Thermochemistry Notes
I. Thermochemistry deals with the
changes in energy that accompany a
chemical reaction. Energy is measured in
a quantity called enthalpy, represented as
H. The change in energy that
accompanies a chemical reaction is
represented as H. Page 519
a. The energy absorbed or released as heat
in a chemical or physical change is
measured in a calorimeter. In one kind of
calorimeter, known quantities of reactants
are sealed in a reaction chamber, which is
immersed in a known quantity of water in
an insulated vessel. Therefore, the energy
given off (or absorbed) during the reaction
is equal to the energy absorbed (or given
off) by the known quantity of water. The
amount of energy is determined from the
temperature change of the known mass of
surrounding water.
Calorimeter
Using the change in temperature, T,
determined from calorimeter one can
use the following equation to
determine the quantity of energy
gained or lost during the reaction or
for a physical change:
J)
q = (cp)(m)(T)
T in kelvin
q represents the energy lost or gained (in
m is the mass of the sample (in g)
cp is the specific heat of a substance at a
given temperature
pg513
• Practice Problem: 1. How much heat energy
is needed to raise the temperature of a 33.0
gram sample of aluminum from 24.0C to
100.C?
• 2. Determine the specific heat of a material if
a 35 g sample absorbed 96 J as it was heated
from 293K to 313K.
• 3. During a chemical reaction carried out in a
calorimeter the temperature of water within the
calorimeter raised from 24.0C to 125C. If
250. grams of water were present calculate
the amount of heat energy the water gained.
b. In thermochemical equations the
quantity of energy released or absorbed as
heat during a reaction is written and is
represented by H.
Example:
2H2(g) + O2(g)  2H2O(l) H= - 571.6 kJ/mol
c. H can be used to determine if the
reaction is exothermic or endothermic. If
the H value of an equation is negative
that represents an exothermic reaction.
(Meaning energy is released, therefore
the energy of the products would be less.)
If the H value of an equation is positive
that represents an endothermic
reaction. Example:
2H2(g) + O2(g)  2H2O(g)
H= - 483.6 kJ/mol
H reactants = 1450.8 kJ/mol
H products = 967.2 kJ/mol
Type of Reaction: Exothermic
2H2O(g)  2H2(g) + O2(g)
H= + 483.6 kJ/mol
H reactants = 967.2 kJ/mol
H products = 1450.8 kJ/mol
Type of Reaction: Endothermic
d. Hess’s law provides a method for
calculating the H of a reaction from
tabulated data. This law states that if two
or more chemical equations are added,
the H of the individual equations may
also be added to find the H of the final
equation. As an example of how this law
operates, look at the three reactions
below.
(1) 2H2(g) + O2(g)  2H2O(l)H = 571.6 kJ/mol
(2)2H2O2(l)  2H2(g) + 2O2(g)H = +375.6 kJ/mol
(3)2H2O2(l)  2H2O(l) + O2(g)H = ? kJ/mol
When adding equations 1 and 2, the 2 mol of
H2(g) will cancel each other out, while only 1
mol of O2(g) will cancel.
(3)2H2O2(l)  2H2O(l) + O2(g)H = ? kJ/mol
Warm-up
• What type of
energy and energy
transfer do you see
in this picture
•Thank you Ms.
Bouwman (from
McNeil High
School) for this
wonderful power
point!
Basic Thermochemistry
Courtesy of lab-initio.com
Energy is the capacity to do work
•
Thermal energy is the energy associated with the random motion of
atoms and molecules
•
Chemical energy is the energy stored within the bonds of chemical
substances
•
Nuclear energy is the energy stored within the collection of neutrons
and protons in the atom
•
Electrical energy is the energy associated with the flow of electrons
•
Potential energy is the energy available by virtue of an object’s position
6.1
Energy
Energy is the capacity to do work, and can take many
forms
 Potential energy is stored energy or the energy
of position
 Kinetic energy is the energy of motion
 Thermal energy (heat) is an outward
manifestation of movement at the atomic level
Heat (Enthalpy) Change, ΔH
Definition: The amount of heat energy released or
absorbed during a process.
Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between two bodies that are
at different temperatures.
Temperature is a measure of the thermal energy.
Temperature = Thermal Energy
900C
400C
greater thermal energy
6.2
Heat
Cup gets cooler while hand
gets warmer
The flow of thermal
energy from one object
to another.
Heat always flows
from warmer to
cooler objects.
Ice gets warmer
while hand gets
cooler
3 Types of Heat Transfer
• Radiation- the transfer of energy by
electromagnetic waves.
• Convection – Transfer of energy by currents
• Conduction – Transfer of energy by touching
objects
Exothermic process is any process that gives off heat –
transfers thermal energy from the system to the
surroundings. 2H (g) + O (g)
2H O (l) + energy
2
2
H2O (g)
2
H2O (l) + energy
Endothermic process is any process in which heat has to be
supplied to the system from the surroundings.
energy + 2HgO (s)
energy + H2O (s)
2Hg (l) + O2 (g)
H2O (l)
6.2
Exothermic Processes
Processes in which energy is released as it proceeds, and
surroundings become warmer
Reactants  Products + energy
Endothermic Processes
Processes in which energy is absorbed as it proceeds,
and surroundings become colder
Reactants + energy  Products
Water phase changes
constant during a phase change.
Temperature remains __________
Phase Change Diagram
Processes occur by addition of energy 
 Processes occur by removal of energy
Thermochemical Calculations
Units for Measuring Heat
The Joule is the SI system unit for measuring
heat:
1 kg  m
1 Joule 1 newton  meter 
2
s
2
The calorie is the heat required to raise the
temperature of 1 gram of water by 1 Celsius
degree
1calorie  4.18 Joules
Specific Heat
The amount of heat
required to raise the
temperature of one
gram of substance by
one degree Celsius.
5
4
6
5
7
4
3
8
3
2
9
2
1
11
6
7
8
9
1
10
Specific Heat (cp, sometimes s, but usually c)
Things heat up or cool down at different rates.
Land heats up and cools down faster than water, and
aren’t we lucky for that!?
Specific heat is the amount of heat required
to raise the temperature of 1 kg (but in Chem
we use g) of a material by one degree (C or K,
they’re the same size).
Cp water = 4184 J / kg C (“holds” its heat)
Cp sand = 664 J / kg C (less E to change)
This is why land heats up quickly during the day and cools quickly at
night and why water takes longer.
Calculations Involving Specific Heat
Q  m  T  c p
OR
Q
cp 
m  T
cp = Specific Heat
Q = Heat lost or gained
T = Temperature change
m = Mass
Specific Heat
The amount of heat required to raise the temperature of one gram
of substance by one degree Celsius.
Substance
Specific Heat (J/g·K)
Water (liquid)
4.18
Ethanol (liquid)
2.44
Water (solid)
2.06
Water (vapor)
1.87
Aluminum (solid)
0.897
Carbon (graphite,solid)
0.709
Iron (solid)
0.449
Copper (solid)
0.385
Mercury (liquid)
0.140
Lead (solid)
0.129
Gold (solid)
0.129
The specific heat (s) of a substance is the amount of heat (q) required to raise the
temperature of one gram of the substance by one degree Celsius.
The heat capacity (C) of a substance is the amount of heat
(q) required to raise the temperature of a given quantity
(m) of the substance by one degree Celsius.
C = ms
Heat (q) absorbed or released:
q = mst
q = Ct
t = tfinal - tinitial
6.4
Specific Heat Capacity
If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat
energy are lost by the Al?
heat gain/lose = q = (c)(mass)(∆T)
where ∆T = Tfinal - Tinitial
q = (0.897 J/g•K)(25.0 g)(37 - 310)K
q = - 6120 J
Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al.
How much heat is given off when an 869 g iron bar cools from 940C to 50C?
s of Fe = 0.444 J/g • 0C
t = tfinal – tinitial = 50C – 940C = -890C
q = mst
= 869 g x 0.444 J/g • 0C x –890C
= -34,000 J
6.4
CALORIMETRY
Calorimetry
The amount of heat absorbed or released during a
physical or chemical change can be measured,
usually by the change in temperature of a known
quantity of water in a calorimeter.
Constant-Volume Calorimetry
qsys = qwater + qbomb + qrxn
qsys = 0
qrxn = - (qwater + qbomb)
qwater = mst
qbomb = Cbombt
Reaction at Constant V
No heat enters or leaves!
6.4
Constant-Pressure Calorimetry
qsys = qwater + qcal + qrxn
qsys = 0
qrxn = - (qwater + qcal)
qwater = mst
qcal = Ccalt
Reaction at Constant P
No heat enters or leaves!
6.4
BOOM! Combustible material ignited at
constant volume! This heats up the
“bomb”, which heats up the water
surrounding it…
First, some heat from reaction warms
the water, which we know the mass
of and “c” for…
qwater = (c)(water mass)(∆T)
THEN, some heat from reaction warms
“bomb,” which has a known specific
heat for the entire apparatus (typically),
so we don’t need the mass…
qbomb = (heat capacity, J/K)(∆T)
Total heat evolved = qtotal = qwater + qbomb
Practice
A sample of iron metal is added to 75.00 grams of
water originally at 35.0°C in a calorimeter. The final
temperature of the metal and water in the calorimeter
is measured to be 95.0°C.
a) Describe the transfer of heat energy that occurs in
the calorimeter.
(b) Assuming no heat is lost to the outside, how many
joules of heat energy are transferred?
ENTHALPY (H)
Phase Change Diagram
D
B
A
E
C
Changing Phase
• Heat of Fusion - heat change for freezing and
melting
• Heat of Vaporization – heat change for
condensation or evaporation
For Water: Heat fusion = 340 J/g
Heat vaporization = 2,300 J/g
Heat = (mass)(heat of fusion or vaporization)
How many joules of heat are necessary to melt
500g of ice at its freezing point?
= 500g * 340J/g
= 170,000 J or 170KJ
Enthalpy (H) is used to quantify the heat flow into or out of
a system in a process that occurs at constant pressure.
H = H (products) – H (reactants)
H = heat given off or absorbed during a reaction
Hproducts < Hreactants
H < 0
Hproducts > Hreactants
H > 0
6.3
∆Hfo, standard molar enthalpy of formation
∆Hfo = Enthalpy change when 1 mol of compound is formed
from the corresponding elements under standard
conditions
H2(g) + 1/2 O2(g) --> H2O(g)
∆Hfo (H2O, g)= -241.8 kJ/mol
Decomposition
• ∆Hfo may also be used to calculate the
decomposition of something
• If …
H2(g) + 1/2 O2(g) --> H2O(g) ∆Hf˚ = - 242 kJ/mol
Then…
H2O(g) --> H2(g) + 1/2 O2(g) ∆Hf˚ = + 242 kJ/mol
Enthalpy Values
• Depend on how the reaction is written and on phases of reactants
and products…
H2(g) + 1/2 O2(g) --> H2O(g)
∆H˚ = -242 kJ
2 H2(g) + O2(g) --> 2 H2O(g)
∆H˚ = -484 kJ
H2O(g) ---> H2(g) + 1/2 O2(g)
∆H˚ = +242 kJ
H2(g) + 1/2 O2(g) --> H2O(liquid)
∆H˚ = -286 kJ
Huh? So what’s that mean?
To convert 1 mol of water to 1 mol each
of H2 and CO requires 131 kJ of
energy.
Since delta H is positive, the “water gas” reaction is
ENDOthermic
6.4
6.5
A problem… Using Standard
Enthalpy Values
Calculate the heat of combustion of methanol,
i.e., ∆Horxn for
∆Horxn =  ∆Hfo (prod) -  ∆Hfo (react)
CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g)
∆Hfo CH3OH = -238.6 KJ/mol
∆Hfo O2 = 0 KJ/mol
∆Hfo CO2 = -393.5
∆Hfo H2O = -285.8
= -238.6 – (-393.5 + -285.8)
= 442.7 KJ/mol
Practice
Given the following heats of formation:
NaCl(s): DH(f)= -400 kJ
H2SO4(l): DH(f)= -800 kJ
Na2SO4(s): DH(f)= -1400 kJ
HCl(g): DH(f)= -90 kJ
a) Find the heat of reaction of the following chemical change:
2NaCl(s) + H2SO4 --> Na2SO4(s) + 2HCl(g)
b) Is this reaction exothermic or endothermic?
Energy Stoichiometry
• Remember…
1. In an equation coefficients mean moles!
2. To go from grams to moles divide by molar mass
3. To go from moles to grams multiply my molar
mass
4. Make sure you use the right energy units KJ vs. J
5. Double check the sign of your energy, is it
endothermic or exothermic
Practice
How much heat is evolved when 266 g of white phosphorus (P4) burn in air?
P4 (s) + 5O2 (g)
266 g P4
x
P4O10 (s)
1 mol P4
x 3013 kJ
123.9 g P4
1 mol P4
H = -3013 kJ
= 6470 kJ
Practice
The burning of magnesium is a highly
exothermic reaction.
2Mg(s) + O2(g) --> 2MgO(s) + 1500 kJ
Use the thermochemical equation to calculate
the energy change in kilojoules when 0.5 mol of
Mg burn in an excess of O2.
You try
• Calculate the change in H when 32g of NO
decomposes?
– ½ N2 (g) + ½ O2 (g)  NO (g)
∆Hf˚ = 90.2 kJ
Using Standard Enthalpy Values
Use ∆H˚’s to calculate enthalpy change for
H2O(g) + C(graphite) --> H2(g) + CO(g)
• When calculating ∆H information from several
equations is needed
And then…
H2O(g) + C(graphite) --> H2(g) + CO(g)
From either givens within the problem or reference
books and tables we can find…
• H2(g) + 1/2 O2(g) --> H2O(g) ∆Hf˚ = - 242 kJ/mol
• C(s) + 1/2 O2(g) --> CO(g)
∆Hf˚ = - 111 kJ/mol
And and then… we add ‘em.
H2O(g) --> H2(g) + 1/2 O2(g) ∆Ho = +242 kJ
C(s) + 1/2 O2(g) --> CO(g)
∆Ho = -111 kJ
----------------------------------------------------------------------H2O(g) + C(graphite) --> H2(g) + CO(g)
∆Honet = +131 kJ
• Positive delta H means?
Warm-Up
• Calculate the change in H when 64g of NO is
formed?
– ½ N2 (g) + ½ O2 (g)  NO (g)
∆Hf˚ = 90.2 kJ
Enthalpy Day 2
Things to remember about Thermochemical Equations
1. The coefficients always refer to the number of moles of a
substance
H2O (s)
H2O (l)
H = 6.01 kJ
2. If you reverse a reaction, the sign of H changes
H2O (l)
H2O (s)
H = -6.01 kJ
3. If you multiply both sides of the equation by a factor n, then H
must change by the same factor n.
2H2O (s)
2H2O (l)
H = 2 x 6.01 = 12.0 kJ
6.3
Thermochemical Equations
4. The physical states of all reactants and products must be
specified in thermochemical equations.
H2O (s)
H2O (l)
H = 6.01 kJ
H2O (l)
H2O (g)
H = 44.0 kJ
6.3
What if you have limited information?
Calculate the enthalpy of the reaction:
2 NO(g) + O2(g) --> 2 NO2(g)
Given the following reactions and enthalpies of formation:
½ N2(g) + O2(g) --> NO2(g)
∆ H = 33.2 kJ
½ N2(g) + ½ O2(g) --> NO(g)
∆ H = 90.2 kJ
2 NO(g) + O2(g) --> 2 NO2(g)
2/2 N2(g) + 2 O2(g) --> 2 NO2(g)
2 NO(g) --> 2/2 N2(g) + 2/2 O2(g)
∆ H = 33.2 kJ *2
+
∆ H = - 90.2 kJ * 2
-114 kJ
You try…
Calculate the enthalpy of the reaction:
4B(s)+3O2(g) --> 2B2O3(s)
Given the following information:
B2O3(s) + 3H2O(g) --> 3O2(g) + B2H6(g)
2B(s) + 3H2(g) --> B2H6(g)
H2(g) + ½ O2(g) --> H2O(l)
H2O(l) --> H2O(g)
∆ H = +2035 kJ
∆ H = +36 kJ
∆ H = -285 kJ
∆ H = +44 kJ
Equilibrium
and
Le Chatelier’s Principle
Chemical Equilibrium
Reversible Reactions:
A chemical reaction in which the
products can react to re-form the
reactants
Chemical Equilibrium:
When the rate of the forward reaction
equals the rate of the reverse reaction
and the concentration of products and
reactants remains unchanged
2HgO(s)  2Hg(l) + O2(g)
Arrows going both directions (  ) indicates equilibrium in a chemical equation
LeChatelier’s Principle
When a system at
equilibrium is placed under
stress, the system will
undergo a change in such
a way as to relieve that
stress.
Henry Le Chatelier
Le Chatelier Translated:
When you take something away from a system at
equilibrium, the system shifts in such a way as to
replace what you’ve taken away.
When you add something to a system at
equilibrium, the system shifts in such a way as to
use up what you’ve added.
LeChatelier Example #1
A closed container of ice and water at
equilibrium. The temperature is raised.
Ice + Energy  Water
right
The equilibrium of the system shifts to the _______
to use up the added energy.
LeChatelier Example #2
A closed container of N2O4 and NO2 at equilibrium.
NO2 is added to the container.
N2O4 (g) + Energy

2 NO2 (g)
The equilibrium of the system shifts to the
left
_______
to use up the added NO2.
LeChatelier Example #3
A closed container of water and its vapor at equilibrium.
Vapor is removed from the system.
water + Energy  vapor
The equilibrium of the system shifts to the
right to replace the vapor.
_______
LeChatelier Example #4
A closed container of N2O4 and NO2 at equilibrium.
The pressure is increased.
N2O4 (g) + Energy  2 NO2 (g)
The equilibrium of the system shifts to the
left
_______
to lower the pressure, because there
are fewer moles of gas on that side of the
equation.
What about this?
3 H2(g) + N2(g) <--> 2 NH3(g)
What happens when…
a) More N2 is added
b) Pressure is increased
c) Temp. is deceased
right/left/no change
right/left/no change
right/left/no change
Or this?
• The system depicted here is maintained at a temperature of
30 °C. If the temperature of the system is doubled, the system
will achieve equilibrium by which of the following responses?
a)
The temperature of the liquid water will exceed the temperature of
the vapor.
b) The temperature of the vapor will exceed the temperature of the
liquid water.
c) A higher percentage of the water vapor in the container will
condense to liquid.
d) A higher percentage of the water will move into the vapor phase.