Download Energy and Energy Changes Heat Transfer and The Measurement

Chapter 6
Energy and Chemical Reactions
Energy and Energy Changes
• Capacity to do work
– chemical, mechanical, thermal,
electrical, radiant, sound, nuclear
• Energy may affect matter
– e.g. raise its temperature, eventually
causing a state change
– All physical changes and chemical
changes involve energy changes
Heat Transfer and The
Measurement of Heat
• SI unit J (Joule)
• calorie
1 calorie = 4.184 J
• English unit = BTU
• Specific Heat
amount of heat required to raise the T of 1g
of a substance by 1oC
SH = units = J/goC
1
Example - Converting Calories to
Joules
Convert 60.1 cal to joules
1 cal = 4.184 joules
4.184 J
60.1cal ×
= 251J
1 cal
Energy and the Temperature of
Matter
• The amount the temperature of an
object increases depends on the
amount of heat added (q).
– If you double the added heat energy the
temperature will increase twice as much.
• The amount the temperature of an
object increases depends on its mass
– If you double the mass it will take twice as
much heat energy to raise the temperature
the same amount.
Heat Transfer and the
Measurement of Heat
• Heat capacity
amount of heat required to raise the T of a
substance by 1oC
• HC = J / oC
2
Heat Transfer and the
Measurement of Heat
• Heat transfer equation
necessary to calculate amounts of heat
amount of heat = amount of substance x
specific heat x ∆T
q = m × SH × ∆T
•Example – Calculate the amt. of heat to
raise T of 200 g of water from 10.0 oC to
55.0 oC
Heat Transfer and the
Measurement of Heat
• Heat transfer equation
necessary to calculate amounts of heat
amount of heat = amount substance x
specific heat x ∆T
q = m × SH × ∆T
? J = 200g H2O×
4.184J
× (55.0o C −10.0o C)
1 g H2O
= 3.76×104 J or 37.6 kJ
Specific Heat Capacity
• Specific Heat (s) is the amount of
energy required to raise the
temperature of one gram of a substance
by one Celsius degree
By definition , the specific heat of water is 4.184
J
g °C
Amount of Heat = Specific Heat x Mass x Temperature Change
Q = SH x m x ∆T
3
Example – Calculate the amount of
heat energy (in joules) needed to raise
the temperature of 7.40 g of water
from 29.0°C to 46.0°C
Specific Heat of Water = 4.184
JJ
gg -°°CC
Mass = 7.40 g
Temperature Change = 46.0°C – 29.0°C = 17.0°C
Q = SH x m x ∆T
J
Heat = 4.184
× 7.40g × 17.0°C = 526 J
g °C
Example – A 1.6 g sample of metal that
appears to be gold requires 5.8 J to raise
the temperature from 23°C to 41°C.
Is the metal pure gold?
Q = SH
× m × ∆T
Q
SH =
m × ∆T
∆ T = 41 ° C - 23 ° C = 18 ° C
5.8 J
J
s =
= 0.20
1.6 g x 18 ° C
g °C
Specific heat of gold is 0.13 g J°C
Therefore the metal cannot be pure gold.
The First Law of
Thermodynamics
•
Thermodynamics is the study of the changes
in energy and transfers of energy that
accompany chemical and physical processes.
1. Will two (or more) substances react when they
are mixed under specified conditions?
2. If they do react, what energy changes and
transfers are associated with their reaction?
3. If a reaction occurs, to what extent does it
occur?
4
The First Law of
Thermodynamics
• Exothermic reactions release energy in the form of
heat.
• For example, the combustion of propane is
exothermic.
C 3H 8 (g) + 5 O 2(g) → 3 CO 2(g) + 4H 2 O ( l ) + 2.22 ×103 kJ
•
The combustion of n-butane is also
exothermic.
2 C 4 H10(g) + 13 O 2(g) → 8 CO 2(g) + 10 H 2 O ( l ) + 5.78 × 103 kJ
The First Law of
Thermodynamics
• Exothermic
reactions
generate specific
amounts of heat.
• This is because
the potential
energies of the
products are
lower than the
potential energies
of the reactants.
The First Law of
Thermodynamics
•
There are two basic ideas of importance for
thermodynamic systems.
1. Chemical systems tend toward a state of
minimum potential energy.
Some examples of this include:
H2O flows downhill.
Objects fall when dropped.
The energy change for these two examples is:
Epotential = mgh
∆Epotential = mg(∆h)
5
The First Law of
Thermodynamics
2. Chemical systems tend toward a state of
maximum disorder.
• Common examples of this are:
– A mirror shatters when dropped and does
not reform.
– It is easy to scramble an egg and difficult to
unscramble it.
– Food dye when dropped into water
disperses.
The First Law of
Thermodynamics
• This law can be stated as, “The combined
amount of energy in the universe is
constant.”
• The first law is also known as the Law of
Conservation of Energy.
– Energy is neither created nor destroyed in
chemical reactions and physical changes.
Some Thermodynamic Terms
• The substances involved in the chemical and
physical changes under investigation are called
the system.
– In chemistry lab, the system is the chemicals inside
the beaker.
• The environment around the system is called the
surroundings.
– The surroundings are outside the beaker.
• The system plus the surroundings is called the
universe.
6
Some Thermodynamic Terms
• The set of conditions that specify all of the
properties of the system is called the
thermodynamic state of a system.
• For example the thermodynamic state could
include:
–
–
–
–
The number of moles and identity of each substance.
The physical states of each substance.
The temperature of the system.
The pressure of the system.
Some Thermodynamic Terms
• The properties of a system that depend only on the state
of the system are called state functions.
– State functions are always written using capital letters.
• The value of a state function is independent of pathway.
• An analog to a state function is the energy required to
climb a mountain taking two different paths.
–
–
–
–
–
E1 = energy at the bottom of the mountain
E1 = mgh1
E2 = energy at the top of the mountain
E2 = mgh2
∆E = E2-E1 = mgh2 – mgh1 = mg(∆h)
Some Thermodynamic Terms
• Notice that the energy change in moving from
the top to the bottom is independent of
pathway but the work required may not be!
• Some examples of state functions are:
– T, P, V, ∆E, ∆H, and S
•
Examples of non-state functions are:
– n, q, w
7
Some Thermodynamic Terms
•
In thermodynamics we are often interested in
changes in functions.
– We will define the change of any function X as:
– ∆X = Xfinal – Xinitial
•
•
If X increases ∆X > 0
If X decreases ∆X < 0.
Enthalpy Change
• Chemistry is commonly done in open
beakers on a desk top at atmospheric
pressure.
– Because atmospheric pressure only changes
by small increments, this is almost at constant
pressure.
• The enthalpy change, ∆H, is the change in
heat content at constant pressure.
– ∆H = qp
Enthalpy Change
• ∆Hrxn is the heat of reaction.
– This quantity will tell us if the reaction
produces or consumes heat.
– If ∆Hrxn < 0 the reaction is exothermic.
– If ∆Hrxn > 0 the reaction is endothermic.
• ∆Hrxn = Hproducts - Hreactants
– ∆Hrxn = Hsubstances produced - Hsubstances consumed
– Notice that this is ∆Hrxn = Hfinal – Hinitial
8
Calorimetry
• A coffee-cup
calorimeter is used to
measure the amount
of heat produced (or
absorbed) in a
reaction at constant P
– This is one method to
measure qP for
reactions in solution.
Calorimetry
• If an exothermic reaction is performed in a
calorimeter, the heat evolved by the reaction is
determined from the temperature rise of the
solution.
– This requires a two part calculation.
 Amount of heat
  Amount of heat
  Amount of heat


 = 
 + 

 released by reaction   absorbed by calorimete r   absorbed by solution 
• Amount of heat gained by calorimeter is called the heat
capacity of the calorimeter or calorimeter constant.
– The value of the calorimeter constant is determined by adding a
specific amount of heat to calorimeter and measuring the
temperature rise.
Calorimetry
• Example 15-1: When 3.425 kJ of heat is
added to a calorimeter containing 50.00 g
of water the temperature rises from
24.00oC to 36.54oC. Calculate the heat
capacity of the calorimeter in J/oC. The
specific heat of water is 4.184 J/g oC.
• This is a four part calculation.
9
Calorimetry
1. Find the temperature change.
∆T = (36.54 - 24.00) 0 C = 12.54 0 C
2. Find the heat absorbed by the water in going
from 24.000C to 36.540C.
q P = mC∆T
(
= (50.00 g) 4184
.
J
. C)
)(1254
0
g0 C
. J ≈ 2623 J
= 262337
Calorimetry
3. Find the heat absorbed by the calorimeter.
Take the total amount of heat added to calorimeter
and subtract the heat absorbed by the water.
3.425 kJ = 3425 J
(3425 J
- 2623 J ) = 802 J
4. Find the heat capacity of the calorimeter.
(heat absorbed by the calorimeter)/(temperature
change)
802 J
= 63.955
12.54 0 C
J
0
C
≈ 64.00
J
0
C
Calorimetry
• Example 15-2: A coffee-cup calorimeter is
used to determine the heat of reaction for
the acid-base neutralization
CH3COOH(aq) + NaOH(aq) → NaCH3COO(aq) + H2O(l)
When we add 25.00 mL of 0.500 M NaOH at
23.000oC to 25.00 mL of 0.600 M
CH3COOH already in the calorimeter at the
same temperature, the resulting
temperature is observed to be 25.947oC.
10
Calorimetry
The heat capacity of the calorimeter has
previously been determined to be 27.8 J/0C.
Assume that the specific heat of the mixture
is the same as that of water, 4.18 J/g0C and
that the density of the mixture is 1.02 g/mL.
Calorimetry
• This is a three part calculation.
1.Calculate the amount of heat given off in the reaction.
temperature change
∆T = (25.947 - 23.000) 0 C = 2.9470 C
heat absorbed by calorimeter
(
)
.
q = 2.9470 C ( 278
J0
) = 819. J
C
Calorimetry
temperature change
∆T = (25.947 - 23.000)0 C = 2.9470 C
heat absorbed by calorimeter
(
)
q = 2.9470C (278
.
J0
) = 819. J
C
mass of solution in calorimeter
102
. g
= 510
. g
(25.00 mL + 25.00 mL)
mL
11
Calorimetry
heat absorbed by solution
q = mC∆T
(
q = (51.0 g ) 418
.
J
)(2.947 C) = 628 J
0
g0C
total amount of heat produced by reaction
q = 81.9 J + 628 J = 709.9 J
Calorimetry
2. Determine ∆H for the reaction under the conditions
of the experiment.
•
We must determine the number of moles of reactants
consumed which requires a limiting reactant calculation.
CH 3COOH ( aq ) + NaOH ( aq ) → NaCH 3COO ( aq ) + H 2 O ( l )
(25.00 mL NaOH )
0.500 mmol NaOH 
×

1 mL NaOH
 1 mmol NaCH 3COO  = 12 .5 mmol NaCH COO
3
 1 mmol NaOH 
Calorimetry
C H 3C O O H
(a q )
+ N aO H
(a q )
→ N a C H 3C O O
(a q )
+ H 2O
(l )
0 .5 0 0 m m o l N a O H 
 ×
( 2 5 . 0 0 m L N a O H )

1 m L N aO H
 1 m m o l N a C H 3 C O O  = 1 2 .5 m m o l N a C H C O O


3


1 m m ol N aO H
(2 5 . 0 0
 0 .6 0 0 m m o l C H 3 C O O H 
m L C H 3 C O O H )
 ×

1 m L C H 3C O O H

 1 m m o l N a C H 3C O O 

 = 1 5 .0 m m o l N a C H 3 C O O
 1 m m o l C H 3C O O H 
12
Calorimetry
3. Finally, calculate the ∆Hrxn based on the limiting
reactant calculation.
∆H rxn
12.5 mmol = 0.0125 mol
709.9 J
=
= 56792 J / mol ≈ 56.8 kJ / mol
0.0125 mol
Thermochemical Equations
• Thermochemical equations are a balanced chemical
reaction plus the ∆H value for the reaction.
C5 H12(l ) + 8 O 2(g) → 5 CO 2(g) + 6 H 2 O ( l ) ∆H orxn = - 3523 kJ
• ∆H < 0 designates an exothermic reaction.
• ∆H > 0 designates an endothermic
reaction
Standard States and Standard
Enthalpy Changes
• Thermochemical standard state conditions
– The thermochemical standard T = 298.15 K.
– The thermochemical standard P = 1.0000 atm.
• Be careful not to confuse these values with STP.
• Thermochemical standard states of matter
– For pure substances in their liquid or solid phase the
standard state is the pure liquid or solid.
– For gases the standard state is the gas at 1.00 atm of
pressure.
• For gaseous mixtures the partial pressure must be 1.00 atm.
– For aqueous solutions the standard state is 1.00 M
concentration.
13
Standard Molar Enthalpies of
Formation, ∆Hfo
• The standard molar enthalpy of formation is defined
as the enthalpy for the reaction in which one mole
of a substance is formed from its constituent
elements.
– The symbol for standard molar enthalpy of formation is
∆ H f o.
• The standard molar enthalpy of formation for MgCl2
is:
Mg (s ) + Cl 2( g ) → MgCl 2(s ) + 6418
. kJ
∆H of MgCl 2 ( s ) = −6418
. kJ / mol
Standard Molar Enthalpies of
Formation, ∆Hfo
• Standard molar enthalpies of formation have been
determined for many substances and are tabulated in
Appendix L
• Standard molar enthalpies of elements in their most
stable forms at 298.15 K and 1.000 atm are zero.
• The standard molar enthalpy of formation for phosphoric acid is
-1281 kJ/mol. Write the equation for the reaction for
which ∆Horxn = -1281 kJ.
P in standard state is P4
Phosphoric acid in standard state is H3PO4(s)
Standard Molar Enthalpies of
Formation, ∆Hfo
3
2
H2( g) + 2 O2( g) + 14 P4(s) → H3PO4(s) + 1281 kJ
∆Hof H3PO4( s ) = −1281 kJ / mol
14
Standard Molar Enthalpies of
Formation, ∆Hfo
• Calculate the enthalpy change for the reaction of
one mole of H2(g) with one mole of F2(g) to form
two moles of HF(g) at 25oC and one atmosphere.
H
2 (g )
+
F2 (g )
→
s td . s ta te s td . s ta te
f o r th is rx n . ∆ H
o
298
= 2 ∆H
f r o m A p p e n d ix K , ∆ H
∆H
o
298
=
2 H F( g )
s td . s ta te
o
f
o
f
= −271 kJ / m ol
( 2 m o l )( − 2 7 1 k J / m o l ) = − 5 4 2 k J
Standard Molar Enthalpies of
Formation, ∆Hfo
• Example 15-6: Calculate the enthalpy
change for the reaction in which 15.0 g of
aluminum reacts with oxygen to form Al2O3
at 25oC and one atmosphere.
Standard Molar Enthalpies of
Formation, ∆Hfo
4 Al ( s) + 3 O2( g) → 2 Al 2 O3( s)
from Appendix K, ∆H of Al2O3 = −1676 kJ / mol
1 mol Al 2 mol Al 2 O3
×
×
27.0 g Al
4 mol Al
−1676 kJ
= −466 kJ
1 mol Al 2 O3
? kJ = 15.0 g Al ×
15
Hess’s Law
• Hess’s Law of Heat Summation states that the
enthalpy change for a reaction is the same
whether it occurs by one step or by any
(hypothetical) series of steps.
– Hess’s Law is true because ∆H is a state function.
• If we know the following ∆Ho’s
[1] 4 FeO(s) + O 2(g) → 2 Fe 2O3(s)
[2] 2 Fe(s) + O 2(g) → 2 FeO(g)
[3] 4 Fe(s) + 3 O 2(g) → 2 Fe 2O3(s)
∆H o = −560 kJ
∆H o = −544 kJ
∆H o = −1648 kJ
Hess’s Law
•
•
For example, we can calculate the ∆Ho for reaction [1] by properly
adding (or subtracting) the ∆Ho’s for reactions [2] and [3].
Notice that reaction [1] has FeO and O2 as reactants and Fe2O3 as a
product.
– Arrange reactions [2] and [3] so that they also have FeO and O2 as
reactants and Fe2O3 as a product.
• Each reaction can be doubled, tripled, or multiplied by half, etc.
• The ∆Ho values are also doubled, tripled, etc.
• If a reaction is reversed the sign of the ∆Ho is changed.
2 x [ − 2 ] 2 (2 F e O (s ) → 2 F e (s ) + O 2 ( g ) )
∆H 0
2( + 544 ) kJ
[ 3 ] 4 F e (s ) + 3 O 2 ( g ) → 2 F e 2 O 3 (s )
[1 ] 4 F e O (s ) + O 2 ( g ) → 2 F e 2 O 3
− 1648 kJ
− 560 kJ
Hess’s Law
• Example 15-7: Given the following
equations and ∆Ho values
o
∆H
( kJ )
[1] 2 N 2( g ) + O 2( g ) → 2 N 2 O ( g )
164.1
[2] N 2( g ) + O 2( g ) → 2 NO ( g )
180.5
[3] N 2( g ) + 2 O 2( g ) → 2 NO 2 ( g )
66.4
calculate ∆Ho for the reaction below.
N 2O
(g )
+ NO
2 (g
)
→ 3 NO
(g )
∆H
o
= ?
You do it!
16
Hess’s Law
• Use a little algebra and Hess’s Law to get the
appropriate ∆Ho values
o (kJ)
∆H∆
∆oH
H(kJ)
(kJ)
o
1 O
N
×[[−−
→
−11
1]]] N
N222OO
O( g( g) )→
→NN
N2(2g()g)++
+1212O
2( g2)( g)
2 O
112 ×
2
×[ 2]
33 ×
2
1 ×
2
[ − 3]
( g)
33
2
N2( g) +
2( g)
33
22
2( g)
-82.05
-82.05
-82.05
O22((gg)) →33NO
NO((gg)) 270.75
270.75
NO2( g) → 12 N2( g) + O2( g)
N2O( g) + NO2( g) → 3 NO( g)
-33.2
155.5
Hess’s Law
• The + sign of the ∆Ho value tells us that the reaction is
endothermic.
• The reverse reaction is exothermic, i.e.,
3 NO ( g ) → N 2 O ( g ) + NO 2 ( g )
∆H o = - 155.5 kJ
Hess’s Law
• Hess’s Law in a more useful form.
– For any chemical reaction at standard conditions, the
standard enthalpy change is the sum of the standard
molar enthalpies of formation of the products (each
multiplied by its coefficient in the balanced chemical
equation) minus the corresponding sum for the
reactants.
∆H 0rxn = ∑ n ∆H f0 products − ∑ n ∆H f0 reactants
n
n
n = stoichiometric coefficients
17
Hess’s Law
Hess’s Law
• Calculate the ∆H o298 for the following reaction
from data in Appendix L.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H 2O(l )
Hess’s Law
• Calculate ∆Ho298 for the following reaction from
data in Appendix L.
C3H8( g) + 5 O2( g) → 3 CO2( g) + 4 H2O( l )
[
][
∆Ho298 = 3∆Hof CO2( g ) + 4∆Hof H2O( l ) − ∆Hof C3H8( g ) + 5∆Hof O2 ( g)
]
= {[ 3(−3935
. ) + 4(−2858
. )] − [ (−1038
. ) + 5(0)]} kJ
18
Hess’s Law
• Calculate ∆Ho298 for the following reaction from
data in Appendix L.
C3H8( g) + 5 O2( g) → 3 CO2( g) + 4 H2O( l )
[
][
∆Ho298 = 3∆Hof CO2( g) + 4∆Hof H2O( l ) − ∆Hof C3H8( g) + 5∆Hof O2 ( g)
]
= {[ 3(−3935
. ) + 4(−2858
. )] − [ (−1038
. ) + 5(0)]} kJ
= -2211.9 kJ
∆Ho298 = − 22119
. kJ, and so the reaction is exothermic.
Hess’s Law
• Application of Hess’s Law and more algebra allows us to
calculate the ∆Hfo for a substance participating in a
reaction for which we know ∆Hrxno , if we also know
∆Hfo for all other substances in the reaction.
• Example 15-9: Given the following information, calculate
∆Hfo for H2S(g).
2 H2S(g) + 3 O2( g) → 2 SO2( g) + 2 H2O(l )
∆Hof
?
0
o
∆H298
= -1124 kJ
- 296.8 - 285.8
(kJ / mol)
You do it!
Hess’s Law
[
{
][
[
∆H o298 = 2∆H of SO2 ( g ) + 2∆H of H 2O ( l ) − 2∆H of H2S( g ) + 3∆H of O 2( g )
]}
]
− 1124 kJ = [ 2( −296.8 ) + 2( −285.8] − 2∆H of H 2S( g ) + 3( 0) kJ
now we solve for ∆H of H 2S( g )
2∆H of H 2S( g ) = −412
. kJ
∆H of H 2S( g ) = −20.6 kJ
19
Bond Energies
• Bond energy is the amount of energy required to
break the bond and separate the atoms in the
gas phase.
– To break a bond always requires an absorption of
energy!
A - B ( g ) + bond energy → A ( g ) + B ( g )
H - Cl ( g ) + 432 kJ mol → H ( g ) + Cl ( g )
Bond Energies
• Table of average bond energies
• Molecule
Bond Energy (kJ/mol)
•
F2
159
243
•
Cl2
•
Br2
192
•
O2 (double bond) 498
946
•
N2 (triple bond)
Bond Energies
• Bond energies can be calculated from other ∆Ho298 values.
20
Bond Energies
• Bond energies can be calculated from
other ∆Ho298 values
• Example 15-9: Calculate the bond energy for
hydrogen fluoride, HF.
H - F( g ) + BE HF → H ( g ) + F( g ) atoms NOT ions
or
H - F( g ) → H ( g ) + F( g )
[
∆H o298 = BE HF
] [
∆H o298 = ∆H of H ( g ) + ∆H of F( g ) − ∆H of HF( g )
]
Bond Energies
H - F( g ) + BE HF → H ( g ) + F( g ) atoms NOT ions
or
H - F( g ) → H ( g ) + F( g )
[
∆ H o298 = BE HF
] [
∆ H o298 = ∆ H of H ( g ) + ∆ H of F( g ) − ∆ H of HF( g )
∆ H o298
]
= [218 .0 kJ + 78.99 kJ ] − [ − 271 kJ ]
∆ H o298 = 568 .0 kJ
← BE for HF
Bond Energies
• Example 15-10: Calculate the average N-H
bond energy in ammonia, NH3.
You do it!
21
Bond Energies
NH 3(g ) → N (g ) + 3 H (g )
[
∆H o298 = 3 BE N -H
][
∆H o298 = ∆H fo N ( g ) + 3 ∆H fo H ( g ) − ∆H fo NH 3( g )
]
∆H o298 = {[(472.7) + 3(218)] − [− 46.11]} kJ
∆H o298 = 1173 kJ
average BE N-H =
1173 kJ
= 391 kJ mol N-H bonds
3
Changes in Internal Energy, ∆E
• The internal energy, E, is all of the energy
contained within a substance.
– This function includes all forms of energy
such as kinetic, potential, gravitational,
electromagnetic, etc.
– The First Law of Thermodynamics states that
the change in internal energy, ∆E, is
determined by the heat flow, q, and the work,
w.
Changes in Internal Energy, ∆E
∆E = E products − E reactants
∆E = q + w
q > 0 if heat is absorbed by thesystem.
q < 0 if heat is absorbed by thesystem.
w > 0 if the surroundings do work on the system.
w < 0 if the system does work on the surroundings.
22
Changes in Internal Energy, ∆E
• ∆E is negative when energy is released by a
system undergoing a chemical or physical
change.
– Energy can be written as a product of the process.
C5H12( l) + 8 O2(g) → 5 CO2(g) + 6 H 2O( l) + 3.516× 103 kJ
∆E = - 3.516× 103 kJ
Changes in Internal Energy, ∆E
 ∆E is positive when energy is absorbed by a
system undergoing a chemical or physical
change.
– Energy can be written as a reactant of the process.
5 CO2(g) + 6 H2O(l) + 3.516× 103 kJ → C5H12( l) + 8 O2(g)
∆E = + 3.516× 103 kJ
Changes in Internal Energy, ∆E
•
Example 15-12: If 1200 joules of heat are added
to a system in energy state E1, and the system
does 800 joules of work on the surroundings,
what is the :
1. energy change for the system, ∆Esys?
∆E = E 2 - E1 = q + w
∆ E = 1200 J + (-800 J)
∆ E = 400 J
∆ E sys = + 400 J
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Changes in Internal Energy, ∆E
2. energy change of the surroundings, ∆Esurr?
∆E surr = −400 J
Changes in Internal Energy, ∆E
3. energy of the system in the new state,
E 2?
∆Esys = E2 - E1
E2 = E1 + ∆Esys
= E1 + 400 J
Changes in Internal Energy, ∆E
• In most chemical and physical changes,
the only kind of work is pressure-volume
work.
• Pressure is force per unit area.
P =
force
F
= 2
area d
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Changes in Internal Energy, ∆E
• Volume is distance cubed.
V = d3
• P∆V is a work term, i.e., the same units are used
for energy and work.
( )
F
P∆V =  2  d3 = F × d ← which is work
d 
Relationship of ∆H and ∆E
• The total amount of heat energy that a
system can provide to its surroundings at
constant temperature and pressure is
given by
∆H= ∆E + P ∆V
– which is the relationship between ∆H and ∆E.
• ∆H = change in enthalpy of system
• ∆E = change in internal energy of system
• P∆V = work done by system
Relationship of ∆H and ∆E
• At the start of Chapter 6 we defined
∆H = qP.
• Here we define ∆H = ∆E + P∆V.
– Are these two definitions compatible?
• Remember ∆E = q + w.
• We have also defined w = -P∆V .
– Thus ∆E = q + w = q -P∆V
• Consequently, ∆H = q- P∆V + P∆V = q
– At constant pressure ∆H = qP.
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Relationship of ∆H and ∆E
• For reactions in which the volume change
is very small or equal to zero.
For small volume changes,
∆V ≈ 0 and P∆V ≈ 0.
Since ∆H = ∆E + P∆V then
∆H ≈ ∆E.
For no volume change,
∆H = ∆E.
Relationship of ∆H and ∆E
• Change in enthalpy, ∆H, or heat
of reaction is amount of heat
absorbed or released when a
reaction occurs at constant
pressure.
• The change in energy, ∆E, is the
amount of heat absorbed or
released when a reaction occurs
at constant volume.
• How much do the ∆H and ∆E for
a reaction differ?
– The difference depends on the
amount of work performed by the
system or the surroundings.
Relationship of ∆H and ∆E
 ∆Ho = -3523 kJ/mol for the combustion of
n-pentane, n-C5H12. Combustion of one
mol of n-pentane at constant pressure
releases 3523 kJ of heat. What are the
values of the work term and ∆E for this
reaction?
C5 H12( l ) + 8 O 2(g) → 5 CO 2(g) + 6 H 2 O ( l )
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Relationship of ∆H and ∆E
• Determine the work done by this reaction.
You do it!
C5 H12( l ) + 8 O 2(g) → 5 CO 2(g) + 6 H 2O ( l )
1442443 14442444
3
8 mol gas
5 mol gas
Since ∆H o = - 3523 kJ/mol, we know that T = 298 K.
w = - P∆V = - (∆n gas )RT
∆n gas = 5 − 8 mol = - 3 mol
w = -(-3 mol)(8.314 J mol K )( 298 K) = 7433 J = 7.433 kJ
Relationship of ∆H and ∆E
• Now calculate the ∆E for this reaction from the
values of ∆H and w that we have determined.
∆H = ∆E + P∆V ∴ ∆E = ∆H - P∆V
since w = - P∆V = 7.433 kJ
then P∆V = - 7.433 kJ
∆E = - 3523 kJ - (-7.433 kJ) = -3516 kJ
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