Transcript Chapter 15 PPT

Chapter 15

Chemical Thermodynamics
1
Chapter Goals
1.
2.
3.
4.
5.
6.
7.
Heat Changes and Thermochemistry
The First Law of Thermodynamics
Some Thermodynamic Terms
Enthalpy Changes
Calorimetry
Themochemical Equations
Standard States and Standard Enthalpy Changes
Standard Molar Enthalpies of Formation, DHfo
2
Chapter Goals
Hess’s Law
9. Bond Energies
10. Changes in Internal Energy, DE
11. Relationship of DH and DE
Spontaneity of Physical and Chemical Changes
8. The Two Aspects of Spontaneity
9. The Second Law of Thermodynamics
10. Entropy, S
11. Free Energy Change, DG, and Spontaneity
12. The Temperature Dependence of Spontaneity
8.
3
The First Law of Thermodynamics


1.
2.
3.
Thermodynamics is the study of the changes in
energy and transfers of energy that accompany
chemical and physical processes.
In this chapter we will address 3 fundamental
questions.
Will two (or more) substances react when they are
mixed under specified conditions?
If they do react, what energy changes and transfers
are associated with their reaction?
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.
C3H8(g)  5 O2(g)  3 CO2(g)  4H 2O(  )  2.22 103 kJ

The combustion of n-butane is also exothermic.
2 C4 H10(g)  13 O2(g)  8 CO2(g)  10 H 2O()  5.78 103 kJ
5
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.
6
The First Law of Thermodynamics

1.

There are two basic ideas of importance for
thermodynamic systems.
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
DEpotential = mg(Dh)
7
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.
8
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.
9
Some Thermodynamic Terms

The substances involved in the chemical and
physical changes under investigation are called the
system.


The environment around the system is called the
surroundings.


In chemistry lab, the system is the chemicals inside the
beaker.
The surroundings are outside the beaker.
The system plus the surroundings is called the
universe.
10
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.
11
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
DE = E2-E1 = mgh2 – mgh1 = mg(Dh)
12
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, DE, DH, and S
Examples of non-state functions are:

n, q, w
13
Some Thermodynamic Terms

In thermodynamics we are often interested in
changes in functions.




We will define the change of any function X as:
DX = Xfinal – Xinitial
If X increases DX > 0
If X decreases DX < 0.
14
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, DH, is the change in
heat content at constant pressure.

DH = qp
15
Enthalpy Change

DHrxn is the heat of reaction.




This quantity will tell us if the reaction produces
or consumes heat.
If DHrxn < 0 the reaction is exothermic.
If DHrxn > 0 the reaction is endothermic.
DHrxn = Hproducts - Hreactants


DHrxn = Hsubstances produced - Hsubstances consumed
Notice that this is DHrxn = Hfinal – Hinitial
16
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.
17
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.
18
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.
19
Calorimetry
1.
Find the temperature change.
DT = 36.54 - 24.00 C = 12.54 C
0
2.
0
Find the heat absorbed by the water in going from
24.000C to 36.540C.
q P  mCDT

= 50.00 g 4184
.
J
12.54 C
0
0
g C
 262337
. J  2623 J
20
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
4.
- 2623 J  = 802 J
Find the heat capacity of the calorimeter.

(heat absorbed by the calorimeter)/(temperature change)
802 J
 63.955 J  64.0 J
C
C
12.54 C
0
0
0
21
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()
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.
22
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.
23
Calorimetry

This is a three part calculation.
1.
Calculate the amount of heat given off in the reaction.
temperature
change
temperature
change
temperature
change
0
00  0 C = 2.947
00
D
T
=
25.947
23.000

23.000 C
C == 2.947
2.947 CC C
DT =  25.947 - 23.000
heat absorbed by calorimeter
heat absorbed by calorimeter
2.947
 819
. J


0 C 27.8
J
q =  2.947 C 27.8 C  819
. J
mass of solution in calorimeter
q =
0
J
0
25.00 mL + 25.00 mL
0
C
102
. g
 510
. g
mL
24
Calorimetry
heat absorbed by solution
q = mCDT

q = 51.0 g  418
.
J
2.947 C  628 J
0
g0 C
total amount of heat produced by reaction
q = 81.9 J + 628 J = 709.9 J
25
Calorimetry
2.
Determine DH 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.
CH33COOH
COOH  aq  ++NaOH
NaOH  aq  
NaCH
NaCH
+H O
CH
3COO
3COO
 aq 
 aq 
 aqaq+ H 2 O2 l   l 
0.500 mmol NaOH 

 25.00 mL NaOH0.500 mmol NaOH   
1 mL NaOH  
25.00 mL NaOH 



1 mL NaOH
 1 mmol NaCH 3COO   12.5 mmol NaCH COO


3


1
mmol
NaCH
COO
1
mmol
NaOH


3
  12.5 mmol NaCH 3COO
  0.600 mmol CH 3COOH 
1
mmol
NaOH
25
.
00
mL
CH
COOH


 
3



1 mL CH 3COOH

 1 mmol NaCH 3COO 

  15.0 mmol NaCH 3COO 26
 1 mmol CH 3COOH 
Calorimetry
3.
Finally, calculate the DHrxn based on the limiting reactant
calculation.
DH rxn
12.5 mmol = 0.0125 mol
709.9 J

 56792 J / mol  56.8 kJ / mol
0.0125 mol
27
Thermochemical Equations

Thermochemical equations are a balanced chemical
reaction plus the DH value for the reaction.

For example, this is a thermochemical equation.
C5 H12( )  8 O 2(g)  5 CO2(g)  6 H 2O (  )  3523 kJ
1 mole


8 moles
5 moles
6 moles
The stoichiometric coefficients in thermochemical
equations must be interpreted as numbers of moles.
1 mol of C5H12 reacts with 8 mol of O2 to produce 5
mol of CO2, 6 mol of H2O, and releasing 3523 kJ is
referred to as one mole of reactions.
28
Thermochemical Equations

This is an equivalent method of writing thermochemical
equations.
C5H12()  8 O2(g)  5 CO2(g)  6 H2O() DHorxn  - 3523 kJ


DH < 0 designates an exothermic reaction.
DH > 0 designates an endothermic reaction
29
Thermochemical Equations

Example 15-3: Write the thermochemical
equation for the reaction in Example 15-2.
You do it!
CH 3COOH aq  + NaOH aq   NaCH 3COO aq  + H 2O l  DH = -56.8 kJ / mol
30
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.
31
Standard Molar Enthalpies of
Formation, DHfo

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 DHfo.
The standard molar enthalpy of formation for MgCl2 is:
Mg  s   Cl 2 g   MgCl 2s   6418
. kJ
o
DH f MgCl 2  s 
 6418
. kJ / mol
32
Standard Molar Enthalpies of
Formation, DHfo



Standard molar enthalpies of formation have been
determined for many substances and are tabulated in Table
15-1 and Appendix K in the text.
Standard molar enthalpies of elements in their most stable
forms at 298.15 K and 1.000 atm are zero.
Example 15-4: The standard molar enthalpy of formation for
phosphoric acid is -1281 kJ/mol. Write the equation for the reaction
for whichDHorxn = -1281 kJ.
P in standard state is P4
Phosphoric acid in standard state is H3PO4(s)
You do it!
33
Standard Molar Enthalpies of
Formation, DHfo
3
2
H2g  2 O2g  14 P4s  H3PO4s  1281 kJ
o
DHf H3PO4 s 
 1281 kJ / mol
34
Standard Molar Enthalpies of
Formation, DHfo

Example 15-5: 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 
std. state std. state

2 H F g 
std. state
for this rxn. D H o298  2 D H of
from A ppendix K , D H of   271 kJ / m ol
D H o298 
 2 m ol   271 kJ / m ol    542 kJ
35
Standard Molar Enthalpies of
Formation, DHfo

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.
You do it!
36
Standard Molar Enthalpies of
Formation, DHfo
4 Al s + 3 O 2g  2 Al 2 O 3s
from Appendix K,
o
DH f Al2O3
 1676 kJ / mol
1 mol Al 2 mol Al 2 O 3
? kJ = 15.0 g Al 


27.0 g Al
4 mol Al
1676 kJ
 466 kJ
1 mol Al 2 O 3
37
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 DH is a state function.
If we know the following DHo’s
1 4 FeO(s)  O 2(g)  2 Fe2O3(s)
2 2 Fe(s)  O 2(g)  2 FeO(g)
3 4 Fe(s)  3 O 2(g)  2 Fe2O3(s)
DH o  560 kJ
DH o  544 kJ
DH  1648 kJ
o
38
Hess’s Law


For example, we can calculate the DHo for reaction [1] by properly adding
(or subtracting) the DHo’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 DHo values are also doubled, tripled, etc.
If a reaction is reversed the sign of the DHo is changed.
2 x [2] 2(2 FeO  s   2 Fe  s   O 2 g  )
 3 4 Fe s   3 O 2 g   2 Fe 2 O 3s 
1
4 FeO  s   O 2 g   2 Fe 2 O 3
DH 0
2( 544 ) kJ
 1648 kJ
 560 kJ
39
Hess’s Law

Example 15-7: Given the following equations
and DHo values
DH o 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 
calculate DHo for the reaction below.
N 2 O g  + NO2 g   3 NOg 
You do it!
66.4
DH o  ?
40
Hess’s Law

Use a little algebra and Hess’s Law to get the
appropriate DHo values
oo
DHDDoH
H(kJ)
(kJ)
(kJ)
-82.05
NN222ggg++12112O
- 82.05
11 NN22OOg g 

-82.05
2O
2O
g22gg
3  2
3 N
33 O
+
 3 NO 270.75


22
22 22gg 
22 22gg 3 NOgg 270.75
1   3 NO
1 N

- 33.2


2
2
2 g 
2 g  + O 2 g 
11 
22
N 2 O  g   NO 2 g   3 NO  g 
155.5
41
Hess’s Law


The + sign of the DHo 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 
DH
o
= -155.5 kJ
42
Hess’s Law

Hess’s Law in a more useful form.

DH
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.
0
rxn
  n DH
n
0
f products
  n DH
0
f reactants
n
n  stoichiome tric coefficien ts
43
Hess’s Law

Example 15-8: Calculate the DH o298 for the
following reaction from data in Appendix K.
C3H 8(g)  5 O 2(g)  3 CO 2(g)  4 H 2 O (  )
44
Hess’s Law

Example 15-8: CalculateDHo298 for the following
reaction from data in Appendix K
H88(g)
 5O 2(g)2
3CO 2(g)2g 4H
O2(O
C33H
+ 4 2H
 )  l
g  3 CO
g + 5 O

  
 
DHo  3DHo
 4DHo
 DH o  5DH o
DH298  3DHf CO2 g  4DHf H2Ol   DHf C3H8 g  5DHf O2 g
DH  3 393.5  4 285.8   103.8  50.0 kJ
0
rxn
0
rxn
0
0
fCO 2(g)
fH 2 O (  )
0
0
fC 3 H 8 (g)
fO 2(g)
 3(3935
. )  4(2858
. )  (1038
. )  5(0) kJ
DH  2219.9 kJ
0
rxn
DH   2219.9 kJ, and so the reaction is exothermic.
0
rxn
45
Hess’s Law


Application of Hess’s Law and more algebra allows us to
calculate the DHfo for a substance participating in a reaction
for which we know DHrxno , if we also know DHfo for all other
substances in the reaction.
Example 15-9: Given the following information, calculate
DHfo for H2S(g).
2 H2Sg + 3 O2g 2 SO2g +2 H2Ol
DHo298 = -1124 kJ
DHof ?
(kJ / mol)
0
-296.8 -285.8
You do it!
46
Hess’s Law


1124 kJ  2(2968
. )  2(2858
.   2DH
DHo298  2DHof SO2g  2DHof H2Ol   2DHof H2Sg  3DHof O2g
o
f H2S g


 3(0) kJ
now we solve for DHof H2Sg
2DHof H2Sg  412
. kJ
DHof H2Sg  206
. kJ
47
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 
48
Bond Energies







Table of average bond energies
Molecule
Bond Energy (kJ/mol)
F2
159
Cl2
243
Br2
192
O2 (double bond)
498
N2 (triple bond)
946
49
Bond Energies

Bond energies can be calculated from otherDHo298 values.
50
Bond Energies


Bond energies can be calculated from
otherDHo298 values
Example 15-10: Calculate the bond energy for hydrogen
fluoride, HF.
H - Fgg 
BEHF

H
atomsNOT
NOTions
ions
 BE
H
Hggg FFggg atoms
HF
HF 
or
or
or
H - Fgg 
Hgg  FFgg
H

o
DDHHo298
BEHFHF
298 BE
HF
 

oo
oo
oo
oo
D
H

D
H

D
H

D
H
DH 298
 DHf HF
 g g 
 gg 
298  DHff H
H gg   DHf f FF
f HF
DH o298   218.0 kJ + 78.99 kJ    271 kJ 
DH o298
 568.0 kJ
 BE for HF
51
Bond Energies

Example 15-11: Calculate the average N-H bond
energy in ammonia, NH3.
You do it!
52
Bond Energies
NH3g   N g  + 3 H g 

DH
o
298
 3 BE N-H
 DH

DH
o
298
 DH
DH
o
298
 (472.7)  3(218)   46.11 kJ
DH
o
298
 1173 kJ
o
f Ng 
average BE N-H
 3 DH
o
f Hg 
o
f NH3 g 
1173 kJ

 391 kJ mol N-H bonds
3
53
Bond Energies

In gas phase reactions DHo values may be
related to bond energies of all species in the
reaction.
o
DH 298
  BE reactants   BE products
54
Bond Energies

Example 15-12: Use the bond energies listed
in Tables 15-2 and 15-3 to estimate the heat of
reaction at 25oC for the reaction below.
CH 4(g)  2 O 2(g)  CO2(g)  2 H 2O(g)
DH o298  [4 BE C-H  2 BE OO ] - [2 BE CO  4 BE O-H ]
55
Bond Energies
CH 4(g)  2 O 2(g)  CO 2(g)  2 H 2 O (g)
DH
o
298
DH
o
298
 [4 BE C-H  2 BE O O ] - [2 BE CO  4 BE O-H ]
 [4 (414)  2 (498)] - [2 (741)  4 (464)] kJ
DH
o
298
 - 686 kJ
56
Changes in Internal Energy, DE

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, DE, is determined by
the heat flow, q, and the work, w.
57
Changes in Internal Energy, DE
D
E=
 E reactants
D
E
DEE
E E products
EE
products
products
reactants
reactants
DEE
D
E q=wwq + w
D
if heat
by by
the system.
qq 00 if
heatisisabsorbed
absorbed
the system.
q  0 if heat is absorbed by the surroundings.
q  0 if heat is absorbed by the surroundin gs.
w  0 if the surroundings do work on the system.
w  0 if the system does work on the surroundings.
58
Changes in Internal Energy, DE

DE 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.
C5 H12(  )  8 O 2(g)  5 CO2(g)  6 H 2O(  )  3.516  103 kJ
DE  - 3.516  103 kJ
59
Changes in Internal Energy, DE

DE 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 H 2O(  )  3.516  103 kJ  C5 H12(  )  8 O 2(g)
DE   3.516  103 kJ
60
Changes in Internal Energy, DE

1.
Example 15-13: 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 :
energy change for the system, DEsys
DE
DE
DE
DE sys
= E 2 - E1 = q + w
= 1200 J + (-800 J)
= 400 J
= + 400 J
61
Changes in Internal Energy, DE
2.
energy change of the surroundings, DEsurr
You do it!
DE surr  400 J
62
Changes in Internal Energy, DE
3.
energy of the system in the new state, E2
DE sys  E 2 - E1
E 2  E1  DE sys
 E1  400 J
63
Changes in Internal Energy, DE


In most chemical and physical changes, the
only kind of work is pressure-volume work.
Pressure is force per unit area.
force
F
P =
 2
area
d
64
Changes in Internal Energy, DE

Volume is distance cubed.
V = d

3
PDV is a work term, i.e., the same units are used for
energy and work.
 
F 3

PDV =  2  d  F  d  which is work
d 
65
Changes in Internal Energy, DE

This movie shows a gas doing work to prove
that PDV is a different way of writing work.
66
Changes in Internal Energy, DE

Using the ideal gas law PV = nRT, we can look at
volume changes of ideal gases at constant T and P
due to changes in the number of moles of gas
present, Dngas.
P V = nR T


P  D V   D n gas R T

Dngas = (number of moles of gaseous products) - (number of
moles of gaseous reactants)
67
Changes in Internal Energy, DE

Work is defined as a force acting through a specified
distance.
w  F  d  -PDV  -Dn gas RT
Thus we can see that w  -Dn gas RT
at constant T and P.

Consequently, there are three possibilities for volume
changes:
68
Changes in Internal Energy, DE
1.
2.
3.
When Then
V2 = V1 PDV = 0
Dngas = 0
V2 > V1 PDV > 0
Dngas > 0
V2 < V1 PDV < 0
Dngas < 0
Examples
CO(g)  H 2 O (g)  H 2 (g)  CO2(g)

 


2 mol gas
2 mol gas
Zn (s)  2 HCl (aq)  ZnCl 2(aq)  H 2(g)
 
0 mol gas
1 mol gas
N 2(g)  3 H 2(g)  2 NH3(g)






4 mol gas
2 mol gas
69
Changes in Internal Energy, DE

Consider the following gas phase reaction at
constant pressure at 200oC.
2 NO  g  + O 2 g   2 NO 2 g 
 


3 mol gas
2 mol gas
V2  V1 thus DV  0 and PDV  0.
Consequent ly, w  - PDV  0.
Work is done on system by surroundin gs.
70
Changes in Internal Energy, DE

Consider the following gas phase reaction at
constant pressure at 1000oC.
PCl5g   PCl3g   Cl2g 
 

1 mol gas
2 mol gas
V2  V1 thus DV  0 and PDV  0.
Consequent ly, w  - PDV  0.
Work is done by the system on the surroundin gs.
71
Relationship of DH and DE

The total amount of heat energy that a system
can provide to its surroundings at constant
temperature and pressure is given by
DH= DE + P DV




which is the relationship between DH and DE.
DH = change in enthalpy of system
DE = change in internal energy of system
PDV = work done by system
72
Relationship of DH and DE


At the start of Chapter 15 we defined
DH = qP.
Here we define DH = DE + PDV.



Remember DE = q + w.
We have also defined w = -PDV .


Are these two definitions compatible?
Thus DE = q + w = q -PDV
Consequently, DH = q- PDV + PDV = q

At constant pressure DH = qP.
73
Relationship of DH and DE

For reactions in which the volume change is
very small or equal to zero.
For small volume changes,
DV  0 and PDV  0.
Since DH  DE  PDV then
DH  DE.
For no volume change,
DH  DE.
74
Relationship of DH and DE



Change in enthalpy, DH, or heat of reaction is amount of heat
absorbed or released when a reaction occurs at constant
pressure.
The change in energy, DE, is the amount of heat absorbed or
released when a reaction occurs at constant volume.
How much do the DH and DE for a reaction differ?

The difference depends on the amount of work performed by the
system or the surroundings.
75
Relationship of DH and DE

Example 15-14: In Section 15-5, we noted
that DHo = -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 DE for this reaction?
C5 H12( )  8 O 2(g)  5 CO 2(g)  6 H 2 O (  )
76
Relationship of DH and DE

Determine the work done by this reaction.
You do it!
C5 H12( )  8 O 2(g)  5 CO 2(g)  6 H 2 O (  )
 

8 mol gas
5 mol gas
Since DH o  - 3523 kJ/mol, we know that T  298 K.
w  - PDV  - (Dn gas )RT
Dn gas  5  8 mol  - 3 mol
w  -(-3 mol)(8.314
J
mol K
)( 298 K)  7433 J  7.433 kJ
77
Relationship of DH and DE

Now calculate the DE for this reaction from the
values of DH and w that we have determined.
You do it!
DH  DE + PDV  DE = DH - PDV
since w = - PDV = 7.433 kJ
then PDV = - 7.433 kJ
DE = - 3523 kJ - (-7.433 kJ) = -3516 kJ
78
Spontaneity of Physical and
Chemical Changes

Spontaneous changes happen without any continuing
outside influences.


For example the rusting of iron occurs spontaneously.


A spontaneous change has a natural direction.
Have you ever seen rust turn into iron metal without man
made interference?
The melting of ice at room temperature occurs
spontaneously.

Will water spontaneously freeze at room temperature?
79
The Two Aspects of Spontaneity

An exothermic reaction does not ensure
spontaneity.



For example, the freezing of water is exothermic
but spontaneous only below 0oC.
An increase in disorder of the system also
does not insure spontaneity.
It is a proper combination of exothermicity
and disorder that determines spontaneity.
80
The Second Law of Thermodynamics


The second law of thermodynamics states, “In
spontaneous changes the universe tends towards a
state of greater disorder.”
Spontaneous processes have two requirements:
1.
2.
The free energy change of the system must be negative.
The entropy of universe must increase.

Fundamentally, the system must be capable of doing useful work
on surroundings for a spontaneous process to occur.
81
Entropy, S



Entropy is a measure of the disorder or
randomness of a system.
As with DH, entropies have been measured and
tabulated in Appendix K as So298.
When:


DS > 0 disorder increases (which favors
spontaneity).
DS < 0 disorder decreases (does not favor
spontaneity).
82
Entropy, S

From the Second Law of Thermodynamics, for a
spontaneous process to occur:
DSuniverse  DSsystem  DSsurroundings  0

In general for a substance in its three states of
matter:
Sgas > Sliquid > Ssolid
83
Entropy, S


The Third Law of Thermodynamics states, “The
entropy of a pure, perfect, crystalline solid at 0 K is
zero.”
This law permits us to measure the absolute values
of the entropy for substances.


To get the actual value of S, cool a substance to 0 K, or as
close as possible, then measure the entropy increase as the
substance heats from 0 to higher temperatures.
Notice that Appendix K has values of S not DS.
84
Entropy, S

Entropy changes for reactions can be determined
similarly to DH for reactions.
DS
o
298
  nS
o
products
n
  nS
o
reactants
n
85
Entropy, S

Example 15-15: Calculate the entropy change
for the following reaction at 25oC. Use
appendix K.

2 NO2(g)  N 2O 4(g)
You do it!
86
Entropy, S
N O
2 NO2(g) 
2 4(g)
DSorxn   n Soproducts   n Soreactants
n
n
 SoN 2O 4(g)  2SoNO2(g)
 (304.2 J mol K )  2(240.0 J mol K )
 175.8 J mol K or - 0.1758 kJ mol K


The negative sign of DS indicates that the system is
more ordered.
If the reaction is reversed the sign of DS changes.

For the reverse reaction DSo298= +0.1758 kJ/K

The + sign indicates the system is more disordered.
87
Entropy, S

Example 15-16: Calculate DSo298 for the
reaction below. Use appendix K.

3 NOg   N 2Og   NO2g 
You do it!
88
Entropy, S
3 NOg  
 N 2 O g   NO2g 
DS
0
298
S
0
N 2O  g 
S
0
NO2  g 
3S
0
NO g 
 219.7  240.0 - 3210.4 J mol K
 172.4 J mol K or - 0.1724 kJ mol K

Changes in DS are usually quite small compared to
DE and DH.

Notice that DS has units of only a fraction of a kJ while
DE and DH values are much larger numbers of kJ.
89
Free Energy Change, DG,
and Spontaneity

In the mid 1800’s J. Willard Gibbs determined the
relationship of enthalpy, H, and entropy, S, that best
describes the maximum useful energy obtainable in
the form of work from a process at constant
temperature and pressure.


The relationship also describes the spontaneity of a
system.
The relationship is a new state function, DG, the
Gibbs Free Energy.
DG = DH - TDS
(at constant T & P)
90
Free Energy Change, DG,
and Spontaneity

The change in the Gibbs Free Energy, DG, is a
reliable indicator of spontaneity of a physical
process or chemical reaction.

DG does not tell us how quickly the process occurs.


Chemical kinetics, the subject of Chapter 16, indicates the rate of
a reaction.
Sign conventions for DG.



DG > 0 reaction is nonspontaneous
DG = 0 system is at equilibrium
DG < 0 reaction is spontaneous
91
Free Energy Change, DG,
and Spontaneity

Changes in free energy obey the same type of
relationship we have described for enthalpy,
DH, and entropy, DS, changes.
DG
0
298
  nDG
n
0
products
  nDG
0
reactants
n
92
Free Energy Change, DG,
and Spontaneity

Example 15-17: Calculate DGo298 for the
reaction in Example 15-8. Use appendix K.
C H  5 O  3 CO  4 H O
3
8(g)
2(g)
2(g)
2
()
You do it!
93
Free Energy Change, DG,
and Spontaneity
CH
3
DG
o
rxn
8(g)
 [3DG
5O
o
f CO 2 (g )
2(g)
 3 CO
 4DG
o
f H 2O(  )
2(g)
4H O
]  [DG
2
o
f C 3 H 8 (g )
()
 5DG
o
f O 2 (g )
 {[3(394.4)  4(237.3)]  [(23.49)  5(0)]}
 2108.9
 DGo298 < 0, so the reaction is spontaneous at
standard state conditions.
 If the reaction is reversed:
kJ

mol
DGo298 > 0, and the reaction is nonspontaneous at
standard state conditions.
94
kJ
]
mol
The Temperature
Dependence of Spontaneity
Free energy has the relationship DG = DH -TDS.
 Because 0 ≤ DH ≥ 0 and 0 ≤ DS ≥ 0, there are four
possibilities for DG.
DH DS DG
Forward reaction spontaneity

<0
<0
>0
>0
>0
<0
>0
<0
Spontaneous at all T’s.
T dependent Spontaneous at low T’s.
T dependent Spontaneous at high T’s.
>0
Nonspontaneous at all T’s.
<0
95
The Temperature
Dependence of Spontaneity
96
The Temperature
Dependence of Spontaneity

Example 15-18: Calculate DSo298 for the following reaction.
In example 15-8, we found that DHo298= -2219.9 kJ, and in
Example 15-17 we found that DGo298= -2108.5 kJ.
CH
5 O .9 
CO.5kJ4 H O
32108
2219
o
3
8(g)
DS

2(g)
2(g)
2
()
o
o
298
K
DG  DH  TDS
J o
DSo  0.374okJ  -374
o
DH  K
DG  TDKS
o
DH  DG
 DS
T
o
o
o
97
The Temperature
Dependence of Spontaneity



DSo298 = -374 J/K which indicates that the
disorder of the system decreases .
For the reverse reaction,
3 CO2(g) + 4 H2O(g) C3H8(g) + 5 O2(g)
DSo298 = +374 J/K which indicates that the
disorder of the system increases .
98
The Temperature
Dependence of Spontaneity

Example 15-18: Use thermodynamic data to estimate
the normal boiling point of water.

H 2 O (  )  H 2 O (g)
Because this is an equlibrium process DG  0.
DH
Thus DH  TDS and T 
DS
99
The Temperature
Dependence of Spontaneity
assume DH@ BP 
DH 
o
o
DH H 2O(g)

o
DH 298
o
DH H 2O( l )
DH   2418
.  ( 2858
. ) J K
o
DH  44.0 kJ@25 C
o
o
100
The Temperature
Dependence of Spontaneity
assume DS @ BP  DS
0
rxn
DS
S
DS
 188.7  69.91 J K
DS
 118.8
0
rxn
0
rxn
0
rxn
S
0
H 2O g 
0
H 2O  
J
K
or 0.1188
kJ
K
101
The Temperature
Dependence of Spontaneity
DH DH
44.0 kJ
T=
 o

370
K
DS DS 0.1188 kJ K
o
o
370 K-273 K=97 C
102
The Temperature
Dependence of Spontaneity

Example 15-19: What is the percent error in
Example 15-18?
% error =
370 - 373 K  100%  0.80% error
373 K
% error of less than 1%!!
103
Synthesis Question

When it rains an inch of rain, that means that
if we built a one inch high wall around a piece
of ground that the rain would completely fill
this enclosed space to the top of the wall.
Rain is water that has been evaporated from a
lake, ocean, or river and then precipitated
back onto the land. How much heat must the
sun provide to evaporate enough water to rain
1.0 inch onto 1.0 acre of land?
1 acre = 43,460
ft2
104
Synthesis Question
1 in  2.54 cm
1 ft  30.5 cm
1 ft   30.5 cm 
 930 cm
2
 930 cm
1 acre  43,460 ft 
2
 1 ft
7
2
 4.04 10 cm
2
2
2

2

7

volume  4.04 10 cm
 1.03 10 cm
8
3
2



2.54 cm 
105
Synthesis Question
 1g 
mass of water  1.03  10 cm  3 
 cm 
 1.03  108 g of water
8
3
 1 mol 

moles of water  1.03  10 g 
 18 g 
 5.71106 mol

DH vaporization of water  44.0 kJ

8

mol

heat sun must supply  5.71106 mol 44.0 kJ
 2.51 10 kJ
8

mol
106
Group Question

When Ernest Rutherford, introduced in Chapter 5,
gave his first lecture to the Royal Society one of the
attendees was Lord Kelvin. Rutherford announced
at the meeting that he had determined that the earth
was at least 1 billion years old, 1000 times older than
Kelvin had previously determined for the earth’s
age. Then Rutherford looked at Kelvin and told him
that his method of determining the earth’s age based
upon how long it would take the earth to cool from
molten rock to its present cool, solid form
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Group Question
was essentially correct. But there was a new, previously
unknown source of heat that Kelvin had not included in his
calculation and therein lay his error. Kelvin apparently
grinned at Rutherford for the remainder of his lecture.
What was this “new” source of heat that Rutherford knew
about that had thrown Kelvin’s calculation so far off?
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End of Chapter 15

Fireworks are beautiful exothermic
chemical reactions.
109