Thermochemistry
Specific Heat Formula
Q  m  T  c p
Q = Energy (heat) lost or gained
cp = Specific Heat
T = Temperature change
m = Mass
Two Types of Thermal Reactions
•Exothermic: Releases Thermal Energy (heat)
sign is –
•Endothermic: Absorbs Thermal Energy (heat)
sign is +
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
Enthalpy = H
In a chemical reaction, Enthalpy (H) is
equal to the energy that flows as heat.
(at a constant pressure)
Example:
When 1 mol of methane gas is burned it
releases 890 kJ of energy
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + heat
H = -890kj = exothermic reaction
Calculate H for a process in which 5.8 g of
methane are burned. H = -890kJ per mol CH4
Rxn: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) +heat
Solution:
Molar mass of CH4 = 16.04 g / mol
5.8 g CH4 1 mol CH4
= 0.36 mol CH4
16.04 g CH4
0.36 mol CH4
-890kJ
1 mol CH4
= -320 kJ
Calculate H when 12.8 g of sulphur dioxide
reacts with excess oxygen to form sulphur
trioxide. H = -198.2kJ per mol SO2
Solution:
Molar mass of SO2 = 64.07 g / mol
12.8 g SO2 1 mol SO2
= 0.1998 mol SO2
64.07 g SO2
0.1998 mol SO2
-198kj
1 mol SO2
= -39.6 kJ
Hydrogen peroxide decomposes according to the
following thermochemical reaction:
H2O2(l) → H2O(l) + 1/2 O2(g)
ΔH = -98.2 kJ
Calculate the change in enthalpy (ΔH) when 4.00
grams of hydrogen peroxide decomposes.
Calculate the change in enthalpy (ΔH) when 36.00
grams of water are created…
The reaction that occurs in hand warmers is:
4Fe(s) + 3O2(g)  2Fe2O3(s) + heat
H of reaction = -1652kJ
How much heat is released when 2.5 grams of
Fe(s) is reacted with excess oxygen?
Law of Concert Tickets!
Miller + 2 Stitts  LadyGaga
- $600
2 Winters + Miller  Justin Bieber + $300
Stitt  2 Elvis + Winter
- $500
Lady Gaga  Justin Bieber + 4 Elvis $_____ ???
Law of Concert Tickets = Hess’s Law
You can add KNOWN equations to solve UNKOWN equations
Two Rules:
FLIP (reverse) the equation – you must FLIP the sign (- +)
MULTIPY / DIVIDE equation – must Multiply / Divide the ΔH!
Example #1: Calculate the enthalpy for this reaction:
2C(s) + H2(g) ---> C2H2(g)
ΔH° = ??? kJ
Given the following thermo chemical equations:
C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(l) ΔH° = -1299.5 kJ
C(s) + O2(g) ---> CO2(g)
ΔH° = -393.5 kJ
H2(g) + (1/2)O2(g) ---> H2O(l)
ΔH° = -285.8 kJ
a) first eq: flip it so as to put C2H2 on the product side
b) second eq: multiply it by two to get 2C
c) third eq: do nothing.
+1299.5 kJ + (-787 kJ) + (-285.8 kJ) = +226.7 kJ
2C(s) + H2(g) ---> C2H2(g)
ΔH° = ??? kJ
C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(l) ΔH° = -1299.5 kJ
C(s) + O2(g) ---> CO2(g)
ΔH° = -393.5 kJ
H2(g) + (1/2)O2(g) ---> H2O(l)
ΔH° = -285.8 kJ
Example #2: Given the following data:
Find the ΔH of the following reaction:
C(s) + O2(g) ---> CO2(g)
SrO(s) + CO2(g) ---> SrCO3(s)
ΔH = -234 kJ
2SrO(s) ---> 2Sr(s) + O2(g)
ΔH = +1184 kJ
2SrCO3(s) ---> 2Sr(s) + 2C(s) + 3O2(g)
ΔH = +2440 kJ
a) first equation - flip it
b) second equation - divide by two
c) third equation - flip it, divide by two
+234 + (+592) + (-1220) = -394
Example #2: Given the following data:
Find the ΔH of the following reaction:
C(s) + O2(g) ---> CO2(g)
SrO(s) + CO2(g) ---> SrCO3(s)
ΔH = -234 kJ
2SrO(s) ---> 2Sr(s) + O2(g)
ΔH = +1184 kJ
2SrCO3(s) ---> 2Sr(s) + 2C(s) + 3O2(g)
ΔH = +2440 kJ
a) first equation - flip it
b) second equation - divide by two
c) third equation - flip it, divide by two
+234 + (+592) + (-1220) = -394
Hess's Law
In a reaction, the change in enthalpy (ΔH)
is the same - regardless if the reaction
occurs in a single step or in several steps.
If a series of reactions are added together,
the net change in ΔH is the sum of the
enthalpy changes for each step.
Rules for using Hess's Law
If the reaction is multiplied (or divided) by
some factor, Δ H must also be multiplied (or
divided) by that same factor.
If the reaction is reversed (flipped), the
sign of Δ H must also be reversed.
Water phase changes & Energy
constant during a phase change.
Temperature remains __________
Phase Change Diagram
Processes occur by addition of energy 
 Processes occur by removal of energy
Phase Diagram
 Represents phases as a function of temperature
and pressure.
 Critical temperature: temperature above which
the vapor can not be liquefied.
 Critical pressure: pressure required to liquefy AT
the critical temperature.
 Critical point: critical temperature and pressure
(for water, Tc = 374°C and 218 atm).
Phase Changes
Effect of Pressure on Boiling Point
Phase Diagram
 Represents phases as a function of temperature and
pressure.
 Critical temperature: temperature above which the
vapor can not be liquefied.
 Critical pressure: pressure required to liquefy AT the
critical temperature.
 Critical point: critical temperature and pressure (for
water, Tc = 374°C and 218 atm).
Phase changes by Name
Water
Carbon dioxide
Phase
Diagram
for Carbon
dioxide
Phase
Diagram
for Carbon
Carbon
Phase Diagram for Sulfur
Reaction Pathway
• Shows the change in energy during a chemical
reaction
Exothermic Reaction
• reaction that
releases
energy
• products have
lower PE
than reactants
energy
released
2H2(l) + O2(l)  2H2O(g) + energy
Endothermic Reaction
• reaction that
absorbs
energy
• reactants have
lower PE
than products
energy
absorbed
2Al2O3 + energy  4Al + 3O2
Latent Heat of Phase Change
Molar Heat of Fusion
The energy that must be absorbed in
order to convert one mole of solid to
liquid at its melting point.
Molar Heat of Solidification
The energy that must be removed in
order to convert one mole of liquid to
solid at its freezing point.
Latent Heat of Phase Change #2
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.
Latent Heat – Sample Problem
Problem: The molar heat of fusion of water is
6.009 kJ/mol. How much energy is needed to convert 60
grams of ice at 0C to liquid water at 0C?
60 g H 2O 1 mol H 2O 6.009 kJ
 20 kiloJoules
18.02 g H 2O 1 mol H 2O
Mass
of ice
Molar
Mass of
water
Heat
of
fusion
Heat of Solution
The Heat of Solution is the amount of heat
energy absorbed (endothermic) or released
(exothermic) when a specific amount of
solute dissolves in a solvent.
Substance
Heat of Solution
(kJ/mol)
NaOH
-44.51
NH4NO3
+25.69
KNO3
+34.89
HCl
-74.84
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
Thermochemistry is the study of heat change in
chemical reactions.
The system is the specific part of the universe that is
of interest in the study. SURROUNDINGS
SYSTEM
open
Exchange: mass & energy
closed
isolated
energy
nothing
6.2
Exothermic process is any process that gives off heat –
transfers thermal energy from the system to the
surroundings.
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
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
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 at constant press
Hproducts < Hreactants
H < 0
Hproducts > Hreactants
6.3
H > 0
Thermochemical Equations
Is H negative or positive?
System absorbs heat
Endothermic
H > 0
6.01 kJ are absorbed for every 1 mole of ice
that melts at 00C and 1 atm.
H2O (s)
H2O (l) H = 6.01 kJ
6.3
Thermochemical Equations
Is H negative or positive?
System gives off heat
Exothermic
H < 0
890.4 kJ are released for every 1 mole of
methane that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g)
H = -890.4 kJ
CO2 (g) + 2H2O (l)
6.3
Thermochemical Equations
•
The stoichiometric coefficients always refer to the
number of moles of a substance
H2O (s)
•
If you reverse a reaction, the sign of H changes
H2O (l)
•
H2O (l) H = 6.01 kJ
H2O (s) H = -6.01 kJ
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
•
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
How much heat is evolved when 266 g of white
phosphorus (P4) burn in air?
P4 (s) + 5O2 (g)
P4O10 (s)
H = -3013 kJ
1 mol P4
3013 kJ
= 6470 kJ
266 g P4 x
x
1
mol
P
123.9 g P4
4
6.3
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
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 –89
= 0-34,000
C
J
6.4
Constant-Volume Calorimetry
qsys = qwater + qbomb + qrx
qsys = 0
qrxn = - (qwater + qbomb)
qwater = mst
qbomb = Cbombt
Reaction at Constant V
H = qrxn
No heat enters or leaves!
H ~ qrxn
6.4
Constant-Pressure Calorimetry
qsys = qwater + qcal + qrxn
qsys = 0
qrxn = - (qwater + qcal)
qwater = mst
qcal = Ccalt
Reaction at Constant P
H = qrxn
No heat enters or leaves!
6.4
Terminology
• Energy, U
– The capacity to do work.
• Work
– Force acting through a distance.
• Kinetic Energy
– The energy of motion.
Prentice-Hall © 2002
General Chemistry: Chapter 7
Slide 49 of 50
Energy
• Kinetic Energy
1
ek =
2
mv2
kg m2
= J
[ek ]
2
s
=
• Work
w = Fd
Prentice-Hall © 2002
[w ]
=
General Chemistry: Chapter 7
kg m m
= J
2
s
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Energy
• Potential Energy
– Energy due to condition, position, or
composition.
– Associated with forces of attraction or
repulsion between objects.
• Energy can change from potential to
kinetic.
Prentice-Hall © 2002
General Chemistry: Chapter 7
Slide 51 of 50
Energy and Temperature
• Thermal Energy
– Kinetic energy associated with random molecular
motion.
– In general proportional to temperature.
– An intensive property.
• Heat and Work
– q and w.
– Energy changes.
Prentice-Hall © 2002
General Chemistry: Chapter 7
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Heat
Energy transferred between a system and its
surroundings as a result of a temperature
difference.
• Heat flows from hotter to colder.
– Temperature may change.
– Phase may change (an isothermal process).
Prentice-Hall © 2002
General Chemistry: Chapter 7
Slide 53 of 50
Units of Heat
• Calorie (cal)
– The quantity of heat required to change the
temperature of one gram of water by one
degree Celsius.
• Joule (J)
– SI unit for heat
1 cal = 4.184
J
Prentice-Hall © 2002
General Chemistry: Chapter 7
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Heat Capacity
• The quantity of heat required to change the
temperature of a system by one degree.
– Molar heat capacity.
• System is one mole of substance.
– Specific heat capacity, c.
• System is one gram of substance
– Heat capacity
• Mass
Prentice-Hall © 2002
q = mcT
q = CT
specific heat.
General Chemistry: Chapter 7
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