Transcript Chapter

Chemistry II
Chapter 17
Free Energy and Thermodynamics
First Law of Thermodynamics
• First Law of Thermodynamics: Energy cannot be
Created or Destroyed
 the total energy of the universe cannot change
 though you can transfer it from one place to another
ΔEuniverse = 0 = ΔEsystem + ΔEsurroundings
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,
 converted to heat, q
 used to do work, w
• Energy conservation requires that the energy change in the system equal the
heat released + work done
ΔE=q+w
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
 conversion of energy to heat which is “lost” by
heating up the surroundings
e.g. if a light bulb is 5 % efficient
95% is lost as heat
Heat Tax
fewer steps
generally results in
a lower total heat
tax
Spontaneous and Nonspontaneous
Processes
• e.g. in physics - will an object fall when dropped?
in chemistry - will iron rust?
• thermodynamics predicts whether a process will proceed under the
given conditions - spontaneous process
 nonspontaneous processes require energy inputs
• In physics sponaneity is determined by looking at the potential
energy change of a system
• In chemistry spontaneity is determined by comparing the free
energy change of the system
 if the system after reaction has less free energy than before the
reaction, the reaction is thermodynamically favorable.
• spontaneity ≠ fast or slow
Comparing Potential Energy
High PE → Low PE
The direction of
spontaneity can
be determined by
comparing the
potential energy
of the system at
the start and the
end.
High ‘Free Energy’ → Low ‘Free Energy’
Reversibility of Process
• any spontaneous process is irreversible
 it will proceed in only one direction
• a reversible process will proceed back and forth between the two end
conditions
 equilibrium
 results in no change in free energy
• if a process is spontaneous in one direction, it must be nonspontaneous in the
opposite direction
Thermodynamics vs. Kinetics
Don’t confuse
spontaneity (direction
and extent of reaction)
with speed
This reaction will require an energy input since
‘free energy’ of reactants < ‘free energy’ products
Diamond → Graphite
Graphite has less FE 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).
Factors Affecting Whether a Reaction Is
Spontaneous
• The two factors that determine spontaneity (free energy) are the
enthalpy and the entropy
• The enthalpy is a comparison of the bond energy of the reactants to
the products.
 bond energy = amount needed to break a bond.
 ΔH
• The entropy factors relates to the randomness/orderliness of a system
 ΔS
• The enthalpy factor is generally more important than the entropy factor
Enthalpy, ΔH
• ΔH related to the internal energy (kJ/mol)
• stronger bonds = more stable molecules
• if products more stable than reactants, energy released
 Exothermic (ΔH = negative)
• if reactants more stable than products, energy absorbed
 Endothermic (ΔH = positive)
Hess’ Law:
Δ H°rxn = Σ(nΔ H°prod) - Σ(nΔ H°react)
Where n = stoichiometric coefficient
Changes in Entropy, ΔS
• entropy change is favorable when the result is a more random system
 ΔS is positive
• Some changes that increase the entropy are:
 reactions whose products are in a more disordered state.
 (solid > liquid > gas)
 reactions which have larger numbers of product molecules than
reactant molecules
 increase in temperature
 solids dissociating into ions upon dissolving
Melting Increases in Entropy
Evaporating
Dissolution
Entropy
• entropy is a thermodynamic function that increases as the number of
energetically equivalent ways of arranging the components increases, a
measure of randomness, S
 S generally J/K
• S = k ln W
 k = Boltzmann Constant = 1.38 x 10-23 J/K
 W is the number of energetically equivalent ways, unitless
• A chemical system proceeds in a direction that increases the entropy of
the universe
• Next we consider W,
W
Energetically Equivalent
States for the Expansion
of a Gas in a vacuum
The following are macro
states
Macrostates → Microstates
These microstates all
have the same
macrostate
So there are 6
different particle
arrangements that
result in the same
This macrostate can be achieved through
macrostate
several different arrangements of the particles
Macrostates and Probability
There is only one possible
arrangement that gives State A and
one that gives State B
There are 6 possible arrangements
that give State C
Therefore State C has higher
entropy than either State A or
State B
The macrostate with the highest
entropy also has the greatest
dispersal of energy, now imagine
thousands of atoms!
Entropy
• Second law of thermodynamics:
For any spontaneous process, the entropy of the universe increases,
ΔSuniv > 0
• Spontaneous processes increase the entropy of the universe
Δ S = Sfinal – Sinitial
• In our flask analogy, ΔS > 0
• Explains why heat travels from a substance at high temp. to a
substance at low temp. but never the opposite
Entropy Change in State Change
• when materials change state, the number of macrostates it
can have changes as well
 for entropy: solid < liquid < gas
 because the degrees of freedom of motion increases
solid → liquid → gas
Entropy Change and State Change
Heat Flow, Entropy, and the 2nd Law
Heat must flow from water to
ice in order for the entropy of
the universe to increase
Is Entropy Increases or Decreasing?
• Iodine sublimes
• A liquid boils
• A solid decomposes into two gaseous products
Heat Transfer and Changes in S of
Surroundings
• the total entropy change of the universe must be
positive for a process to be spontaneous
 for reversible process ΔSuniv = 0,
 for irreversible (spontaneous) process Δ Suniv > 0
ΔSuniverse = ΔSsystem + Δ Ssurroundings
• if the entropy of the system decreases, then the
entropy of the surroundings must increase by a larger
amount
 when ΔSsystem is negative, ΔSsurroundings is positive
 e.g. when water freezes
• the increase in ΔSsurroundings often comes from the heat
released in an exothermic reaction
Temperature Dependence of ΔSsurroundings
• 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
 the higher the original temperature, the less effect addition or
removal of heat has
Ssurroundings 
 H system
T
Temperature Dependence of ΔSsurroundings
• Example:
2N2 (g) + O2 (g) → 2N2O (g)
ΔH = +163.2 kJ/mol
(i) Calculate the entropy change in the surroundings at 25 ºC
ΔSsurr. = -ΔHsys. / T = -163.2 kJ/298K = -0.548 kJ/K = -548 J/K
(ii) Determine the sign of the entropy change
3 mols on LHS, 2 mols on RHS so -ΔSsys
(iii) Determine the sign of Δ Suniverse , is reaction spontaneous?
Δ Suniverse = ΔSsystem + Δ Ssurroundings
-ve + -ve = -ve
Not spontaneous
Gibbs Free Energy, ΔG
• Enthalpy (ΔHsys) and entropy (ΔSsurr) are related by:
ΔSsurr. = -ΔHsys. / T
• Also: Δ Suniverse = ΔSsystem + Δ Ssurroundings
• Combining these: ΔSuniverse = ΔSsystem -ΔHsys. / T
• Now ΔSuniv can be calculated with the ‘system’ only
• Multiplying by –T:
-TΔSuniverse = -TΔSsystem -ΔHsys.
• This is now called the Gibbs free Energy expression (remember we
multiplied by a –ve)
Gibbs Free Energy, ΔG
• maximum amount of energy from the system available to do work on the
surroundings
G = H – T∙S
ΔGsys = Δ Hsys – T Δ Ssys
The change in Gibbs free energy for a process at constant T and P
is ~ -ΔSuniv
Δ Gsys = – T Δ Suniverse
• ΔG (Gibbs free energy) being –ve is a criterion for spontaneity
• there is a decrease in free energy of the system that is released into the
surroundings; therefore a process will be spontaneous when ΔG is
negative (i.e. ΔS +ve, since we mult. by - T)
Gibbs Free Energy, ΔG
• process will be spontaneous when ΔG is negative
ΔGsys = Δ Hsys – T Δ Ssys

Three possibilities:
(1) Δ G will be negative - spontaneous
 Δ H is negative and ΔS is positive
 exothermic and more random
 ΔH is negative and large and ΔS is negative but small
 Δ H is positive but small and ΔS is positive
 high temperature favors spontaneity, nonspon. at low T
(2) ΔG will be +ve when ΔH is +ve and ΔS is −ve - nonspontaneous
 never spontaneous at any temperature
(3) when ΔG = 0 the reaction is at equilibrium
Which of the following is spontaneous?
ΔHº
TΔSº
TΔSº
ΔHº
ΔHº
TΔSº
ΔHº
TΔSº
ΔG, ΔH, and ΔS
Ex. 17.3a – The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has
ΔH = +95.7 kJ and Δ S = +142.2 J/K at 25°C.
Calculate Δ G and determine if it is spontaneous.
Given:
Find:
Concept
Plan:
ΔH = +95.7 kJ, Δ S = 142.2 J/K, T = 298 K
ΔG, kJ
T, ΔH, ΔS
ΔG
G  H  TS
Relationship
s: G  H  TS
Solution:
  95.7  103 J   298 K   142.2 KJ


 5.33  104 J
Answer: Since ΔG 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
ΔH = +95.7 kJ and ΔS = +142.2 J/K.
Calculate the minimum temperature it will be spontaneous.
Given:
Find:
Concept Plan:
ΔH = +95.7 kJ, ΔS = 142.2 J/K, ΔG < 0
T, K
ΔG, ΔH, ΔS
G  H  TS
Relationships:
Solution:
T
3
J
K
 95.7 10 J   T
 142.2 
3
J
K
T  673 K
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
Entropy
• entropy is a thermodynamic function that increases as the number of
energetically equivalent ways of arranging the components increases, a
measure of randomness, S
 S generally J/K
• S = k ln W
 k = Boltzmann Constant = 1.38 x 10-23 J/K
 W is the number of energetically equivalent ways, unitless
• A chemical system proceeds in a direction that increases the entropy of
the universe
• Next we consider W,
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
 therefore, every substance that is not a
perfect crystal at absolute zero has some
energy from entropy
 therefore, the absolute entropy of
substances is always +ve
Standard Entropies
• S° (J/mol.K)
• Extensive property (depends on amount)
• 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
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 (l)
70.0
H2O (g)
188.8
Relative Standard Entropies
Molar Mass
• the larger the molar mass, the
larger the entropy
Relative Standard Entropies
Allotropes
• the less constrained the
structure of an allotrope
is, the larger its entropy
Relative Standard Entropies
Molecular Complexity
• larger, more complex molecules
have larger entropy than single
atoms
• more available energy states
(translational, rotational,
vibrational) allowing more
dispersal of energy through the
states
Substance
Molar
Mass
S°,
(J/mol∙K)
Ar (g)
39.948
154.8
NO (g)
30.006
210.8
Relative Standard Entropies
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
Standard Entropy of Reaction, ΔSrxn
• Calculated similar to ΔHrxn

 
S  n pS products  nrSreactants
• Where n = stoichiometric coefficient

Substance
S, J/molK
NH3(g)
192.8
O2(g)
205.2
NO(g)
210.8
H2O(g)
Given: standard entropies from Appendix IIB
188.8
Ex. 17.4 –Calculate ΔS for the reaction
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(l)
Find: ΔS, J/K
Concept
Plan:
Relationship
s:
Solution:
SNH3, SO2, SNO, SH2O,

 
S  n pS products  nrSreactants

 
S  n pS products  nrS reactants


ΔS
 [4(S NO( g ) )  6(S H 2 O( g ) )]  [4(S NH 3 ( g ) )  5(SO 2 ( g ) )]
 [4(210 .8 K )  6(188 .8 K )]  [4(192 .8 K )  5(205 .2 K )]
J
J
J
J
 178 .8 K
J
Check:
ΔS is +, as you would expect for a reaction with more gas
product molecules than reactant molecules
Calculating ΔGrxn
• At 25 C and at temperatures other than 25 C:
 assuming the change in ΔHoreaction and Δ Soreaction are small over a
limted temp. range
Δ Grxn = Δ Hrxn – T Δ Srxn
Ex. 17.6 – The reaction SO2(g) + ½ O2(g)  SO3(g) has
ΔHrxn = -98.9 kJ and ΔSrxn = -94.0 J/K at 25°C.
Calculate ΔG at 125C and determine if it is spontaneous.
Given:
Find:
Concept
Plan:
ΔH = -98.9 kJ, ΔS = -94.0 J/K, T = 398 K
ΔG, kJ
T, ΔH, ΔS
ΔG
G   H  TS
Relationship
s:
Solution:
G   H  TS



  98.9  103 J  398 K   94.0 K
J

 61.5  103 J  61.5 kJ
Answer: Since ΔG is -, the reaction is spontaneous at
this temperature, though less so than at 25C
Calculating ΔGrxn with ΔG f
• at 25C:
ΔGorxn = ΣnpΔGof(products) - ΣnrΔGof(reactants)
• Like ΔHf°, ΔGf° = 0 for an element in its standard state:
• Example: ΔGf° (O2(g)) = 0
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
52
Gf, kJ/mol
-50.5
0.0
-394.4
-228.6
163.2
Substance
CH4(g)
O2(g)
CO2(g)
H2O(g)
O3(g)
Ex. 17.7 –Calculate ΔG at 25C for the reaction
CH4(g) + 8 O2(g)  CO2(g) + 2 H2O(g) + 4 O3(g)
Given: standard free energies of formation from Appendix IIB
Find: ΔG, kJ
Concept
Plan:
ΔGf of prod & react

 

G  n p Gf products  nr Gf reactants
Relationship
Solution:
s:
G   n p G f products  nr G f reactants

ΔG
 

 [( G f CO 2 )  2(G f H 2O)  (G f O3 )]  [( G f CH 4 )  8(G f O 2 )]
 [( 394 .4 kJ )  2(228 .6 kJ )  (163 .2 kJ) ]  [( 50.5 kJ )  8(0.0 kJ )]
 148 .3 kJ
ΔG Relationships
• if a reaction can be expressed as a series of reactions, the sum of the
ΔG values of the individual reaction is the ΔG of the total reaction
 ΔG is a state function
• if a reaction is reversed, the sign of its ΔG value reverses
• if the amounts of materials is multiplied by a factor, the value of the
ΔG is multiplied by the same factor
 the value of ΔG of a reaction is extensive
ΔG°rxn ≠ Reaction Rate
• Although ΔG°rxn can be used to predict if a reaction will be
spontaneous as written, it does not tell us how fast a reaction will
proceed.
• Example:
C(s, diamond) + O2(g) CO2(g)
But diamonds are forever….
ΔG°rxn= -397 kJ<<0
ΔG°rxn≠ rate
Why Free Energy is ‘Free’
• If ΔG –ve: change in Gibbs free energy = max amount of available
‘free’ energy to do work, should be called ‘Gibbs available energy’
• For many reactions ΔG < ΔH e.g.
C(s) + 2H2(g) →CH4(g)
ΔHrxn = -74.6 kJ, ΔSºrxn = -80.8 J/K, ΔGºrxn = -50.5 kJ
• Exothermic reaction but only half as much available to do work
• Entropy is –ve, so some of emitted heat must inc. entropy of the
universe in order to make ΔGºrxn -ve
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 theoretically a
reversible reaction
Real Reactions
• in a real reaction, some of the free energy is “lost” as heat
 if not most
• therefore, real reactions are irreversible
Energy is lost – when the
battery does work on the
motor some energy is lost in
the wires as heat (resistance
to e- flow)
ΔG under Nonstandard Conditions

Δ G = Δ 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, ΔG = ΔG + RTlnQ
Q is the reaction quotient

at equilibrium ΔG = 0
ΔG = ─RTlnK

ΔG under Nonstandard Conditions


Spilled water spontaneously evaporates even though ΔG◦rxn for the
vaporization of water is +ve..why?
We do not live at std. conditions
ΔG◦rxn = +8.56 kJ/mol
Q = PH2O
ΔG rxn = -ve
Non-std.
Conditions
Under std.
conditions water
does not vaporize
ΔG rxn = ΔG◦rxn
= +ve
Spontaneous in
reverse direction
Water condenses
Example – ΔG nonstandard
•
Calculate ΔG 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
= 1.2 x 10-7
(99.0)3
ΔH°rxm = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J
ΔS°rxn = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K
ΔG° = -92380 J - (700 K)(-198.2 J/K)
ΔG° = +46400 J
ΔG = ΔG° + RTlnQ
ΔG = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10-7)
ΔG = -46300 J = -46 kJ
Free Energy and Equilibrium:
Relating ΔG°rxn to K
• Standard free energy change of a reaction and the equilibrium constant are
related
• K becomes larger as ΔG° becomes more –ve
• If ΔG° is large and –ve then reaction has a large equil. constant, if ΔG° is large
and +ve reaction has a small equil. Constant, favors reactants at equilibrium
• ΔG = ΔG + RTlnQ and at equilibrium Q = K, ΔG = 0
ΔG°rxn = -RT lnK
Free Energy and Equilibrium:
Relating ΔG°rxn to K
• ΔGrxn = ΔGrxn + RT lnQ and at equilibrium Q = K, ΔGrxn = 0
ΔG°rxn = -RT lnK
• When K < 1, lnK is –ve and ΔG°rxn is +ve
 When Q = 1, reaction is spontaneous in reverse direction
• When K > 1, lnK is +ve and ΔG°rxn is -ve
 When Q = 1, reaction is spontaneous in forward direction
• When K = 1, lnK is 0 and ΔG°rxn is 0
 Reaction is at equilibrium under std. conditions
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)
ΔH° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J
ΔS° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K
ΔG° = -92380 J - (700 K)(-198.2 J/K)
ΔG° = +46400 J
ΔG° = -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