Ch. 10: Thermochemistry
Dr. Namphol Sinkaset
Chem 200: General Chemistry I
I. Chapter Outline
I. Introduction
II. Energy
III. 1st Law of Thermodynamics
IV. Quantifying Heat and Work
V. Reactions at Constant V or P
VI. Thermochemical Equations/Hess’s Law
VII. Formation Reactions
VIII. Using Bond Energies
IX. Born-Haber Cycle
I. Chemistry and Energy
• We have said before that chemistry is
all about energy – energy lost, gained,
and transferred.
• Things always seek the lowest energy
state which is equivalent to seeking
highest stability.
• We start to quantify energy in this
chapter.
II. What is Energy?
• Energy is the capacity to do work.
II. Kinetic Energy
• Kinetic energy involves motion.
 A moving car possesses kinetic energy.
• Thermal energy is the energy
associated w/ temp of an object.
 A type of KE because it arises from
motions of atoms/molecules in the
substance.
II. Potential Energy
• Potential energy is the energy associated
with the position or composition of an
object.
 An object raised above the floor possess PE
due to its position in a gravitational field.
• Chemical energy is the energy
associated with relative positions of e-’s
and nuclei in atoms and molecules.
II. Summary of Energy
II. Exchanging Energy
• Objects possess energy
(sometimes both kinetic
and potential).
• Objects exchange
energy through one of
two ways:
 work: force acting
through a distance
 heat: flow of energy
caused by temp.
difference
II. Units of Energy
• The SI unit of energy is the joule (J), 1 J
= 1 kg·m2/s2.
• An older unit is the calorie (cal), defined
as the energy needed to raise the temp.
of 1 g H2O by 1 °C.
 Note that 1 cal = 4.184 J
• Nutritional Calorie is actually the kcal.
II. Energy Conservation
• The law of conservation of energy
states that energy cannot be created or
destroyed.
• But, we know it can be transferred via
work or heat.
• Energy can also change forms.
II. Energy Transformations
II. System and Surroundings
• We define points of
view to observe where
energy goes.
• system: part of
universe being studied
• surroundings:
everything else with
which the system can
exchange energy
II. Changes are Equal but
Opposite
• If the system loses
energy (-), then the
surroundings must
gain energy (+).
• If the system gains
energy (+), then the
surroundings must
lose energy (-).
III. 1st Law of Thermodynamics
• You already know this law by another name:
Law of Conservation of Energy.
• This can be restated as the total energy of the
universe is constant, i.e. there’s never a
change in Euniverse.
III. The Internal Energy
• Say we have a system comprised of
reactants in solution.
• Every particle moves constantly and
feels different attractions/repulsions.
 There is KE from movement and PE from
position.
• The sum of KE and PE for all particles
in a system is called the internal energy
(E).
III. State Functions
• Internal energy (E) is a state function.
 The value of a state function depends only on the
state of the system, not on how it arrived at that state.
• Elevation vs. distance to get to that elevation.
III. State Functions
• The E of a system does not depend on
how it got there. Same goes for T, P, V.
 “Independent of path.”
 “It is what it is.”
• State functions usually given uppercase variables.
• Changes in state functions are easily
calculated by subtracting initial value
from final value (final – initial).
III. Changes in Internal Energy
• When a rxn occurs, internal energy of
products is usually different than internal
energy of reactants.
• The change can be easily calculated by
taking the difference between the two.
III. Energy Diagram
• Internal energy changes resulting from a
reaction can be displayed in an energy
diagram.
• For C(s) + O2(g)  CO2(g), we have:
III. Where Does the Energy Go?
• We see that as C(s) and O2(g) transform
into CO2(g), energy is lost.
• If reaction is the system, then energy
must flow to the surroundings.
 ∆𝐸𝑠𝑦𝑠 = −∆𝐸𝑠𝑢𝑟𝑟
III. How Does the Energy Go?
• The total change in internal energy is equal to
the energy transferred via heat and/or work.
 ∆𝐸 = 𝑞 + 𝑤
• The sign convention for q and w is very
important!!
III. Sign Conventions
III. Sample Problem
• Identify each energy exchange as heat or
work and determine whether the sign of
heat or work (relative to the system) is
positive or negative.
a) An ice cube (system) melts and cools the
surrounding beverage.
b) A metal cylinder (system) is rolled up a ramp.
c) Steam (system) condenses on skin, causing a
burn.
III. Sample Problem
• A cylinder and piston assembly (defined
as the system) is warmed by an
external flame. The contents of the
cylinder expand, doing work on the
surroundings. If the system absorbs
559 J of heat and does 488 J of work
during the expansion, what is value of
ΔE?
IV. Measuring Energy Changes
• To find ΔE, we need to measure or
calculate both q and w.
• Heat is related to temperature change.
• The 2 types of w that are most relevant
are electrical work and pressure-volume
work.
IV. Temperature vs. Heat
• Temperature is the measure of thermal
energy within a sample of matter.
• Heat is the transfer of thermal energy.
• Thermal energy always flows from hot
to cold until thermal equilibrium is
reached.
IV. Measuring Heat
• Intuitively, we know temp. is related to
heat, and different substances require
different amounts of heat to change
their temp.
 q = C × ΔT
• heat capacity (C): the amount of heat
needed to change the temp of a
substance by 1 °C
 What’s wrong with this?
IV. Specific Heat Capacity
• specific heat capacity (Cs): the amount of
heat required to change the temp. of 1 g of a
substance by 1 °C
• Can also have molar heat capacity.
IV. Sample Problem
• Find the heat transferred (in kJ) when
5.50 L of ethylene glycol in a car
radiator cools from 37.0 °C to 25.0 °C if
the density of ethylene glycol is 1.11
g/mL, and its specific heat capacity is
2.42 J/g·K.
IV. Sample Problem
• A 55.0-g aluminum block (Cs= 0.903
J/g·°C) initially at 27.5 °C absorbs 725 J
of heat. What is the final temperature of
the aluminum?
IV. Thermal Energy Transfer
• As we know, heat flow from hotter
substance to colder substance until both
substances have same temperature.
• Assuming no heat loss, then the heat
lost by one must equal the heat gained
by the other.
 𝑞𝑠𝑦𝑠 = −𝑞𝑠𝑢𝑟𝑟
IV. Common Thermal Transfer
• Many problems can
be written based on
this set up.
• Remember the
negative sign!
• Expand each side
into specific q
expressions.
IV. Sample Problem
• A 32.5-g cube of aluminum (Cs= 0.903
J/g·°C) initially at 45.8 °C is submerged
into 105.3 g of water (Cs= 4.18 J/g·°C)
at 15.4 °C. What is the final
temperature of both substances at
thermal equilibrium, assuming no heat
loss?
IV. Measuring Work
IV. Sample Problem
• When fuel is burned in a cylinder
equipped with a piston, the volume
expands from 0.255 L to 1.45 L against
a constant pressure of 1.02 atm. In
addition, 875 J is emitted as heat. What
is ΔE for the burning of the fuel? Note
that 1 L·atm = 101.3 J.
V. Measuring ΔE
• With the two equations for q and w, we
can measure ΔE for reactions.
• Simplifications can be made if specific
conditions are chosen.
V. Reactions at Constant V
V. Calorimetry
• Ignite the reaction.
• Measure ΔT.
• Obtain qcal.
 qcal = Ccal × ΔT
• Relate back to qrxn.
 qcal = -qrxn
• Relate back to ΔErxn
 qrxn = ΔErxn
V. Sample Problem
• When 1.550 g of liquid hexane (C6H14)
undergoes combustion in a bomb
calorimeter, the temperature rises from
25.87 °C to 38.13 °C. Find ΔErxn for the
reaction in kJ/mole hexane. Note that
the heat capacity of the bomb
calorimeter is 5.73 kJ/°C.
V. Reactions at Constant P
• At constant P, we introduce a variable called
enthalpy (H) that doesn’t necessitate that we
calculate or measure q and w separately.
• Enthalpy is defined as H = E + PV.
• Typically we are interested in ΔH…
V. More Useful ΔH Equation
• If only P-V work can be done under constant P…
V. More Useful ΔH Equation
• We see that enthalpy change is the heat
gained/lost under constant pressure conditions.
• ΔH measures just heat; ΔE measures all energy.
• Much easier to measure ΔH than ΔE, and
typically they are close in value because few
reactions do P-V work.
V. Constant P Calorimetry
• The important
relationships to
remember are:
qsystem= -qsurroundings
qrxn= -qsoln
qrxn = ΔHrxn
V. Heats of Reaction
• H is a state function, so ΔH is simply Hfinal – Hinitial.
• The enthalpy of a reaction is known as the heat of
reaction (ΔHrxn).
• Reactions can be either exothermic (-ΔHrxn) or
endothermic (+ΔHrxn).
V. Sample Problem
• 50.0 mL of 1.00 M NaOH is added to
25.0 mL of 1.25 M H2SO4 in a coffee
cup calorimeter. If the initial temp. is
25.00 °C, and the final temp. is 33.83
°C, what is ΔHrxn in kJ? Use d = 1.00
g/mL for all solutions and Cs = 4.184
J/g·°C for the combined solution.
VI. Thermochemical Equations
• A thermochemical equation is a
balanced equation that includes the
heat of reaction.
• The value of ΔHrxn is stoichiometrically
related to the reactants and products.
2H2O(l)  2H2(g) + O2(g)
ΔHrxn = 572 kJ
VI. Sample Problem
• How many grams of carbon dioxide are
formed when the following reaction is used
to produce 1.5 × 103 kJ of heat?
2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(g)
ΔHrxn = -5316 kJ
VI. Rules for Manipulating
Thermochemical Equations
1) If equation is multiplied by a constant
factor, then ΔHrxn is multiplied by the
same factor.
2) If equation is reversed, then ΔHrxn
changes sign.
3) If equation can be written as a sum of a
series of steps, then ΔHrxn for the overall
is the sum of the ΔHrxn’s of each step.
VI. Hess’s Law
• The last rule is
Hess’s law.
• Hess’s Law: the
enthalpy change of
an overall process is
the sum of the
enthalpy changes of
its individual steps
VI. Sample Problem
• Calculate ΔHrxn for the reaction 2NO2(g)
+ ½ O2(g)  N2O5(g) given the
information below.
N2O5(g)  2NO(g) + 3/2 O2(g)
NO(g) + 1/2 O2(g)  NO2(g)
ΔH = 223.7 kJ
ΔH = -57.1 kJ
VI. Sample Problem
• Calculate ΔHrxn for the reaction Ca(s) +
½ O2(g) + CO2(g)  CaCO3(s) using the
information below.
Ca(s) + ½ O2(g)  CaO(s)
CaCO3(s)  CaO(s) + CO2(g)
ΔH = -635.1 kJ
ΔH = 178.3 kJ
VII. Standard Heats of Reaction
• Thermodynamic variables vary a little with
conditions, so we need to define a set of
standard conditions.
• Standard conditions are defined at 1 atm and
25 °C.
• Products and reactants at these conditions
are said to be in their standard states. Note
that solutions must be 1 M.
• The “not” symbol is used to indicate standard
conditions, e.g. ΔH°rxn.
VII. Formation Reactions
• A formation reaction is one that creates
exactly 1 mole of a substance from its
elements.
• Of course, standard heats of formation
occur at 1 atm and 25 °C.
Na(s) + ½ Cl2(g)  NaCl(s)
ΔH°f = -411.1 kJ/mole
VII. Using ΔH°f’s
• Standard heats of formation are very
powerful; they allow calculation of
ΔH°rxn for anything using a Hess’s Law
type of calculation.
ΔH°rxn = (sum ΔH°f products) – (sum ΔH°f reactants)
VII. Visual Example
VII. Standard Heats of Formation
VII. Sample Problem
• Calculate ΔH°rxn for C2H2(g) + 5/2 O2(g)
 2CO2(g) + H2O(g) given that the
standard heats of formation for
acetylene, carbon dioxide, and steam
are 227.0 kJ/mole, -393.5 kJ/mole, and
-241.8 kJ/mole, respectively.
VIII. BE’s and Chemical
Change
• In a reaction involving covalent
compounds, bonds are broken and new
bonds are formed.
 Relative strengths of bonds determine
whether rxn is exo or endo.
• We can calculate heats of reaction
based on the energy difference between
bonds made and bonds broken.
VIII. Bond Energies
• Breaking bonds requires energy…
 A-B(g)  A(g) + B(g)
ΔH˚break = BEA-B
• Forming bonds releases energy...
 A(g) + B(g)  A-B(g)
ΔH˚form = -BEA-B
• Different bonds have different levels of
attraction and thus different BE’s.
VIII. Estimating Rxn Enthalpies
VIII. Calculating w/ BE’s
•
A reaction can be considered a two-step
process:
1) Heat absorbed to break bonds
2) Heat released to make bonds
•
We can sum these to get the heat of
reaction.
VIII. Bond Energy Table
VIII. Sample Problem
• Calculate ΔH˚rxn for the reaction shown
below. Use the indicated bond enthalpies.
IX. Ionic Bond Formation
• Recall that we envision the creation of
ionic bond as an e- transfer from the
metal to the nonmetal.
• However, if we look at the energies
involved in these two steps, we see
something puzzling.
IX. Ion Formation for NaCl
• We break up the e- transfer process into
two steps and add up the energies.
Na(g)  Na+(g) + e-
IE1 = 496 kJ/mole
Cl(g) + e-  Cl-(g)
EA = -349 kJ/mole
Na(g) + Cl(g)  Na+(g) + Cl-(g)
IE1 + EA = 147 kJ/mole
IX. Other Energies
• For most ionics, this is typical; ion
formation is an endothermic process.
• So why do ionics form at all?
• There must be a huge exothermic
process to offset these endothermic
processes.
• The strong +/- attractions that are
formed are the source of this exo step.
IX. Lattice Energy
• lattice energy: the energy associated with
forming a crystalline lattice of alternating
cations and anions from gaseous ions.
IX. Born-Haber Cycle
• Although lattice energy is a critical
component of ionic bond formation, it
cannot be measured directly.
• We use Hess’s Law in the Born-Haber
Cycle.
IX. Born-Haber Cycle for NaCl
IX. Calculating Lattice Energy
• Since enthalpy is independent of path,
the following equation applies.
ΔH˚f = ΔH˚atom + ΔH˚BE + ΔH˚IE + ΔH˚EA + ΔH˚lattice
IX. Sample Problem
• Calculate the lattice energy for CaCl2 given
the following enthalpy values: heat of
sublimation for Ca = 179.3 kJ/mole, Cl2 bond
energy = 243 kJ/mole, 1st ionization energy
Ca = 589.8 kJ/mole, 2nd ionization energy Ca
= 1145.4 kJ/mole, 3rd ionization energy Ca =
4912.4 kJ/mole, electron affinity Cl = -349
kJ/mole, heat of formation CaCl2 = -795.8
kJ/mole.