Thermochemistry
Chapter 16
Thermochemistry

The study of energy
involved during
chemical reactions.
Energy sources:
~ chemical
~ nuclear
~ solar
~ geothermal
~ wind/water




Heat: the energy of motion of
molecules
Kinetic molecular theory:
substances are composed of
particles that are continually
moving and colliding with other
particles.
Kinetic energy: energy of motion
Potential energy: stored energy
Temperature:


transfer of heat to
a substance
because of faster
molecular
movement (as
long as there is no
phase change)
Measured with a
thermometer


A temperature change is explained as a
change in kinetic energy
Temperature depends on the quantity of
heat (q) flowing out or in of the
substance.
Heat (q)
q=mc ∆t




q=heat
m=mass
∆t=change in temperature (tf-ti)
c=specific heat capacity (J/(g oC)
Specific heat capacity is the quantity of heat
required to raise the temperature of a
unit mass of a substance by one degree
Celsius.
Table of Specific Heats
Law of conservation of energy



∆E universe = O
The total energy of the universe is
constant, it is not created or
destroyed, however it can be
transferred from one substance to
another.
∆E universe = ∆E system + ∆E surroundings
First Law of thermodynamics




Any change in energy of a system is
equivalent by an opposite change in
energy of the surroundings.
∆E system = - ∆E surroundings
According to this law, any energy
released or absorbed by a system
will have a transfer of heat, q.
So, q system = -q surroundings
Sample Problem


15 g of ice was added to 60g of
water. The Ti of water was 26.5 oC,
the final temperature of the mixture
was 9.7 oC. How much heat was
lost by the water?
q=mc ∆t
q=(60g) (4.18 J/g oC) (9.7-26.5 oC)
q= - 4213.44 J
Calculate the amount of heat needed
to increase the temperature of 250g
of water from 20oC to 56oC.
q = m x c x (Tf - Ti)
q = 250g x 4.18 J/g oC x (56oC 20oC)
q = 37 620 J = 38 kJ


Calculate the specific heat capacity of
copper given that 204.75 J of energy
raises the temperature of 15g of copper
from 25oC to 60oC.
q = m x c x (Tf - Ti)
q = 204.75 J
m = 15g
Ti = 25 oC
Tf = 60 oC
204.75 J = 15 g x c x (60oC - 25oC)
204.75 J= 15 g x c x 35oC
204.75 J= 525 g oC x c
c = 204.75 J ÷ 525 g oC = 0.39 J/ g oC
Watch flash about heat flow
http://www.mhhe.com/physsci/chemi
stry/animations/chang_7e_esp/enm1
s3_4.swf
Assignment

p.636 #7-10
Enthalpy (H)




Total kinetic and
potential energy
(internal energy) of a
system under constant
pressure.
System: area where
reaction takes place
Its particles average
motion define its
properties.
Surroundings:
outside of the system
Open-system


both matter and
energy can freely
cross from the
system to the
surroundings and
back.
Ex: an open test
tube
Closed-System


energy can cross
the boundary, but
matter cannot.
Ex: a sealed test
tube
Isolated-System


neither matter nor
energy can cross
between the
system and the
surroundings.
Ex: The universe

there are no
surroundings to
exchange matter or
energy with (as far
as we know!)

The internal energy of a reactant or
product cannot be measured, but their
change in enthalpy (heat of reaction)
can.
∆ H = Hproducts – Hreactants
A change in enthalpy occurs during phase
changes, chemical reactions and nuclear
reactions.
∆H
system
=q
surroundings
If energy is released out of the
system then it is considered to be
an exothermic enthalpy change
(exit).
If energy is absorbed into the system
then it is considered to be an
endothermic enthalpy change
(enter).
Endothermic Reactions
Endothermic Reactions
Method 2:
2 HgO
(s)
 2 Hg
(l)
+ O2(g) ΔH= +181.67 kJ
Method 3:
2 HgO
(s)
+ 181.67 kJ  2 Hg
(l)
+ O2(g)
Exothermic Reactions
Exothermic Reactions
Method 2:
4Al(s)+ 3O2(g) 2 Al2O3(g) ΔH=-1675.7 kJ
Method 3:
4 Al(s) + 3O2(g) 2 Al2O3(g) +1675.7 kJ
Calorimeter



Instrument used to measure
amount of energy involved in a
chemical reaction.
It is equivalent to an isolated or
closed system. (nothing may enter
or exit the system)
The energy change is not measured
within the system, but the energy
transferred to its surroundings.
A basic calorimeter


Two styrofoam cups nestled within
one another (insulation), then filled
with a specific quantity of water.
A chemical reaction or phase
change takes place inside and a
thermometer is placed within to
measure any change in temperature
that occurs to the system.
Assumptions
1.
2.
3.
It is an isolated or closed system
and there is no heat transfer
between the calorimeter and its
surroundings.
The amount of heat absorbed or
released by the calorimeter itself
is too small to influence
calculations.
Any dilute solutions involved in the
reaction are treated as if they are
water.
Bomb calorimeter




Cannot use our basic design for
combustion reactions.
Used in research for ∆H for fuels,
oils, food, explosives…
Larger and more sophisticated
The reaction container is strong
enough to with stand an explosion,
hence the name “bomb”



Have fixed components, like volume of
water, thermometer…
Heavily insulated or vacuum insulated so
no convection or conduction can occur
affecting the enthalpy of the system.
Use the equation: q= -CcalΔt
A Bomb Calorimeter
Problems


Pg 638 # 5-7
Handout Questions 1-10
Standard Molar Enthalpy of Formation


Quantity of energy released (-) or
absorbed (+) when one mole of a
compound is formed directly from
its elements at standard
temperature and pressure.
We use a table to find them. (refer
Table E13, p.848 for values)


Unit for ΔHf: kJ/mol
Watch your states!
Calculating enthalpy changes



Amount of a substance reacting
matters, so can use q= nΔH.
Remember n=amount of moles.
If you are given a mass (g) and
molar mass (g/mol), then you can
solve for n by dividing mass by
molar mass. (review from chem 11
stoichiometry section)
Sample Problem:


Show the formation reaction of
methanol.
If I had 10.5 g of CH3OH(l), how
much energy would be released?
C(s) + 2H2(g) + 1/2O2(g)  CH3OH(l)
ΔHf = -239 kJ/mol
mm= (12.01g) +(4 x 1.01g) + (16.00g) = 32.05 g/mol
q = nΔH = (10.5 g / 32.05 g/mol) (-239 kJ/mol) = -78.3 KJ
On Your Own:

I have 125.6 g of NaOH(s) formed
at standard temperature and
pressure, how much energy will be
released?
Answer: - 1336.4 kJ
Standard Molar Enthalpy of
Combustion



Energy changes involved with
combustion reactions of one mole of
a substance.
Remember that these reactions are
only measured once cooled to 25oC
Combustion is a reaction with
oxygen as a reactant
Practice Problem:

Write the combustion reaction for
methane (CH4)


ΔHcomb for CH4 is -965.1 kJ/mol
How much energy is released when
12.8 g of CH4 is combusted?
CH4(g) + 2O2 (g)  2H2O(l) + CO2 (g)
ΔHcomb = -965.1 kJ/mol
q=nΔH = (12.8 g/ 16.05 g/mol) (-965.1 kJ/mol) = - 769.7 kJ
Combustion reaction:

Always form water and carbon dioxide.
Ex: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

Remember your alkanes: CnH2n+2


Meth: C=1, Eth: C=2, Prop: C=3, But:
C=4, Pent: C=5, Hex: C=6…
Practice


Pg 643 # 15-17
Pg 645 # 19,21,23
Enthalpy and changes of state
A change in matter without
any change in chemical
composition of the
system
*** Always involves energy
changes, but not
temperature changes.
There is constant
temperature during a
phase change. Why?
Because…
No change in kinetic energy, however we
know that energy is entering the
system because the bonds that are
holding the molecules together are
being broken or altered. This increases
Ep of the molecules
Just think!

We know that
there are energy
changes during
phase changes,
because when we
sweat our body is
trying to cool itself
down using
evaporation!
Phase changes: (surroundings)
(s) to (l) fusion
endo
T
(l) to (g) vaporization
endo
T
(g) to (l) condensation exo
T
(l) to (s) solidifying
exo
T
(g) to (s) sublimation
exo
T
(s) to (g) sublimation
endo
T
Latent Heat of Phase Change
ΔH melt = - ΔH freez
Molar Heat of Fusion or melting
The energy that must be absorbed in
order to convert one mole of solid to
liquid at its melting point.
Molar Heat of Freezing or solidifying
The energy that must be removed in
order to convert one mole of liquid to
solid at its freezing point.
ΔH vap = - ΔH cond
Molar Heat of Vaporization
The energy that must be absorbed in
order to convert one mole of liquid to
gas at its boiling point.
Molar Heat of Condensation
The energy that must be removed in
order to convert one mole of gas to
liquid at its condensation point.
Molar Enthalpy for changes of state


You can find tables of these values
listed
Unit: kJ/mol
Latent Heat – Sample Problem
Problem: The molar heat of fusion of water is
6.009 kJ/mol. How much energy is needed to convert
60g of ice at 0C to liquid water at 0C?
60 g H 2O 1 mol H 2O 6.009 kJ
 20.00 kiloJoules
18.02 g H 2O 1 mol
Mass
of ice
Molar
Mass of
water
Heat
of
fusion
Chemical reactions:

If a reaction is exothermic, energy
is released to the surroundings
(temp increases) and chemical
potential energy of the system
decreases

∆ H is a negative number

If a reaction is endothermic, energy
is absorbed from the surroundings
(temp. decreases) and the chemical
potential energy of the system
increases

Δ H is a positive number

It requires 10-100 times more
energy for a chemical reaction to
take place than a phase change
because a chemical reaction
requires stronger ionic and covalent
bonds to be broken, but a phase
change only requires intermolecular
bonds to be broken.
Manipulating the Enthalpy Change Term

The value of Δ H given as kJ/ mol refers to kJ
per 1 mole of reactant or product as written in
the equation.
N2(g) + 3H2(g) 2NH3(g)

Δ H = - 92.4 kJ
92.4 kJ of energy is released for every 1 mole of
N2(g), 3 moles of H2(g), 2 moles of NH3(g)
produced.




How much energy is released if only 1
mole of ammonia (NH3) gas is produced?
92.4 kJ of energy is released in the
production of 2 moles of ammonia gas
Half as much energy will be released if
only half the amount of ammonia gas is
produced
½ x 92.4 = 46.2 kJ of energy will be
released in the production of 1 mole of
ammonia.

How much energy is released if 10
moles of nitrogen (N2) gas and 30
moles of hydrogen (H2) gas is used
in the reaction?
92.4 kJ of energy is released for
every 1 mole of N2(g)
 10 times as much energy will be
released if 10 times the amount of
reactants are used
 10 x 92.4 = 924kJ of energy will
be released

Problems
Pg 648-649
# 26,28-29
 Handout #16-18

Heat Curve: Temperature vs. Time
Total energy change of a system

When a given amount of energy is
available to heat a substance,
sometimes there is more then one
change of state involved as the
substance goes from an initial to a
final temperature.

1.
This requires energy calculations
for changing the temperature in a
certain state and then calculations
for a change of state.
To calculate the energy involved in
a change of temperature of a
substance without a change of
state, we use q=mc ∆t
2.
To calculate the energy involved in
a change of state we would use:
q = nΔHrxn
Sample Problem #1

Calculate the total energy needed or
released during the cooling of 1000
g of liquid iron (c= 0.82 J/g oC) to
solid iron (c=0.52 J/g oC), where
the initial temperature is 1700 oC to
80 oC.


The enthalpy of solidification is -15
KJ/mol.
The melting point of iron is 1535 oC.
Heating Curve
1700
1535
q1
q2
Temp
q3
(oC)
80
Time (min)

ΔE = cooling liquid + phase change from l to s
+ solid cooling
= q1 + q2 + q3
= q (cooling (l)) + n Δ Hsolid+q(cooling (s))
= mc Δt + n Δ Hsolid + mc Δt
= [(1000g)(0.82 J/g C)(1535-1700)] +
[(1000g/55.85 g/mol)(-15 KJ/mol) ] + [
(1000g)( 0.52 J/g C) (80-1535)]
=-135300 J + (-268.6 KJ) + (-756600 J)
= -1160 KJ
Sample Problem #2

How much energy is required to raise the
temperature of 50.0 g of water (ice) at
-15.0°C to a temperature of 175°C?

c H2O(l) = 4.18 J/g C
c H2O(g)and (s) = 2.01 J/g C
∆Hmelt = 6.02 kJ/mol
∆Hvap = 40.7 kJ/mol
Melting point = 0 oC

Boiling point = 100 oC





First, start by drawing your own heat
curve.

the energy required to change the
temperature of the ice from -15.0°C to
0°C, q1; the energy involved in the
change of state as the ice melts at 0°C,
q2; the energy involved to heat the liquid
water from 0°C to 100°C, q3; the energy
involved in the change of state as the
water boils, q4; and finally the energy
required to heat water vapour at 100°C to
175°C, q5.
175
oC
Heat Curve
100 oC
Temp
(oC)
0 oC
- 15 oC
Time (min)






q1 = mcΔt = 50.0 g x (2.06 J/g•°C) x 15.0°C
= 1540 J or 1.54 kJ
q2 = nΔHfus = 50.0 g / (18.01 g/mol) x 6.03
kJ/mol = 16.7 kJ
q3 = mcΔt = 50.0 g x (4.18 J/g•°C) x
100.0°C = 20 900 J or 20.9 kJ
q4 = nΔHvap = 50.0 g/(18.01 g/mol) x 40.8
kJ/mol = 113 kJ
q5 = mcΔt = 50.0 g x (2.02 J/g•°C) x 75.0°C
= 7 580 J or 7.58 kJ
qtotal = q1 + q2 + q3 + q4 + q5 = 1.54 kJ +
16.7 kJ + 20.9 kJ + 113 kJ + 7.58 kJ = 160
kJ
Practice:


Pg 655 # 30-34
Handout #1 (finish it, #19&20)