Chapter 5
Thermochemistry
(The study of the relationships between
chemical reactions and energy changes
involving heat)
&
Chapter 8 Section 8
Strength of Covalent Bonds
Energy
• The ability to do work or transfer
heat.
Work
• Energy used to
move an object over
some distance.
• w = F  d,
where w is work, F
is the force, and d is
the distance over
which the force is
exerted.
Heat
• Heat is energy
transferred between
substances due to a
difference in
temperature (from
higher temp to lower
temp)
Potential Energy
Energy an object possesses by virtue of its
position or chemical composition.
PE = mgh
(mass x gravity x height)
Kinetic Energy
Energy an object possesses by virtue of its
motion.
1
KE =  mv2
2
(½ x mass x velocity2)
(true except at 0 K)
Transferal of Energy Example:
a) The potential energy of this ball of
clay is increased when it is moved
from the ground to the top of the wall.
Transferal of Energy
a) The potential energy of this ball of
clay is increased when it is moved
from the ground to the top of the wall.
b) As the ball falls, its potential energy is
converted to kinetic energy.
Transferal of Energy
a) The potential energy of this ball of
clay is increased when it is moved
from the ground to the top of the wall.
b) As the ball falls, its potential energy is
converted to kinetic energy.
c) When it hits the ground, its kinetic
energy falls to zero (since it is no
longer moving); some of the energy
does work on the ball, the rest is
dissipated as heat.
Units of Energy
• The SI unit of energy is the joule (J).
kg m2
1 J = 1 
s2
• An older, non-SI unit is still in widespread
use: The calorie (cal).
1 cal = 4.184 J
1 food Calorie = 1kcal
System and Surroundings
• System - the
molecules we want to
study (here, the
hydrogen and oxygen
molecules).
• Surroundings everything else (here,
the cylinder and
piston).
First Law of Thermodynamics
a.k.a. Law of Conservation of
Energy
• Energy is neither created nor destroyed.
• The total energy of the universe is a constant;
if the system loses energy, it must be gained by
the surroundings, and vice versa.
Internal Energy
The internal energy of a system is the sum of
all kinetic and potential energies of all
components of the system; we call it E.
Usually we don’t measure E, we just measure
the change in E (ΔE)
Internal Energy
By definition, the change in internal energy,
E, is the final energy of the system minus the
initial energy of the system:
E = Efinal − Einitial
(We don’t actually use this equation….we just need to know the change is
between two states!)
• If E is positive (E > 0, Efinal > Einitial)
The system absorbed energy from the surroundings.
Energy of the system increases
Energy of the surroundings decreases
• If E is negative (E < 0, Efinal < Einitial )
The system released energy to the
surroundings.
 Energy of the system decreases
Energy of the surroundings increases
Changes in Internal Energy
• When energy is
exchanged between
the system and the
surroundings, it is
exchanged as either
heat (q) or work (w).
E = q + w.
E, q, w, and Their Signs
State Functions
• Depend only on the present state of the
system, not on the path by which the system
arrived at that state.
• In the system below, the water could have reached
room temperature from either direction.
 E depends only on Einitial and Efinal and is a
state function.
State Functions
• However, q and w are
not state functions.
• Whether the battery is
shorted out or is
discharged by running
the fan, its E is the
same.
 But q and w are different
in the two cases.
State Functions
• Generally…..
Quantities we look at the change in (E) are
state functions
Quantities we don’t look at changes in (q &
w) are NOT state functions.
P-V Work
When a chemical
reaction occurs,
commonly the only
work done is a change
in volume of a gas
pushing on the
surroundings (or being
pushed on by the
surroundings).
PV Work
We can measure the work done by the gas if
the reaction is done in a vessel that has been
fitted with a piston.
w = −PV
Example #1
Hydrogen & Oxygen gas are in a cylinder
with a movable piston. When ignited, the
system loses 1150J of heat to the
surroundings. The reaction also causes the
piston to rise (from the expanding gases)
and 480J of work is done as the piston is
pushed against the atmosphere. What is the
change in internal energy of the system?
Heat is transferred out of the system so q is
negative (q=-1150J)
Work is done by the system, so w is negative
(w = -480J)
ΔE = q + w = -1150J + -480J = -1630J
1630J of energy has been transferred from the
system to the surroundings
Enthalpy at Constant Pressure
• If a process takes place at constant pressure
(as the majority of processes we study do) and
the only work done is this pressure-volume
work, we can account for heat flow during the
process by measuring the enthalpy of the
system.
• Enthalpy is the internal energy plus the
product of pressure and volume:
H = E + PV
Enthalpy at Constant Pressure
• When the system changes at constant
pressure, the change in enthalpy, H, is
H = (E + PV)
• This can be written
H = E + PV
Enthalpy at Constant Pressure
• Since E = q + w and w = −PV, we can
substitute these into the enthalpy
expression:
H = E + PV
H = (q+w)− w
H = qp (p means constant pressure)
• So, at constant pressure the change in
enthalpy is the heat gained or lost.
Important!
We use H to describe chemical reactions because
most reactions happen at constant pressure, and
at constant pressure H= q and q is (relatively)
easy to find!
Often for chemical reactions, PΔV is very small
and ΔE ≈ ΔH
ΔH and sign:
Endothermic – ΔH is positive (heat transferred
to the system from the surroundings)
Phase changes: s l  g
Exothermic – ΔH is negative (heat transferred
from the system to the surroundings)
Phase changes: g l  s
Thermochemical Equations
• Must specify physical states of products and reactants
 Enthalpy depends on state!
• Must specify enthalpy change
Example:
CaO(s) + CO2(g)  CaCO3(s)
H298 = -178kJ
H, is called the enthalpy of reaction, or the heat of reaction
(HRXN).
(Temp at which the reaction occurs is sometimes included)
How to we figure out H?
Method #1 to find Δ𝐻 :
Calorimetry
When Δ𝐻=q, you can calculate q like you did in
your first year class when given the proper
information!
qp = m Tcp
qp =ccT
qlost = qgained
Specific heat of water (and most aqueous sol’ns):
4.184J/gºC or 4.184J/g K
*On AP Equation
Sheet! (q=mc T)
Constant Pressure Calorimetry
By carrying out a
reaction in aqueous
solution in a simple
calorimeter such as this
one, we can indirectly
measure the heat
change for the system
by measuring the heat
change for the water in
the calorimeter.
Constant Volume (Bomb)
Calorimetry
Reactions can be
carried out in a
sealed “bomb”
calorimeter, such as
this one, to measure
the heat absorbed
by the water.
Bomb Calorimetry
• Because the volume
in the bomb
calorimeter is
constant, what is
measured is really the
change in internal
energy, E, not H.
• For most reactions,
the difference is very
small.
Temperature changes are converted to heat energy by using heat capacities.
Heat Capacity (C) - The amount of energy required to raise the
temperature of a substance by 1 K (J/K)
Molar Heat Capacity (Cm) – The amount of energy required to
raise the temperature of 1 mole of a substance by 1 K
(J/molK)
Specific Heat Capacity (or simply Specific Heat) (Cs) - The
amount of energy required to raise the temperature of 1 g of a
substance by 1 K. (J/gK)
Calorimeter Constant (cc) – The heat capacity of a
calorimeter(J/K)
*temperatures may also be in °C
Example #2
The specific heat of ammonia is 4.381 J/gK.
Calculate the heat required to raise the
temperature of 1.50g of ammonia from 213.0K
to 218.0K.
q = mΔTcp
q = (1.50g)(218.0K – 213.0K )(4.381J/gK )
q = 33 J
Example #3
2.15 g of methane is combusted (burned) to form CO2 and
H2O. All of the heat evolved is used to raise the
temperature of 100.0 g of water by 17.8 K. Calculate the
heat (q) of combustion per mol of methane.
q = (100.0 g) (17.8 K) (4.184 J/g K) = 7450 J = 7.45 kJ
7.45 kJ
16.0 g CH4
---------X ----------- = 55.4 kJ/mol
1 mole CH4
2.15 g CH4
Example # 4
50.00 mL of 0.750 M HCl solution and 50.00 mL
of 0.875 M KOH solution are combined in a
calorimeter. The temperature of the
calorimeter and both solutions goes from 25.0
oC to 29.6 oC. Assume that the specific heat
of each solution is the same as water, 4.184
J/g oC and that the density of each solution is
1.00 g/mL. The heat capacity of the
calorimeter is 125 J/oC. Calculate the heat of
reaction per mole of water formed.
HCl(aq) + KOH(aq)  H2O(l) + KCl(aq)
Hrxn= q= energy change of solutions + energy change of calorimeter
= (m)( T)(cp) + (cc)( T)
(100.00g)(4.6oC)(4.184 J/goC) + (125 J/oC)(4.6oC)
1900 J + 580 J
2500 J or 2.5 kJ
since this reaction is exothermic: Hrxn = - 2500J or - 2.5 kJ
We want the heat of reaction per mole of water!
Do an excess-limiting problem:
.0500L
.0500L
.750M
.875M
? mol
HCl(aq) + KOH(aq)  H2O(l) + KCl(aq)
. 0500𝐿 𝐻𝐶𝑙 ∗
.750 𝑚𝑜𝑙 𝐻𝐶𝑙
1 𝐿 𝐻𝐶𝑙
.0500𝐿 𝐾𝑂𝐻 ∗
∗
.875𝑚𝑜𝑙 𝐾𝑂𝐻
1𝐿 𝐾𝑂𝐻
1𝑚𝑜𝑙 𝐻2 𝑂
1𝑚𝑜𝑙 𝐻𝐶𝑙
∗
= 0.0375 𝑚𝑜𝑙 𝐻2 𝑂
1𝑚𝑜𝑙 𝐻2 𝑂
1𝑚𝑜𝑙 𝐾𝑂𝐻
= 0.0438 𝑚𝑜𝑙 𝐻2 𝑂
0.0375 moles of H2O are formed in the reaction
- 2.5 kJ
-------------- = - 67 kJ / mol H2O
0.0375 mol
Some more about Enthalpy…
1. Enthalpy is an extensive property (it depends
on amount).
2. H for a reaction in the forward direction is
equal in size, but opposite in sign, to H for the
reverse reaction.
3. H for a reaction depends on the state of the
products and the state of the reactants.
(Enthalpy: gases > liquids > solids)
ΔH & Stoichiometry
Enthalpy is an extensive property so changing the
stoichiometry changes the value of ΔH
CaO(s) + CO2(g)  CaCO3(s)
H = -178kJ
2CaO(s) + 2CO2(g)  2CaCO3(s)
H = -356kJ
½ CaO(s) + ½ CO2(g)  ½ CaCO3(s) H = -89.0kJ
Example #5
How much heat is released when 4.50g of methane gas
is burned in a constant pressure system?
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ΔH = -802 kJ
For every 1 mol of methane burned 802 kJ is released
1mol CH 4
802kJ
4.50 g CH 4 *
*
 226kJ
16.0g CH 4 1mol CH 4
Method #2 to find Δ𝐻 :
Hess’s Law
Hess’s law states that
“If a reaction is carried
out in a series of
steps, H for the
overall reaction will be
equal to the sum of
the enthalpy changes
for the individual
steps.”
Example #6
Given:
C(s) + O2(g)  CO2(g)
ΔH = -393.5 kJ
CO(g) + ½ O2(g)  CO2(g) ΔH = -283.0 kJ
Find ΔH for the following:
C(s) + ½ O2(g)  CO(g)
C(s) + O2(g)  CO2(g)
CO(g) + ½ O2(g)  CO2(g)
ΔH = -393.5 kJ
ΔH = -283.0 kJ
½
C(s) + O2(g)  CO2(g)
ΔH = -393.5 kJ
+ CO2(g)  CO(g) + ½ O2(g) ΔH = +283.0 kJ
C(s) + ½ O2(g)  CO(g)
ΔH = -110.5 kJ
Example #7
Given:
C2H2(g) + 5/2 O2(g)  2CO2(g) + H2O(l) ΔH = - 1299.6 kJ
C(s) + O2(g)  CO2(g)
Δ H = - 393.5 kJ
H2(g) + ½ O2(g)  H2O(l)
Δ H = - 285.9 kJ
Find:
2 C(s) + H2(g)  C2H2(g)
2 CO2(g) + H2O(l)  C2H2(g) + 5/2 O2(g)
Δ H = + 1299.6 kJ
2 C(s) + 2 O2(g)  2 CO2(g) Δ H = - 787.0 kJ
H2(g) + ½ O2(g)  H2O(l) Δ H = - 285.9 kJ
2C(s) + H2(g)  C2H2(g)
ΔH = 226.7 kJ
Enthalpy (Heat) of Formation
An enthalpy of formation, Hf, is defined as the
enthalpy change for the reaction in which a
compound is made from its constituent elements
in their elemental forms.
H2(g) + ½ O2(g)  H2O(g)
ΔHf = -242.3 kJ/mol
Hf depends on physical state, pressure and
temperature
Standard Enthalpy of Formation
Standard enthalpies of formation,H f , are measured under
standard conditions (25°C and 1.00 atm pressure).
The standard enthalpy of formation is the heat of formation
of one mole of the compound in its standard state formed
from its constituent elements in their standard states
The standard state of formation of any element in its most
stable form is by convention taken to be zero
• values are tabulated in Appendix C of your textbook
(page 1059)
Method #3 to find Δ𝐻
The change in enthalpy, H, is the enthalpy of the
products minus the enthalpy of the reactants:
ΔH°RXN = Σ nΔ H°f products –
Σ mΔ H°f reactants
n and m are the stoichiometric coefficients of the
chemical equation
**On AP Equation sheet! (without m & n)
Example #8
(a) Calculate the standard enthalpy change
for the combustion of 1 mole of benzene,
C6H6(l) to CO2(g) and H2O(l) (b) If the
ΔH°rxn for the combustion of 1 mole of
propane is – 2220 kJ, compare the quantity
of heat produced by combustion of 1.00 g
propane C3H8(g) to that produced by 1.00 g
of benzene
a) C6H6(l) + 15/2 O2(g)  6 CO2(g) + 3 H2O(l)
Δ H°rxn = Σ n Δ H°f (prod.) - Σ m Δ H°f (reactants)
using appendix C
Δ H°rxn = [6(-393.5 kJ) + 3(-285.8 kJ)] –
[(49.04 kJ) + 15/2(0 kJ)]
= - 3267 kJ
(b)
Propane: C3H8(g):
−2220 𝑘𝐽 1 𝑚𝑜𝑙𝑒
𝑘𝐽
∗
= 50.5 (𝑟𝑒𝑙𝑒𝑎𝑠𝑒𝑑)
1 𝑚𝑜𝑙𝑒
44.0𝑔
𝑔
Benzene C6H6(l):
−3267 𝑘𝐽 1 𝑚𝑜𝑙𝑒
𝑘𝐽
∗
= 41.9 (𝑟𝑒𝑙𝑒𝑎𝑠𝑒𝑑)
1 𝑚𝑜𝑙𝑒
78.0𝑔
𝑔
Example #9
The standard enthaply change for the reaction
CaCO3(s)  CaO(s) + CO2(g)
is 178.1 kJ
Given ΔH°f CaO(s) as -635.5 kJ and Δ H°f
CO2(g) as -393.5 kJ
calculate Δ H°f CaCO3(s) (without looking it up
in the table)
Δ H°rxn = Δ H°f CaO + Δ H°f CO2 - Δ H°f CaCO3
178.1 kJ = -635.5 kJ – 393.5 kJ - Δ H°f CaCO3
solving for Δ H°f CaCO3 gives -1207.1 kJ
which matches the value in the table in
Appendix C 
Method #4 to find Δ𝐻
• The energy required to break a bond between
two atoms is called the bond enthalpy.
• We use the letter D to represent bond enthalpies.
Ex: D(H-Br) = 366 KJ/mol
Δ Hrxn =
Σ(bond enthalpies of bonds broken) –
Σ(bond enthalpies of bonds formed)
Example #10
• Calculate the enthalpy of the reaction
when methane gas and chlorine gas
react to form methyl chloride (CH3Cl)
and hydrogen chloride.
CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)
CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)
Bonds Broken:
4* D(C-H) + D(Cl-Cl)
Bonds Formed:
3*D(C-H) + D(C-Cl) + D(H-Cl)
Δ Hrxn = 4(413) + 242 – 3(413) – 328 - 431
= 1650 + 242 – 1240 – 328 - 431
= -110 KJ