Thermochemistry
Chapters 6 and 16
Units of Work (F x d)

When you change pressure or
volume of a gas, you are doing
WORK. If you hold pressure
constant (like in most chem
labs) the following equation hold
true

W = -PDV
Units of Work
The sign for work is similar to
the sign for heat
 “-“ system does work on the
surroundings (and the system
therefore expands)
 “+” system has work done on it
by the surroundings (and the
system therefore shrinks)

First Law of Thermodynamics

Energy can be converted from
one form to another but cannot
be created nor destroyed.


Electrical energy to nuclear energy
THE TOTAL ENERGY OF THE
UNIVERSE IS CONSTANT
First Law of Thermodynamics
For all reactions, the energy
change of the SYSTEM is equal
to the SYSTEM’S WORK plus the
SYSTEM’S HEAT FLOW.
 Ex What is the energy change in
a system that does 38.9 j of
work when it loses 16.2 joules of
heat?

First Law of Thermodynamics

So the energy change of a
system is related to both
1. Work the system does/has done
 2. Heat exchanged in/out of the
system

delta E (-) system loses energy
 delta E (+) system gains energy

Enthalpy, H, (Heat Content)
Energy of a system at a given
pressure and volume. Very
difficult to measure in practice,
but it is VERY EASY to measure
the CHANGE in enthalpy, delta H
 DH= q at constant pressure (in
most chem labs)

First Law of Thermodynamics

DH is most commonly measured
heat unit in chemistry labs, and
is often used interchangeably
with q (since pressure rarely
changes within the walls of a
chemistry lab)
First Law of Thermodynamics
DH = Hproducts - Hreactants
(final enthalpy – initial enthalpy)
 DH is often called “the heat of
reaction”

Endothermic reactions



Enthalpy of products > enthalpy of
reactants
DH is positive
Energy is absorbed
Exothermic reactions



Enthalpy of reactants > enthalpy of
products
DH is negative
Energy is released
Spontaneity


Spontaneous reactions will tend
toward conditions of lower enthalpy
More negative DH values
Enthalpy Stoichiometry and Calorimetry

Ex
CH4 + 2O2  CO2 + 2H2O
DH = -1308 kJ
 Notice that the given DH is for the
whole reaction and is therefore a
stoichiometric quantity
 How much energy will be released
when 27.00 g of H2O are
produced?
Standard Enthalpy of Formation DHfo

def Enthalpy change when one mole
of a compound is made FROM ITS
ELEMENTS in their standard states
under standard conditions.
Standard Enthalpy of Formation DHfo

Standard States


for elements the form that the element
exists in at 25oC and 1 atm (don’t
confuse with STP from gases)
for compounds
gases—at 1.0 atm
 solutions – at 1.0 M


a PURE liquid or solid – the actual pure
liquid or solid
Standard Enthalpy of Formation DHfo

DHfo[C2H5OH(l)] = -279kJ/mol


Means that when the reaction below is
carried out, 279 kJ of energy are
released
2C(graphite)+ 3H2(g)+ ½ O2(g)  C2H5OH(l)
Standard Enthalpy of Formation DHfo

We begin studying DHfo by finding
values for several compounds; find
these empirical values on page A-21
of your textbook.



Al2O3 (s)
Ti (s)
SO42-(aq)
Using DHfo to find STANDARD
ENTHALPHY CHANGES for reactions


We use this “simple” equation (we’ll
call it the “state function” equation
all year, since we’ll see lots of
similar equations come by based on
this idea)
DHreaction =SDHfoproducts - SDHforeactants
Using DHfo to find STANDARD
ENTHALPHY CHANGES for reactions
“The enthalpy change for a reaction
is equal to the sum of the heats of
formation of the products minus the
sum of the heats of formation of the
reactants”
 Ex Find DHo for the following
reaction if DHofCH3Br = -36kJ/mol
2C(graphite) + 3H2(g) + Br2(g)  2CH3Br(l)

Standard Enthalpy of Combustion



The enthalpy change when one
mole of a substance is completely
burned in oxygen under standard
conditions
Usually negative
Ex:
DHC2H6o = -1565 kJ/mol
C2H6(g) + 3½O2(g)  2CO2(g) + 3H2O(l)
Hess’s Law

Legal def The DH of a reaction is the
same whether the reaction occurs in
one step or in a series of steps

This is also the definition of a “state
function”, meaning the path is
unimportant for the quantity being
studied
Hess’s Law
 Interpreted
def If you can
combine a series of chemical
reactions mathematically to
get another reaction, you
can also combine the DH’s.
Hess’s Law

3 tricks to doing Hess’s Law
calculations
a. You can reverse any equation,
but change the sign of the DH
 b. You can multiply the entire
equation by a number, but do
the same to the DH
 c. You can do A and B together.

Hess’s Law

Ex You plan to attempt reaction
to create a diamond from
graphite. What heat of reaction,
DH, should you expect, given
these known reactions?
Cgraphite(s) + O2(g)  CO2(g) DH = -394kJ
Cdiamond(s) + O2(g)  CO2(g) DH = -396kJ
Hess’s Law
Solution Write a TARGET
equation first
 By trial and error, we must find
species and get them on the
same side of the equation as the
target.

Hess’s Law

Ex. Find DH for
 2 B (s) + 3H2 (g)  B2H6(g)
given the following.
2B(s) + 3/2 O2(g)  B2O3(s) DH = -1273kJ
B2H6(g) + 3O2(g)  B2O3(s) + 3H2O(g) DH = -2035kJ
H2(g) + ½ O2(g)  H2O(l) DH = -286kJ
H2O(l)  H2O(g) DH = 44kJ
Bond Enthalpies (Review)

The strength of the bond in a
diatomic molecule is given by the
bond dissociation energy (BDE)



Example: H2  2H BDE = +436 kJ
To break a bond, energy must be
put in (endothermic)
When making a bond, energy is
released (exothermic)
Bond Enthalpies
DH = bonds broken – bonds made
 Example – calculate the enthalpy
for the reaction below
CH3CH=CH2 + Br2  CH2BrCHBrCH3

Measuring Heat Flow

Can be done in 2 ways
a. Constant lab pressure (most
common) q = DH
“coffee-cup calorimetry”
b. Constant volume q NOT = DH
Measuring Heat Flow

The amount of heat energy
transferred depends on 3 things
a. what you have (specific heat)
b. how much you have (mass)
c. temperature change you allow to
occur (Dt)
Measuring Heat Flow

Specific Heat Capacity (usually just
called “specific heat”) is the
AMOUNT of heat energy required to
raise 1 gram of ANY SUBSTANCE
1.00 0C
Measuring Heat Flow



“Calorie” is the specific heat JUST
FOR WATER
Notice the table 6.1 on page 248
lists SH capacities in (J)/(g 0C)
The LOWER the number, the EASIER
heat flows IN or OUT of a
substance. Water has a high SH.
Measuring Heat Flow

q = (SH) (m) (Dt) for all
calorimetry, whether bomb or
coffee-cup
Measuring Heat Flow
Example Coffee-Cup Calorimetry
Given
NaOH(s) NaOH(aq) DH = -43 kJ/mol


If 10.0 g NaOH is added to 1.00 L H2O
(SH 4.184 J/goC) at 25.00oC in a
coffee-cup calorimeter, what will be the
final temperature? (Density of the final
solution is 1.05 g/ml, and the solute
does not increase solution volume)
Measuring Heat Flow

Notice, you should always
remember that water is generally in
the SURROUNDINGS (system +
surroundings = universe). What
happens to the system is OPPOSITE
of what happens to the
surroundings.
Measuring Heat Flow

Bomb Calorimetry Bomb calorimeters are
expensive. Each has a known amount of
a known chemical added with a known
heat of reaction. Since you know
everything, this is used to find the “heat
capacity” of the calorimeter. That heat
capacity is then used in all subsequent
calorimetry measurements using that
bomb (the heat capacity is actually
determined at the factory, then stamped
onto the side of the bomb, so you know
when you buy it).
Measuring Heat Flow

Ex Bomb Calorimetry

The heat of combustion (DH) of glucose
C6H12O6, is 2800 kJ/mo. 5.00 g of
glucose are burned with excess oxygen
at constant volume (this tells you it’s a
bomb). The bomb temperature
increased 2.4oC.
Heating Curves

A temperature vs time curve that
shows the physical states at all
temperature points and AT
CONSTANT PRESSURE. Consider
one mole of water as an example
Heating curve for water.
Example

ex how much energy does it take to
convert 150 grams of ice at –20oC
to steam at 120oC? (SH of ice is
2.1 J/goC; steam is 1.8 J/goC;
water 4.184 J/goC ) DHfice = - 6.0
kJ/mol)
Spontaneous Processes and
Entropy

First law



“Energy can neither be created nor
destroyed"
The energy of the universe is constant
Spontaneous Processes





Processes that occur without outside
intervention
Products are favored at the stated
conditions
Spontaneous processes may be fast or
slow
Many forms of combustion are fast
Conversion of diamond to graphite is
slow
Spontaneous Processes and
Entropy

Non-spontaneous processes


Reactant are favored at the stated
conditions
Generally needs outside intervention to
occur
Entropy (S)






A measure of the randomness or disorder
The driving force for a spontaneous process
is an increase in the entropy of the
universe
Entropy is a thermodynamic function
describing the number of arrangements
that are available to a system
Nature proceeds toward the states that
have the highest probabilities of existing
THE MORE ORDERED A COMPOUND IS, THE
LESS ENTROPY IT HAS!
The unit for entropy is J/K mol
Positional Entropy

The probability of occurrence of a
particular state depends on the
number of ways (microstates) in
which that arrangement can be
achieved
Ssolid < Sliquid < Sgas
An example

Predict the sign of DS of the
following reactions
MgO(s) + H2O(l)  Mg(OH)2(s)
Na2CO3(s)  Na2O(s) + CO2(g)
Calculating Entropy Change in a
Reaction


DS = SnpSoproducts - SnrSoreactants
Entropy is an extensive property (a
function of the number of moles)
Generally, the more complex the
molecule, the higher the standard
entropy value
DS Example


What is the entropy change when you burn
glucose, C6H12O6?
The positive change makes sense, since
you are producing 12 moles of gas from 6
moles of gas and one mole of solid
(POSITIVE DS = more randomness,
NEGATIVE DS = less randomness. Cleaning
your room has a negative DS.)
Second Law of Thermodynamics



"In any spontaneous process there is
always an increase in the entropy of the
universe"
"The entropy of the universe is
increasing"
For a given change to be spontaneous,
DSuniverse must be positive
DSuniverse = DSsystem + DSsurroundings
Third Law of Thermodynamics
 The
entropy of a perfect crystal
at absolute zero is zero (and
this is impossible to attain)
General Rules
 Reactions
that increase the
number of moles of a LESS
ordered state are more favored
 Things that made a mess are
more favorable because there
are more ways to make
something messy than there
are ways to keep it clean.
Entropy
DSsurroundings = - DH/T
 This means that the amount of
messiness you produce in the
surroundings is directly related to
how exothermic the system is and
inversely related to temperature
 So the higher the temperature,
the less the effect on entropy.

Driving Forces

We see that there are 2 driving forces in a
reaction system’s spontaneity.


Will it become more random,
thereby increasing the randomness
of the universe (+DS of a system
favors spontaneity)
Will it release energy to
surroundings, thereby increasing the
randomness of the universe (-DH for
a system favors spontaneity)
Standard Free Energy Change


DGo is the change in free energy
that will occur if the reactants in
their standard states are converted
to the products in their standard
states
DGo cannot be measured directly
The more negative the value for DGo, the
farther to the right the reaction will
proceed in order to achieve equilibrium
 Equilibrium is the lowest possible free
energy position for a reaction

Calculating Free Energy
Method #1


For reactions at constant
temperature:
DGo = DHo - TDSo
DH, DS, DG and Spontaneity

DG = DH - TDS
H is enthalpy, T is Kelvin temperature
Value of
DH
Value of
TDS
Value of
DS
Sponteneity
Negative
Positive
Positive
Spontaneous
Positive
Negative
Negative
Nonspontaneous
Negative
Negative
???
Spontaneous if the
absolute value of DH is
greater than the
absolute value of TDS
(low temp)
Positive
Positive
???
Spontaneous if the
absolute value of TDS is
greater than the
absolute value of DH
Calculating Free Energy
Method #2
 An
adaptation of Hess's Law:
Cdiamond(s) + O2(g)  CO2(g) DGo = -397 kJ
Cgraphite(s) + O2(g)  CO2(g) DGo = -394 kJ
CO2(g)  Cgraphite(s) + O2(g)
DGo = +394 kJ
Cdiamond(s)  Cgraphite(s)
DGo = -3 kJ
Calculating Free Energy
Method #3

Using standard free energy of
formation (DGfo):
DGo = SnpDGof(products) – SnrDGof(reactants)

DGfo of an element in its standard
state is zero
Example
Calculate DGo for the following
reaction: N2(g) + 3H2(g)  2NH3(g)

The Dependence of Free Energy
on Pressure


Enthalpy, H, is not pressure
dependent
Entropy, S

Entropy depends on volume, so it also
depends on pressure
Slarge volume > Ssmall volume

Slow

pressure
> Shigh pressure
Free Energy and Work


The maximum possible useful work
obtainable from a process at constant
temperature and pressure is equal to the
change in free energy
The amount of work obtained is always
less than the maximum
Henry Bent's First Two Laws of
Thermodynamics
 First law: You can't win, you can only
break even
 Second law: You can't break even
