Transcript Energy

Thermochemistry
The study of energy and its transformations
Definitions
When examining chemical systems or reactions, in
order to keep track of energy changes, we
consider the system and its surroundings.
 The system is where we put our focus.
Typically, it is the reactants and products.
 The surroundings include everything else in the
universe. Often, we just consider the immediate
surroundings, such as the reaction vessel.
Energy Changes
Chemical reactions and physical changes
typically involve a transfer of energy. If a
process such as melting ice requires energy, the
reverse process of freezing water releases the
same amount of energy.
Definitions
If a reaction results in the evolution of heat,
energy flows out of the system and into the
surroundings. These reactions are exothermic.
The energy lost by the system must be equal
to the energy gained by the surroundings.
Definitions
If a reaction requires heat, energy flows from
the surroundings into the system. These
reactions are endothermic.
The energy gained by the system must be
equal to the energy lost by the surroundings.
Physical Changes and Energy
Changing the state of a substance involves
an energy change. In melting or boiling a
substance, the attractive forces between the
atoms or molecules must be overcome, and heat
is required. This process is endothermic.
When a substance cools and condenses or
freezes, heat is given off, and the process is
exothermic.
Chemical Reactions and Energy
In chemical reactions, the energy changes
result from the breaking and the formation of
chemical bonds.
Bond breaking always requires energy.
Bond making always releases energy.
Chemical Reactions and Energy
Bond breaking always requires energy.
Bond making always releases energy.
In exothermic reactions, more energy is
released in forming the products than is used in
breaking apart the reactants.
Types of Systems
Systems can be open (both energy and matter
are exchanged, closed (only energy is exchanged)
and isolated (neither energy nor matter is
exchanged with the surroundings.)
Definitions - Energy

Chemical systems contain both kinetic energy and
potential energy.
Energy is the capacity to do work or to
produce heat.
An example of both is the combustion of
gasoline. The gaseous products expand and do
work (moving the pistons of an engine) and the
reaction also produces heat.
Definitions - Energy
Energy is the capacity to do work or to produce
heat.
Heat and work are ways that objects can
exchange energy.
Definitions - Energy
Kinetic energy is the energy of motion, and it
depends upon the mass of the object and its
velocity. Since molecules, especially those of
gases, are in motion, they posess kinetic energy.
Definitions- Energy
Potential energy is energy due to position or
composition.
Chemical energy is potential energy due to
composition. For example, gasoline and oxygen
have the potential to produce energy if they
react.
Internal Energy
The internal energy (E, or U) of a system is the
sum of the kinetic and potential energies of all
of the particles of the system.
It is generally not possible to determine the
internal energy of a system, but we can measure
changes in internal energy.
Internal energy is changed by the flow of
work and/or heat.
Internal Energy
Internal Energy is a state function. That is, it
depends solely on the present state of the
system, and not how it may have gotten to a
particular state. A state function is independent
of pathway.
Internal Energy
Internal energy (E or U) is a state function, and
depends only on the state of the system, and not
how it got to that state.
Internal Energy & the
st
1
Law
The First Law of Thermodynamics states
that:
Energy can be converted from one form to
another, but cannot be created or destroyed.
It is not possible to measure the total energy
of a system, but it is possible to determine
changes in energy.
Internal Energy
Since energy may flow to or from the
surroundings and the system, we are concerned
with energy changes rather than the absolute value
of the internal energy.
ΔE = Efinal – Einitial
(some texts use the symbol U for internal energy)
The
st
1
Law of Thermodynamics
Since the energy change of the system is
equal and opposite to the energy change of the
surroundings,
ΔEsystem= –Δesurroundings
or
ΔUsystem= –ΔUsurroundings
The
st
1
Law of Thermodynamics
The implication of the first law of
thermodynamics is that:
The energy of the universe is constant.
Work and Heat
Energy is the capacity to do work or transfer
heat. The internal energy of a system changes
when heat is exchanged, or when the system
does work on its surroundings, or when the
surroundings do work on the system.
ΔE = q + w
or
ΔU = q + w
Energy, Work and Heat
As the system loses energy to the
surroundings, it can do so by losing heat (q),
and/or doing work (w). As a result,
ΔE = heat + work
ΔE = q + w
ΔE = q -PΔV
Work
Chemical reactions can be harnessed to do
electrical work (as with batteries), or expansion
work (such as expanding gases in an internal
combustion engine).
Our focus will be on expansion work.
Work
Expansion work results when a reaction
produces more gaseous products than reactants,
and thus pushes back the atmosphere as the
reaction proceeds.
If the volume of the system contracts during
reaction (more gaseous reactants than products),
work is done by the surroundings on the system.
Expansion Work
Usually we just consider the volumes of
gases in a chemical reaction.
2 H2O(g)  2 H2(g) + O2(g)
Since 2 moles of gaseous reactants produce 3
moles of gaseous products, the system expands,
and does work in pushing back the atmosphere.
Expansion Work
work = Force x Distance
Pressure = Force/Area, or
Force = Pressure(Area)
work = Pressure(Area) x Distance
work = Pressure(Area) x ∆h
work = Pressure (length x width) x ∆h
work = P ∆V
Expansion Work
work = P ∆V
If gases are produced by a reaction and the
volume expands, the system is doing work on
the surroundings. The sign, when considering
the system, must be negative. So,
work = - P ∆V
Energy, Work and Heat
ΔE = q –PΔV
Solving for q, the heat change,
q = ΔE + PΔV
The heat change, q, will vary with the
reaction conditions. Reactions in an open vessel
are performed at constant pressure, those in a
sealed vessel are performed at constant volume.
Reactions at Constant Volume
q = ΔE + PΔV
At constant volume, expansion work isn’t
possible, and the heat change, qv, equals the
change in internal energy.
indicates constant
qv = ΔE
volume
or
qv = ΔU
Energy, Work and Heat
q = ΔE + PΔV
At constant pressure, q becomes qp, and
qp = ΔE + PΔV
or
qp = ΔU + PΔV
denotes constant pressure
Energy, Work and Heat
q = ΔE + PΔV
At constant pressure, q becomes qp, and
qp = ΔE + PΔV
or
qp = ΔU + PΔV
denotes constant pressure
Energy and Enthalpy
qp = ΔE + PΔV
Since an open vessel is such a common
apparatus, the heat transferred at constant
pressure is given its own name, the enthalpy
change, ΔH.
qp = ΔE + PΔV = ΔH
Enthalpy
Enthalpy is a state function, and is
independent of reaction pathway.
ΔH = Hfinal-Hinitial
ΔH = Hproducts-Hreactants
Standard Conditions
Many reactions are categorized by their
standard enthalpy change, ΔHo. The degree sign
indicates standard conditions.
Standard conditions specify that the
reactants and products are in the same molar
amounts represented by the coefficients in the
balanced chemical reaction.
Standard Conditions
In a given experiment, the quantities of
reactants and enthalpy change will vary, but the
standard enthalpy change is reported based on molar
quantities.
The enthalpy of a reaction will also vary with
the physical states of reactants or products as
well as temperature and pressure.
Standard Conditions
A thermodynamic standard state refers to a
specific set of conditions. The standard is used
so that values of enthalpy changes can be
directly compared.
The standard state is the most stable form of
a substance at 1 atm pressure and 25oC.
Standard Conditions

Standard conditions are indicated using a degree
symbol ( o ). Standard conditions for
thermochemical data differ from the standard
conditions used in the gas laws.
1. All gases have a pressure of exactly 1 atm.
2. Pure substances are in the form that they
normally exist in at 25oC and 1 atm pressure.
3. All solutions have a concentration of exactly
1M.
Standard Conditions
For example, since oxygen is a diatomic gas
at 25oC, the standard state of oxygen is O2(g) at
a pressure of 1 atm.
Thermochemical Equations
Chemical reactions (or changes of state) may
be written with their enthalpy of reaction. For
example,
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
ΔH=–890.4 kJ/mol
The enthalpy change is for the reaction
written, assuming molar quantities. 890.4 kJ of
heat is produced when a mole of CH4(g) is
burned.
Thermochemical Equations
If the reaction is reversed, 890.4 kJ is
required to produce each mole of CH4(g).
CO2(g) + 2 H2O(l) CH4(g) + 2 O2(g)
ΔH=+890.4 kJ/mol
If the coefficients in a balanced reaction are
multiplied by an integer, the value of ∆H is
multiplied by the same integer.
Calorimetry
Calorimetry is the science of measuring heat.
It typically involves measuring temperature
changes as a substance loses or gains heat.
Since substances vary in how much their
temperature changes as heat is lost or gained, it
is important to know the heat capacity (C) of
substances involved in the reaction.
Heat Capacity (C)
The heat capacity of a substance is the amount
of heat absorbed, usually in joules, per 1 degree
(C or K) increase in temperature. The amount
(mass) of the substance also determines the
amount of heat lost or gained.
C = heat absorbed = _q_
increase in temp. ΔT
Heat Capacity
The specific heat capacity is for a gram of a
substance. It has the units J/oC-g or J/K-g.
The molar heat capacity is for a mole of a given
substance. It has the units J/oC-mol or
J/K-mol.
Calorimeter Constant
In measuring heat changes during a reaction,
any heat absorbed or lost be the calorimeter (the
apparatus itself) must be considered. If this
amount of heat is significant, the calorimeter
constant may be provided or measured. This is
the heat capacity of the specific apparatus used,
and is expressed in J or kJ per degree change in
temperature (K or oC).
Coffee Cup Calorimetry
A simple device for
determining heat changes
of aqueous reactions at
constant pressure is a
coffee cup calorimeter.
Since the contents are
open to the atmosphere,
the pressure,
atmospheric pressure,
remains constant during
the reaction.
Coffee Cup Calorimetry
The heat change for the reaction, qp, is equal
to the enthalpy change for the reaction.
If heat is given off, it goes towards warming
up the contents of the calorimeter and toward
warming up the calorimeter walls, thermometer,
stirrer, etc.
Coffee Cup Calorimetry
qreaction = qcontents + qcal
qcontents = (mass of solution) (ΔTsoln)Csoln
Csoln is the specific heat capacity of the
reaction mixture. If solutions are aqueous and
fairly dilute, the specific heat capacity of water,
4.18J/oC-g, may be used.
Coffee Cup Calorimetry
qreaction = qcontents + qcal
qcal = Ccal (ΔT)
Ccal is the calorimeter heat capacity. It
includes the heat needed to warm up the walls,
thermometer and stirrer of the calorimeter,
along with any heat loss due to leaks.
In many simple calculations, Ccal is assumed
to be negligible, and may be ignored.
Obtaining ΔH of Reaction
The enthalpy change of a reaction, ΔH, can
be obtained from qp.
First, a sign must be assigned. If the
temperature increased during the reaction, the
reaction is exothermic, and q is negative.
If the temperature decreased during the
reaction, the reaction is endothermic, and q is
positive.
Obtaining ΔH of Reaction
The enthalpy change of a reaction, ΔH, can be
obtained from qp.
Also, qp is for a specific quantity of reactants.
Typically, ΔHrxn is for molar quantities of
reactants. To calculate ΔHrxn from qp, you must
calculate the heat change per mole of reactant.
Problem: Calorimetry

A coffee cup calorimeter contains 125. grams of
water at 24.2oC. A 10.5 g sample of KBr, also at
24.2oC, is added. After dissolving, the mixture
reaches a final temperature of 21.1oC. Calculate
∆Hsoln in joules/gram and kJ/mol. Assume the
specific heat of the solution is 4.18 J/g-oC, and
no heat is transferred to or from the calorimeter
or surroundings.
Constant Volume Calorimetry

Certain reactions, notably combustion reactions,
do not lend themselves to open vessels. These
reactions are usually carried out in a sealed
reaction vessel called a bomb calorimeter.
The bomb calorimeter is a rigid steel
container that is sealed after the reactants have
been added. The reaction takes place once an
electrical current is sent through an ignition wire
to the reaction mixture.
Bomb Calorimetry
The steel bomb is
immersed in an
insulated bath
containing either
water or mineral oil.
As the combustion
reaction releases heat,
the heat is transferred
to the bath.
Bomb Calorimetry
Reaction vessel
Ignition wire
O2 inlet
gaskets
Bomb Calorimetry
Once the bomb has been charged with
reactants, it is placed in the water or oil bath
until it reaches a constant temperature.
A current is sent through the ignition wire,
and the combustion reaction takes place. The
heat given off by the reaction is evident from
the increase in temperature of the water/oil
bath.
Bomb Calorimetry
The heat generated by the reaction warms up
the contents of the calorimeter (the bomb,
thermometer, container walls, stirrer) and the
water (or oil) bath.
Usually, the calorimeter constant, which is
the heat capacity of the entire apparatus, is
provided, or determined by combusting a
substance with a known energy of combustion.
Problem: Bomb Calorimetry

The energy released by combustion of benzoic
acid is 26.42 kJ/g. The combustion of .1584g of
benzoic acid increases the temperature of a
bomb calorimeter by 2.54 oC.
a) Calculate the calorimeter constant.
Problem: Bomb Calorimetry
b) 0.2130 g of vanillin (C8H8O3) is burned in
the same calorimeter with a temperature increase
of 3.25oC. Calculate the energy of combustion
of vanillin in kJ/g and kJ/mol.
Hess’s Law
Enthalpy is a state function. This means that a
change in enthalpy depends solely on the initial
and final states (products and reactants), and is
independent of the reaction pathway.
Hess’s Law
Hess’s Law is a method that combines
related chemical reactions and their enthalpy
changes. Since enthalpy changes are
independent of pathway, as long as the net
reaction matches the reaction of interest, the
sum of the enthalpy changes will yield ΔH for
the desired reaction.
Hess’s Law
Hess’s Law
There are a few basic rules in applying Hess’s Law:
1. If a reaction is reversed, the sign of ΔH is also
reversed.
2. If the coefficients in a balanced reaction are
multiplied by an integer, the value of ∆H is
multiplied by the same integer.
Problem: Hess’s Law
Calculate ∆Ho for the reaction:
C6H4(OH)2(aq) + H2O2(aq) C6H4O2(aq) + 2 H2O(l)
Using:
1)C6H4(OH)2(aq) C6H4O2(aq) + H2(g) ∆Ho = +177.4 kJ
2) H2 (g) + O2(g) H2O2(aq)
∆Ho = –191.2 kJ
3) H2 (g) + 1/2 O2(g)  H2O(g)
∆Ho = –241.8 kJ
4) H2O(g)  H2O(l)
∆Ho = –43.8 kJ
Standard Enthalpies of Formation
A formation reaction involves combining
elements, in their standard states, to form one
mole of a compound.
A table of standard enthalpies of formation
(∆Hfo) is in appendix of the text. The ∆Hfo
values of most common compounds have been
determined and tabulated.
Standard Enthalpies of Formation
CaCO3(s) has a ∆Hfo of -1207 kJ/mol. This
is the enthalpy change for the reaction:
Ca(s) + C(graphite) + 3/2 O2(g)  CaCO3(s)
Fractional coefficients are acceptable since
all quantities are molar, and a formation reaction
produces one mole of a compound.
Standard Enthalpies of Formation
∆Hfo values can be used to calculate the
standard enthalpy changes for many reactions.
In an application of Hess’s Law, it is as if the
reactants are decomposed into their elements,
and then the elements are recombined into the
desired products. Since enthalpies of reaction
are independent of pathway, this provides an
accurate way to calculate enthalpies of reactions.
Standard Enthalpies of Formation
CH4(g) + 2O2(g)  CO2(g) + 2 H2O(l)
Hess’s Law and ∆Hf
o
∆Hrxno = ∑ ∆Hfo(products) - ∑ ∆Hfo(reactants)

Problem: Use standard heats of formation to
calculate ∆Horxn for the combustion of methane
(CH4) :
CH4(g) + 2 O2(g)  2 H2O(g) + CO2(g)
Bond Dissociation Energies
Bond dissociation energies can be used to
estimate the enthalpy of a reaction. The
enthalpy change will approximately equal the
energy of bonds broken – energy of bonds
formed.
ΔHo ≈D(bonds broken) –D(bonds formed)
Bond Energies
Since bond energies are average values for a
variety of molecules, they will only provide an
estimate of the enthalpy change for a specific
reaction. The following relationship can also be
used to estimate enthalpies of reaction.
ΔHo ≈D(reactant bonds) –D(product bonds)
Bond Energies - Problem
Use average bond enthalpies to estimate the
enthalpy of reaction for:
N2(g) + 3 H2(g)  2 NH3(g)
Energetics of Ionic Bonds
The lattice energy is a measure of the strength
of ionic bonds within a specific crystal structure.
It is usually defined as the energy change when a
mole of a crystalline solid is formed from its
gaseous ions.
M+(g) + X-(g)  MX(s)
Lattice Energy
Your text uses a different convention,
defining the lattice energy as the energy change
when a mole of a crystalline solid is converted to
its gaseous ions.
MX(s)  M+(g) + X-(g)
This approach results in positive values for
lattice energy, and it requires energy to break
apart and vaporize the ions in the solid.
Lattice Energy
Lattice energies cannot be measured directly,
so they are obtained using Hess’ Law. They will
vary greatly with ionic charge, and, to a lesser
degree, with ionic size.
Starting with the elements is their standard
states, we know the enthalpies of sublimation,
ionization, bond dissociation, electron affinity,
and the enthalpy of formation for the ionic
solid.
1/2 bond
energy of Cl2
Electron
Affinity of Cl
Ionization
energy of K
∆Hsub of K}
∆Hf of
KCl
–Lattice Energy
of KCl
Ionic charge has a
huge effect on
lattice energy.
Ionic vs. Covalent Compounds
Ionic Compounds
 Crystalline solids (made of
ions)
 High melting and boiling
points
 Conduct electricity when
melted
 Many soluble in water but
not in nonpolar liquid
Covalent Compounds
 Gases, liquids, or solids
(made of molecules)
 Low melting and boiling
points
 Poor electrical conductors
in all phases
 Many soluble in nonpolar
liquids but not in water
Properties of NaCl and CCl4