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Chapter 5
Energy & Work
Thermodynamics
Drill
Objectives
•
•
•
•
iWBAT
Distinguish between temperature and heat
Use work and heat values to solve for energy
Distinguish between the system and the
surroundings
Definition
• Thermodynamicsis the study of energy transformations.
Chemical Reactions
• Chemical reactions involve not just the
conversion of reactants into products, but also
involve an energy change in the form of heat—
heat released as the result of a reaction, or heat
absorbed as a reaction proceeds.
• Energy changes accompany all chemical
reactions and are due to rearranging of chemical
bonding.
Making Bonds
• Addition of energy is always a requirement for
the breaking of bonds but the breaking of bonds
in and of itself does not release energy.
• Energy release occurs when new bonds are
formed.
Bond Energy
• If more energy is released when new bonds form
than was required to break existing bonds, then
the difference will result in an overall release of
energy.
• If, on the other hand, more energy is required to
break existing bonds than is released when new
bonds form, the
difference will result overall in energy being
absorbed.
Overall Reaction Energy
• Whether or not an overall reaction releases or
requires energy depends upon the final balance
between the breaking and forming of chemical
bonds.
Energy is...
•
•
•
•
the ability to do work or produce heat.
conserved.
made of heat and work.
a state function. (Energy is a property that is
determined by specifying the condition or “state” (e.g.,
temperature, pressure, etc.) of a system or substance.)
• independent of the path, or how you get from
point A to B.
Thermodynamics
State functions are properties that are determined by the state
of the system, regardless of how that condition was achieved.
, pressure, volume, temperature
Examples: change in
energy or enthalpy
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
6.7
Energy
• While the total internal energy of a system (E)
cannot be determined, changes in internal
energy (E) can be determined.
• The change in internal energy will be the amount
of energy exchanged between a system and its
surroundings during a physical or chemical
change.
Δ E = E final - E initial
Definitions
• Work is a force acting over a distance.
• Heat is energy transferred between objects
because of temperature difference.
(Heat is not a property of a system or substance and is
not a state function. Heat is a process—the transfer of
energy from a warm to a cold object.)
System vs Surroundings
• The Universe is divided into two halves.
 system and the surroundings.
• In a chemistry setting, a system includes all
substances undergoing a physical or chemical
change.
• The surroundings would include everything else
that is not part of the system.
Heat
• Most commonly, energy is exchanged between a
system and its surroundings in the form of heat.
• Heat will be transferred between objects at
different temperatures.
• Thermochemistry is the study of thermal energy
changes.
Definition of Heat
• Heat energy (or just heat) is a form of energy
which transfers among particles in a substance
(or system) by means of kinetic energy of those
particle. In other words, under kinetic theory,
the heat is transfered by particles bouncing into
each other.
• http://physics.about.com/od/glossary/g/heat.htm
Definition of Temperature
• Temperature
is a measurement of the average kinetic energy
of the molecules in an object or system and can
be measured with a thermometer or a
calorimeter. It is a means of determining the
internal energy contained within the system.
Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between two
bodies that are at different temperatures.
Temperature is a measure of the thermal energy.
Temperature = Thermal Energy
900C
400C
greater thermal energy
6.2
Heat vs Temperature
• Note that temperature is different from heat,
though the two concepts are linked.
Temperature is a measure of the internal energy
of the system, while heat is a measure of how
energy is transferred from one system (or body)
to another. The greater the heat absorbed by a
material, the more rapidly the atoms within the
material begin to move, and thus the greater the
rise in temperature.
• http://physics.about.com/od/glossary/g/temperature.htm
Exo vs Endo
• Exothermic reactions release energy to the
surroundings.
• Endothermic reactions absorb energy from the
surroundings.
Potential energy
CH 4 + 2O 2  CO 2 + 2H 2 O + Heat
CH 4 + 2O 2
Heat
CO 2 + 2 H 2 O
N 2 + O 2 + heat  2NO
Potential energy
2NO
Heat
N2 + O2
Three Parts
Every energy measurement has three parts:
1. A unit ( Joules of calories).
2. A number.
3. a sign to tell direction.
negative - exothermic
positive- endothermic
Surroundings
System
Energy
DE <0
Surroundings
System
Energy
DE >0
Same rules for heat and work
• Heat given off is “negative”.
• Heat absorbed is “positive”.
• Work done by the system on the surroundings is
negative.
• Work done on the system by the surroundings is
positive.
First Law of Thermodynamics
• The energy of the universe is constant.
• It is also called the:
• Law of conservation of energy.
q = heat
w = work
• In a chemical system, the energy exchanged
between a system and its surroundings can be
accounted for by heat (q) and work (w).
DE = q + w
• Take the system’s point of view to decide
signs.
DE = q + w
Thermodynamics
DE is the change in internal energy of a system
q is the heat exchange between the system and the surroundings
w is the work done on (or by) the system
w = -PDV when a gas expands against a constant external pressure
6.7
Enthalpy and the First Law of Thermodynamics
DE = q + w
At constant pressure, q = DH and w = -PDV
DE = DH - PDV
DH = DE + PDV
6.7
Conservation of Energy
• Energy exchanged between a system and its
surroundings can be considered to off set one
another.
• The same amount of energy leaving a system will
enter the surroundings (or vice versa), so the
total amount of energy remains constant.
Metric Units
• The SI (Metric System) unit for all forms of
energy is the joule (J).
Heat and Work
• DE = q + w
• - q is exothermic
• +q is endothermic
-q = -∆H
• -w is done “by” the system
• +w is done “on” the system
• Note:
• ∆H stands for enthalpy which is the heat of
reaction
Practice Problem
• A gas absorbs 28.5 J of heat and then performs
15.2 J of work. The change in internal
• energy of the gas is:
(a) 13.3 J
(b) - 13.3 J
(c) 43.7 J
(d) - 43.7 J
(e) none of the above
Answer
• (b) E = q + w
• 28.5 J - 15.2 J = + 13.3 J
Practice Problem
Which of the following statements correctly
describes the signs of q and w for the following
exothermic process at 1 atmosphere pressure and
370 Kelvin?
H2O(g) → H2O(l)
(a) q and w are both negative
(b) q is positive and w is negative
(c) q is negative and w is positive
(d) q and w are both positive
(e) q and w are both zero
Answer
• (c) An exothermic indicates q is negative and the
gas is condensing to a liquid so it is exerting less
pressure on its surroundings indicating w is
positive.
What is work?
• Work is a force acting over a distance.
w= F x Dd
P = F/ area
d = V/area
w= (P x area) x D (V/area)= PDV
• Work can be calculated by multiplying pressure
by the change in volume at constant pressure.
• Use units of liter•atm or L•atm
Pressure and Volume Work
• Work refers to a force that moves an object over
a distance.
• Only pressure/volume work
(i.e., the expansion/contraction of a gas) is of
significance in chemical systems and only when
there is an increase or decrease in the amount of
gas present.
Work needs a sign
• If the volume of a gas increases, the system has
done work on the surroundings.
• work is negative
• w = - PDV
• Expanding work is negative.
• Contracting,
surroundings do work on the system W is positive.
• 1 L•atm = 101.3 J
Example
• When, in a chemical reaction, there are more
moles of product gas compared to
reactant gas, the system can be thought of as
performing work on its surroundings (making w
< 0) because it is “pushing back,” or moving
back the atmosphere to make room for the
expanding gas.
When the reverse is true, w > 0.
Compressing and Expanding Gases
• Compressing gas
▫ Work on the system is positive
▫ Work is going into the system
• Expanding gas
▫ Work on the surroundings is negative
▫ Work is leaving the system
Clarification Info
• If the reaction is performed in a rigid container,
there may be a change in pressure,
but if there is no change in volume, the
atmosphere outside the container didn’t “move”
and without movement, no work is done by or
on the system.
• If there is no change in volume (V= 0), then no
work is done by or on the system (w= 0) and the
change in internal energy will be entirely be due
to the heat involved ( ΔE = q).
Examples
• What amount of work is done when 15 L of gas
is expanded to 25 L at 2.4 atm pressure?
• If 2.36 J of heat are absorbed by the gas above.
what is the change in energy?
• How much heat would it take to change the gas
without changing the internal energy of the
gas?
Enthalpy
Enthalpy
• The symbol for Enthalpy is “H”
• H = E + PV (that’s the definition)
• DH = DE + PDV (at constant pressure)
Enthalpy (H) is used to quantify the heat flow into or out of a
system in a process that occurs at constant pressure.
DH = H (products) – H (reactants)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
DH < 0
Hproducts > Hreactants
DH > 0
6.3
DH = DE + PDV
Using
DH = DE + PDV
• the heat at constant pressure qp can be
calculated from:
• DE = qp + w
(if w = - PDV then…)
• DE = qp – PDV (now rearrange)
• qp = DE + P DV = DH
Examples of Enthapy Changes
• KOH(s) → K(aq) + OH-1 (aq)
ΔHsolution = - 57.8 kJ mol1
• C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔHcombustion = -2221kJ/mol
• H2O(s) → H2O(l)
ΔHfusion = 6.0 kJ/mol
• Fe2O3(s) + 2Al(s) → Al2O3(s) + 2Fe(s)
ΔHreaction - 852 kJ/mol
• Ca(s) + O2(g) H2(g) → Ca(OH)2(s)
ΔHformation - 986 kJ/mol1
3 Methods
•
•
There are a variety of methods for calculating
overall enthalpy changes that you should be
familiar with.
The three most common are the:
1. the use of Heats of Formation
2.
Hess’s Law
3.
the use of Bond Energies
Heat of Reaction
• To compare heats of reaction for different
reactions, it is necessary to know the
temperatures at which heats of reaction are
measured and the physical states of the
reactants and products.
• Look in the Appendix of the textbook to find
Standard Enthapy tables.
Standard Enthalpy of Formation
• Measurements have been made and tables
constructed of Standard Enthalpies of Formation
with reactants in their “standard states”.
• Use the symbol DHºf
• Standard state is the most stable physical state of
reactants at:
1 atmosphere pressure
specified temperature—usually 25 °C
1 M solutions
• For solids which exist in more than one allotropic
form, a specific allotrope must be specified.
• Because there is no way to measure the absolute
value of the enthalpy of a substance, must I
measure the enthalpy change for every reaction
of interest?
• Establish an arbitrary scale with the standard
enthalpy of formation (DH0) as a reference
point for all enthalpy expressions.
Standard enthalpy of formation (DH0)fis the heat change
that results when one mole of a compound is formed from
its elements at a pressure of 1 atm.
The standard enthalpy of formation of any element in its
most stable form is zero.
DH0f (O2) = 0
DH0 (O3) = 142 kJ/mol
f
DH0f (C, graphite) = 0
DH0 (C, diamond) = 1.90 kJ/mol
f
6.5
6.5
ΔH°formation
• It is important to recognize that the
ΔH°formation (abbreviated as ΔH°f) is really just
the heat of reaction for a chemical change
involving the formation of a compound from its
elements in their standard states.
Standard Enthalpies of Formation
• The standard heat of formation is the amount
of heat needed to form 1 mole of a compound
from its elements in their standard states.
• See the table in the Appendix
• Remember: For an element the value is 0
Equation Practice
• You need to be able to write the equation
correctly before solving the problem.
Try…
• What is the equation for the formation of NO2 ?
Try writing the equation.
Practice Answer
• ½N2 (g) + O2 (g)  NO2 (g)
• You must make one mole to meet the definition.
The standard enthalpy of reaction (DH0 rxn
) is the enthalpy of a
reaction carried out at 1 atm.
aA + bB
cC + dD
0
DHrxn
= [ cDH0f (C) + dDH0f (D) ] - [aDH0f (A) + bDH0f (B) ]
0
DHrxn
= S nDH0f (products) - S mDHf0 (reactants)
6.5
• Hess’s Law: When reactants are converted to
products, the change in enthalpy is the same
whether the reaction takes place in one step or in
a series of steps.
• (Enthalpy is a state function. It doesn’t matter
how you get there, only where you start and
end.)
Since we can manipulate the
equations
• We can use heats of formation to figure out
the heat of reaction.
• Lets do it with this equation.
• C2H5OH +3O2(g)  2CO2 + 3H2O
• which leads us to this rule.
( DH of products) - ( DH of reactants) = DH o
Hess’s Law
• Definition:
• When a reaction may be expressed as the
algebraic sum of other reactions, the enthalpy
change of the reaction is the algebraic sum of
the enthalpy changes for the combined
reactions.
Hess’ Law
• Enthalpy is a state function.
• It is independent of the path.
• We can add equations to come up with the
desired final product, and add the DH values.
• Two rules to remember:
• If the reaction is reversed the sign of DH is
changed
• If the reaction is multiplied, so is DH
Enthalpy
• As enthalpy is an extensive property, the
magnitude of an enthalpy change for a chemical
reaction depends upon the quantity of material
that reacts.
• This means:
if the amount of reacting material in an
exothermic reaction is doubled, twice the
quantity of heat energy
will be released.
Thermochemical Equations
• The stoichiometric coefficients always refer to the number
of moles of a substance
H2O (s)
H2O (l)
DH = 6.01 kJ
• If you reverse a reaction, the sign of DH changes
H2O (l)
H2O (s)
DH = -6.01 kJ
• If you multiply both sides of the equation by a factor n,
then DH must change by the same factor n.
2H2O (s)
2H2O (l)
DH = 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)
DH = 6.01 kJ
H2O (l)
H2O (g)
DH = 44.0 kJ
How much heat is evolved when 266 g of white
phosphorus (P4) burn in air?
P4 (s) + 5O2 (g)
266 g P4 x
1 mol P4
123.9 g P4
P4O10 (s)
DH = -3013 kJ
3013 kJ
= 6470 kJ
x
1 mol P4
6.3
For the oxidation of sulfur dioxide gas
• SO2(g) + ½O2(g) → SO3(g)
ΔH° = - 99 kJ/mol
• Doubling the reaction results in:
• 2SO2(g) + O2(g) → 2SO3(g)
ΔH° = - 198 kJ/mol
Notice that if you double the reaction, you must
double the ΔH° value.
Sign Change of ΔH
• 2SO2(g) + O2(g) → 2SO3(g)
ΔH° = - 198 kJ/mol
• If the reaction is written as an endothermic
reaction:
• 2SO3(g) → 2SO2(g) + O2(g)
ΔH° = + 198 kJ/mol
Tips for Hess’s Law Problems
• It is always a good idea to begin by looking for
species that appear as reactants and products
in the overall reaction.
• This will provide a clue as to whether a reaction
needs to be reversed or not.
• Second, consider the coefficients of species that
appear in the overall reaction.
• This will help determine whether a reaction
needs to be multiplied before the overall
summation.
Calculate the standard enthalpy of formation of CS2 (l)
given that:
C(graphite) + O2 (g)
CO2 (g) DH0 rxn= -393.5 kJ
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
SO2 (g)
DH0 rxn
= -296.1 kJ
CO2 (g) + 2SO2 (g) DH0 rxn
= -1072 kJ
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic)
CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
C(graphite) + O2 (g)
2S(rhombic) + 2O2 (g)
+ CO2(g) + 2SO2 (g)
6.5
CO2 (g) DH0 rxn= -393.5 kJ
2SO2 (g) DH0 rxn
= -296.1x2 kJ
CS2 (l) + 3O2 (g) DH0 rxn
= +1072 kJ
C(graphite) + 2S(rhombic)
CS2 (l)
0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ
DHrxn
Example
C(s) + O2(g) → CO2(g)
ΔH°f = - 394 kJ/mol
2H2(g) + O2(g) → 2H2O(l)
ΔH°f = - 572 kJ/mol
CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ΔH°f = + 891 kJ/mol
----------------------------------------------------C(s) + 2H2(g) → CH4(g)
ΔH°f = - 75 kJ/mol
Example
C(s) + O2(g) → CO2(g)
ΔH°f = - 394 kJ/mol
2H2(g) + O2(g) → 2H2O(l)
ΔH°f = - 572 kJ/mol
CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ΔH°f = + 891 kJ/mol
----------------------------------------------------C(s) + 2H2(g) → CH4(g)
ΔH°f = - 75 kJ/mol
GivenLaw Example
Hess’s
O 2 (g) + H 2 (g)  2OH(g) DHº= +77.9kJ
O 2 (g)  2O(g) DHº= +495 kJ
H 2 (g)  2H(g) DHº= +435.9kJ
Calculate DHº for this reaction
O(g) + H(g)  OH(g)
• Given
Example
5
O 2 (g)  2CO 2 (g) + H 2 O( l)
2
DHº= -1300. kJ
C(s) + O 2 (g)  CO 2 (g)
DHº= -394 kJ
1
H 2 (g) + O 2 (g)  H 2 O(l)
calculate
2
DHº= -286 kJ
DHº for this reaction
C 2 H 2 (g) +
2C(s) + H 2 (g)  C 2 H 2 (g)
Benzene (C6H6) burns in air to produce carbon dioxide and
liquid water. How much heat is released per mole of
benzene combusted? The standard enthalpy of formation
of benzene is 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
0
DHrxn
= S nDH0f (products) - S mDHf0 (reactants)
0
DHrxn
= [ 12DHf0 (CO2) + 6DH0f (H2O)] - [ 2DHf0 (C6H6)]
0
DHrxn
= [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ
-5946 kJ
= - 2973 kJ/mol C6H6
2 mol
6.5
Problems to Try
• Try 12-3 Practice Problems
H (kJ)
O2 NO2
-112 kJ
180 kJ
N2 2O2
NO2
68 kJ
• Collegeboard. (2007-2008). Professional Development workshop materials: Special
focus thermochemistry.
http://apcentral.collegeboard.com/apc/public/repository/5886-3_Chemistry_pp.ii88.pdf