741 j/K
Q=741 x 3.46
ΔT=3.46˚C
Q=2563.86
Answer:
The heat evolved was 2560 J
Or 2.56 kJ
#6
Data:
Formula (a): n=m/M
m(hydrazine)=25.4g
n=25.4
M(hydrazine)=
n=0.7937
2(14)+4(1)
= 32 g/mol
For part (a):
n
(hydrazine)
b)
=0.794mol
g
/ 32
g/mol
mol
(a) N2H4 + 2Cl2 4HCl + N2 ΔH=-420.0kJ
If one mol of hydrazine produces -420.0 kJ
then 0.794 mol will produce:
0.794 mol • -420.0 kJ/mol
= - 333.5 kJ
n mol
1.45mol
N2H4 + 2Cl2  4HCl + N2 H=-420.0kJ
1
For part (b): n
2
(hydrazine)
4
1
=(1.45)(1) / 4 = 0.3625 mol
So: 0.3625mol • -420 kJ/mol = -152 kJ
#7 (Use Hess’s Law)

Looking for: 5CO2 + Si3N43SiO + 2N2O + 5 CO
CO2 + SiO  SiO2 + CO
Rx3: 3SiO2 + 3CO
/  3CO
/ 2 + 3SiO
58CO2 + Si3N4  3SiO2 +2N2O+8CO
5

5CO2 + Si3N4  3SiO +2N2O +5CO
H=-520.9
H=+1562.7
H=+461.05
H=+2023.75
Answer: the enthalpy change for the
reaction is… ΔH=+2024 kJ
#8
ΔH
f
C3H8(g)=-103.9 kj/mol
CO2(g)=-393.5 kj/mol
H2O(g) = -241 kJ/mol
O2 = 0 kJ/mol
Lesson 1 Summary
• Thermochemistry
– studies heat changes in chemical reactions
• Molecules have three types of motion
– Vibrational (mostly in solids)
– Rotational (mostly in liquids)
– Translational (mostly in gases)
• A phase diagram shows the transition of
states
– Use a phase diagram to find melting point,
boiling point, heat of fusion, heat of
evaporation, specific heat capacity.
Lesson 2 Summary
• Exothermic reactions release heat
• Endothermic reactions absorb heat
• “Hidden heat” is called enthalpy (H)
– It is a form of potential energy
– It cannot be measured directly, but can be calculated
by experiment.
• In an exothermic reaction, enthalpy decreases
• In and endothermic reaction, enthalpy increases.
• The enthalpy change or “heat of reaction” is the
amount of heat absorbed or released (ΔH)
– Is negative for exothermic reactions, positive for
endothermic ones.
Lesson 3 summary
• A Calorimeter is a device used to
find the amount of heat absorbed or
lost in a reaction.
• The calorimeter formula is Q=mcΔT
• Molar heat of reaction is the heat
released or absorbed by one mole of
a reactant: H=Q/n