Chapter 17 Free Energy and Thermodynamics

Transcription

Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 17
Free Energy
and
Thermodynamics
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2008, Prentice Hall
First Law of Thermodynamics
• you can’t win!
• First Law of Thermodynamics: Energy
cannot be Created or Destroyed
9the total energy of the universe cannot change
9though you can transfer it from one place to another
• ∆Euniverse = 0 = ∆Esystem + ∆Esurroundings
Tro, Chemistry: A Molecular Approach
2
First Law of Thermodynamics
• Conservation of Energy
• For an exothermic reaction, “lost” heat from the system
•
•
•
goes into the surroundings
two ways energy “lost” from a system,
9converted to heat, q
9used to do work, w
Energy conservation requires that the energy change in
the system equal the heat released + work done
9 ∆E = q + w
9 ∆E = ∆H + P∆V
∆E is a state function
9internal energy change independent of how done
3
Energy Tax
• you can’t break even!
• to recharge a battery with 100 kJ of
•
useful energy will require more than
100 kJ
every energy transition results in a
“loss” of energy
9 conversion of energy to heat which is
“lost” by heating up the surroundings
4
Heat Tax
fewer steps
generally results
in a lower total
heat tax
5
Thermodynamics and Spontaneity
• thermodynamics predicts whether a process will
proceed under the given conditions
9spontaneous process
¾nonspontaneous processes require energy input to go
• spontaneity is determined by comparing the free
energy of the system before the reaction with the free
energy of the system after reaction.
9if the system after reaction has less free energy
than before the reaction, the reaction is
thermodynamically favorable.
• spontaneity ≠ fast or slow
6
Comparing Potential Energy
The direction of
spontaneity can
be determined by
comparing the
potential energy
of the system at
the start and the
end.
7
Reversibility of Process
• any spontaneous process is irreversible
9 it will proceed in only one direction
• a reversible process will proceed back and forth
between the two end conditions
9 equilibrium
9 results in no change in free energy
• if a process is spontaneous in one direction, it must be
nonspontaneous in the opposite direction
8
Thermodynamics vs. Kinetics
9
Diamond → Graphite
Graphite is more stable than diamond, so the conversion of
diamond into graphite is spontaneous – but don’t worry,
it’s so slow that your ring won’t turn into pencil lead in
your lifetime (or through many of your generations).
10
Factors Affecting Whether a
Reaction Is Spontaneous
• The two factors that determine the thermodynamic
favorability are the enthalpy and the entropy.
• The enthalpy is a comparison of the bond energy
of the reactants to the products.
9bond energy = amount needed to break a bond.
9 ∆H
• The entropy factors relates to the
randomness/orderliness of a system
9 ∆S
• The enthalpy factor is generally more important
than the entropy factor
11
•
•
•
•
•
•
•
Enthalpy
related to the internal energy
∆H generally kJ/mol
stronger bonds = more stable molecules
if products more stable than reactants, energy released
9exothermic
9 ∆H = negative
if reactants more stable than products, energy absorbed
9endothermic
9 ∆H = positive
The enthalpy is favorable for exothermic reactions and
unfavorable for endothermic reactions.
Hess’ Law ∆H°rxn = Σ(∆H°prod) - Σ(∆H°react)
12
Substance
∆H°
kJ/mol
Substance
∆H°
kJ/mol
Al(s)
Br2(l)
C(diamond)
CO(g)
Ca(s)
Cu(s)
Fe(s)
H2(g)
H2O(g)
HF(g)
HBr(g)
I2(s)
N2(g)
NO(g)
Na(s)
S(s)
0
0
+1.88
-110.5
0
0
0
0
-241.82
-268.61
-36.23
0
0
+90.37
0
0
Al2O3
Br2(g)
C(graphite)
CO2(g)
CaO(s)
CuO(s)
Fe2O3(s)
H2O2(l)
H2O(l)
HCl(g)
HI(g)
I2(g)
NH3(g)
NO2(g)
O2(g)
SO2(g)
-1669.8
+30.71
0
-393.5
-635.5
-156.1
-822.16
-187.8
-285.83
-92.30
+25.94
+62.25
-46.19
+33.84
0
-296.9
Entropy
• entropy is a thermodynamic function that increases as the
number of energetically equivalent ways of arranging the
components increases, S
•
S generally J/mol
• S = k ln W
9 k = Boltzmann Constant = 1.38 x 10-23 J/K
9 W is the number of energetically equivalent ways, unitless
• Random systems require less energy than ordered systems
14
W
Energetically
Equivalent States for
the Expansion of a
Gas
15
Macrostates → Microstates
These microstates
all have the same
macrostate
So there are 6
macrostate can be achieved through
differentThis
particle
several different
arrangements of the particles
arrangements
that
result in the same
macrostate
16
Macrostates and Probability
There is only one possible
arrangement that gives State A
and one that gives State C
There are 6 possible
arrangements that give State B
Therefore State B has
higher entropy than either
State A or State B
The macrostate with the
highest entropy also has the
greatest dispersal of energy
17
Changes in Entropy, ∆S
• entropy change is favorable when the result is a more
•
random system.
9 ∆S is positive
Some changes that increase the entropy are:
9reactions whose products are in a more disordered
state.
¾(solid > liquid > gas)
9reactions which have larger numbers of product
molecules than reactant molecules.
9increase in temperature
9solids dissociating into ions upon dissolving
18
Increases in Entropy
19
The 2nd Law of Thermodynamics
• the total entropy change of the universe must be
positive for a process to be spontaneous
9 for reversible process ∆Suniv = 0,
9 for irreversible (spontaneous) process ∆Suniv > 0
• ∆Suniverse = ∆Ssystem + ∆Ssurroundings
• if the entropy of the system decreases, then the
entropy of the surroundings must increase by a
larger amount
9 when ∆Ssystem is negative, ∆Ssurroundings is positive
• the increase in ∆Ssurroundings often comes from the
heat released in an exothermic reaction
20
Entropy Change in State Change
• when materials change state, the number of
macrostates it can have changes as well
9for entropy: solid < liquid < gas
9because the degrees of freedom of motion increases
solid → liquid → gas
21
Entropy Change and State Change
22
Heat Flow, Entropy, and the 2nd Law
Heat must flow from
water to ice in order for
the entropy of the
universe to increase
23
Temperature Dependence of ∆Ssurroundings
• when a system process is exothermic, it adds heat to
•
•
the surroundings, increasing the entropy of the
surroundings
when a system process is endothermic, it takes heat
from the surroundings, decreasing the entropy of the
surroundings
the amount the entropy of the surroundings changes
depends on the temperature it is at originally
9 the higher the original temperature, the less effect addition or
removal of heat has
∆Ssurroundings =
− ∆H system
T
24
Gibbs Free Energy, ∆G
• maximum amount of energy from the system
available to do work on the surroundings
G = H – T·S
∆Gsys = ∆Hsys – T∆Ssys
∆Gsys = – T∆Suniverse
∆Greaction = Σ n∆Gprod – Σ n∆Greact
• when ∆G < 0, there is a decrease in free
energy of the system that is released into the
surroundings; therefore a process will be
spontaneous when ∆G is negative
25
Ex. 17.2a – The reaction C3H8(g) + 5 O2(g) → 3 CO2(g) + 4
H2O(g) has ∆Hrxn = -2044 kJ at 25°C.
Calculate the entropy change of the surroundings.
Given:
Find:
Concept Plan:
∆Hsystem = -2044 kJ, T = 298 K
∆Ssurroundings, J/K
T, ∆H
Relationships:
∆Ssurr =
Solution:
− ∆H sys
∆Ssurr =
∆Ssurr
T
= 6.86 kJ
K
− ∆H sys
∆S
T
− (− 2044 kJ )
=
298 K
= 6.86 × 103 KJ
Check: combustion is largely exothermic, so the entropy of
the surrounding should increase significantly
Free Energy Change and Spontaneity
27
Gibbs Free Energy, ∆G
• process will be spontaneous when ∆G is negative
• ∆G will be negative when
•
•
9 ∆H is negative and ∆S is positive
¾exothermic and more random
9 ∆H is negative and large and ∆S is negative but
small
9 ∆H is positive but small and ∆S is positive and
large
¾or high temperature
∆G will be positive when ∆H is + and ∆S is −
9never spontaneous at any temperature
when ∆G = 0 the reaction is at equilibrium
28
∆G, ∆H, and ∆S
29
Ex. 17.3a – The reaction CCl4(g) → C(s, graphite) + 2 Cl2(g) has
∆H = +95.7 kJ and ∆S = +142.2 J/K at 25°C.
Calculate ∆G and determine if it is spontaneous.
Given:
Find:
Concept Plan:
∆H = +95.7 kJ, ∆S = 142.2 J/K, T = 298 K
∆G, kJ
T, ∆H, ∆S
∆G
∆G = ∆H − T∆S
Relationships:
Solution: ∆G = ∆H − T∆S
= (+ 95.7 × 103 J ) − (298 K ) + 142.2 KJ
(
)
= +5.33 × 10 4 J
Answer: Since ∆G is +, the reaction is not spontaneous at
this temperature. To make it spontaneous, we need
to increase the temperature.
Ex. 17.3a – The reaction CCl4(g) → C(s, graphite) + 2 Cl2(g) has
∆H = +95.7 kJ and ∆S = +142.2 J/K.
Calculate the minimum temperature it will be spontaneous.
Given:
Find:
Concept Plan:
∆H = +95.7 kJ, ∆S = 142.2 J/K, ∆G < 0
Τ, Κ
∆G, ∆H, ∆S
∆G = ∆H − T∆S
Relationships:
Solution:
T
3
J
K
(+ 95.7 ×10 J ) < T
(+ 142.2 )
3
J
K
673 K < T
∆G = (∆H − T∆S) < 0
3
(+ 95.7 ×10 J ) − (T )(+ 142.2 ) < 0
(+ 95.7 ×10 J ) < (T )(+ 142.2 )
J
K
Answer: The temperature must be higher than 673K for the
reaction to be spontaneous
The 3rd Law of Thermodynamics
Absolute Entropy
• the absolute entropy of a substance
•
is the amount of energy it has due to
dispersion of energy through its
particles
the 3rd Law states that for a perfect
crystal at absolute zero, the absolute
entropy = 0 J/mol·K
9 therefore, every substance that is not a
perfect crystal at absolute zero has some
energy from entropy
9 therefore, the absolute entropy of
substances is always +
32
Standard Entropies
• S°
• extensive
• entropies for 1 mole at 298 K for a particular
state, a particular allotrope, particular molecular
complexity, a particular molar mass, and a
particular degree of dissolution
33
Substance
S°
J/mol-K
Substance
S°
J/mol-K
Al(s)
Br2(l)
C(diamond)
CO(g)
Ca(s)
Cu(s)
Fe(s)
H2(g)
H2O(g)
HF(g)
HBr(g)
I2(s)
N2(g)
NO(g)
Na(s)
S(s)
28.3
152.3
2.43
197.9
41.4
33.30
27.15
130.58
188.83
173.51
198.49
116.73
191.50
210.62
51.45
31.88
Al2O3(s)
Br2(g)
C(graphite)
CO2(g)
CaO(s)
CuO(s)
Fe2O3(s)
H2O2(l)
H2O(l)
HCl(g)
HI(g)
I2(g)
NH3(g)
NO2(g)
O2(g)
SO2(g)
51.00
245.3
5.69
213.6
39.75
42.59
89.96
109.6
69.91
186.69
206.3
260.57
192.5
240.45
205.0
248.5
Relative Standard Entropies
States
• the gas state has a larger entropy than the liquid
state at a particular temperature
• the liquid state has a larger entropy than the
solid state at a particular temperature
Substance
S°,
(J/mol·K)
H2O (g)
70.0
H2O (l)
188.8
35
Molar Mass
• the larger the molar mass,
the larger the entropy
• available energy states
more closely spaced,
allowing more dispersal
of energy through the
states
36
Allotropes
• the less
constrained the
structure of an
allotrope is, the
larger its entropy
37
Molecular Complexity
• larger, more complex
molecules generally
have larger entropy
• more available energy
states, allowing more
dispersal of energy
through the states
Molar
S°,
Substance
Mass (J/mol·K)
Ar (g)
154.8
39.948
NO (g)
30.006
210.8
38
Dissolution
• dissolved solids
generally have larger
entropy
• distributing particles
throughout the mixture
Substance
S°,
(J/mol·K)
KClO3(s)
143.1
KClO3(aq)
265.7
39
Substance
S°, J/mol⋅K
NH3(g)
192.8
O2(g)
205.2
NO(g)
210.8
H2O(g)
standard entropies from Appendix IIB
188.8
Ex. 17.4 –Calculate ∆S° for the reaction
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l)
Given:
Find: ∆S, J/K
Concept Plan:
Relationships:
Solution:
S°NH3, S°O2, S°NO, S°H2O,
(
) (
∆So = Σn pSo products − ΣnrSo reactants
(
) (
∆So = Σn pSo products − ΣnrSo reactants
)
)
∆S
= [4(So NO( g ) ) + 6(So H 2O( g ) )] − [4(So NH3 ( g ) ) + 5(SoO 2 ( g ) )]
J
J
J
J
= [4(210.8 K ) + 6(188.8 K )] − [4(192.8 K ) + 5(205.2 K )]
J
= 178.8 K
Check: ∆S is +, as you would expect for a reaction with
more gas product molecules than reactant molecules
Calculating ∆G°
• at 25°C:
∆Goreaction = ΣnGof(products) - ΣnGof(reactants)
• at temperatures other than 25°C:
9assuming the change in ∆Horeaction and ∆Soreaction is
negligible
∆G°reaction = ∆H°reaction – T∆S°reaction
41
Substance
∆G°f
kJ/mol
Substance
∆G°f
kJ/mol
Al(s)
Br2(l)
C(diamond)
CO(g)
Ca(s)
Cu(s)
Fe(s)
H2(g)
H2O(g)
HF(g)
HBr(g)
I2(s)
N2(g)
NO(g)
Na(s)
S(s)
0
0
+2.84
-137.2
0
0
0
0
-228.57
-270.70
-53.22
0
0
+86.71
0
0
Al2O3
Br2(g)
C(graphite)
CO2(g)
CaO(s)
CuO(s)
Fe2O3(s)
H2O2(l)
H2O(l)
HCl(g)
HI(g)
I2(g)
NH3(g)
NO2(g)
O2(g)
SO2(g)
-1576.5
+3.14
0
-394.4
-604.17
-128.3
-740.98
-120.4
-237.13
-95.27
+1.30
+19.37
-16.66
+51.84
0
-300.4
42
∆G°f, kJ/mol
-50.5
0.0
-394.4
-228.6
163.2
Substance
Ex. 17.7 –Calculate ∆G° at 25°C for the
reaction
CH4(g)
O2(g)
CO2(g)
H2O(g)
O3(g)
CH4(g) + 8 O2(g) → CO2(g) + 2 H2O(g) + 4 O3(g)
Given: standard free energies of formation from Appendix IIB
Find: ∆G°, kJ
Concept Plan:
Relationships:
Solution:
(
∆G°f of prod & react
(
) (
∆G°
∆G o = Σn p ∆G of products − Σnr ∆G of reactants
) (
∆G o = Σn p ∆G of products − Σnr ∆G of reactants
)
)
= [(∆G of CO 2 ) + 2(∆G of H 2O) + (∆G of O3 )] − [(∆G of CH 4 ) + 8(∆G of O 2 )]
= [(−394.4 kJ) + 2(−228.6 kJ) + (+163.2 kJ)] − [(−50.5 kJ) + 8(0.0 kJ)]
= −148.3 kJ
Ex. 17.6 – The reaction SO2(g) + ½ O2(g) → SO3(g) has
∆H° = -98.9 kJ and ∆S° = -94.0 J/K at 25°C.
Calculate ∆G° at 125°C and determine if it is spontaneous.
Given:
Find:
Concept Plan:
∆H° = -98.9 kJ, ∆S° = -94.0 J/K, T = 398 K
∆G°, kJ
T, ∆H°, ∆S°
∆G o = ∆Ho − T∆So
Relationships:
Solution:
∆G°
∆G o = ∆Ho − T∆So
(
)
(
= − 98.9 × 103 J − (398 K ) − 94.0 K
J
)
= −61.5 × 103 J = −61.5 kJ
Answer: Since ∆G is -, the reaction is spontaneous at this
temperature, though less so than at 25°C
∆G Relationships
• if a reaction can be expressed as a series of reactions,
the sum of the ∆G values of the individual reaction is
the ∆G of the total reaction
9 ∆G is a state function
• if a reaction is reversed, the sign of its ∆G value
•
reverses
if the amounts of materials is multiplied by a factor, the
value of the ∆G is multiplied by the same factor
9 the value of ∆G of a reaction is extensive
45
Free Energy and Reversible Reactions
• the change in free energy is a theoretical limit as
to the amount of work that can be done
• if the reaction achieves its theoretical limit, it is
a reversible reaction
46
Real Reactions
• in a real reaction, some of the free energy is
“lost” as heat
9if not most
• therefore, real reactions are irreversible
47
∆G under Nonstandard Conditions
⋅ ∆G = ∆G° only when the reactants and products
are in their standard states
⋅ there normal state at that temperature
⋅ partial pressure of gas = 1 atm
⋅ concentration = 1 M
⋅ under nonstandard conditions, ∆G = ∆G° + RTlnQ
⋅ Q is the reaction quotient
⋅ at equilibrium ∆G = 0
⋅ ∆G° = ─RTlnK
48
49
Example - ∆G
• Calculate ∆G at 427°C for the reaction below if the
PN2 = 33.0 atm, PH2= 99.0 atm, and PNH3= 2.0 atm
N2(g) + 3 H2(g) → 2 NH3(g)
Q=
PNH32
PN21
x
PH23
=
(2.0 atm)2
(33.0
atm)1
(99.0)3
= 1.2 x 10-7
∆H° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J
∆S° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K
∆G° = -92380 J - (700 K)(-198.2 J/K)
∆G° = +46400 J
∆G = ∆G° + RTlnQ
∆G = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10-7)
∆G = -46300 J = -46 kJ
50
51
Example - K
• Estimate the equilibrium constant and position of
equilibrium for the following reaction at 427°C
N2(g) + 3 H2(g) ⇔ 2 NH3(g)
∆H° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J
∆S° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K
∆G° = -92380 J - (700 K)(-198.2 J/K)
∆G° = +46400 J
∆G° = -RT lnK
+46400 J = -(8.314 J/K)(700 K) lnK
lnK = -7.97
K = e-7.97 = 3.45 x 10-4
since K is << 1, the position of equilibrium favors reactants
52
Temperature Dependence of K
• for an exothermic reaction, increasing the
temperature decreases the value of the
equilibrium constant
• for an endothermic reaction, increasing the
temperature increases the value of the
equilibrium constant
∆Horxn  1  ∆Sorxn
ln K = −
 +
R T
R
53

Chapter 17 Free Energy and Thermodynamics

Transcription

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