Transcript Patino-CHM2046-Chapter17

Chapter 17
Free Energy
and
Thermodynamics
Goals
Entropy (S, S) and spontaneity
Free energy; G, Go
G, K, product- or reactant-favored
Review: H (Enthalpy) and the 1st Law of
Thermodynamics
Chemical Equilibria (ch. 14, etc)
First Law of Thermodynamics
• First Law of Thermodynamics: Energy
cannot be created or destroyed
the total energy of the universe cannot change
it can be transfered from one place to another
 Euniverse = 0 = Esystem + Esurroundings
system = reactants & products
surroundings = everything else
(the transfer of energy from one to the other does
not change the energy of the universe)
3
First Law of Thermodynamics
For an exothermic reaction, heat from the system
goes into the surroundings
• two ways energy can be “lost” from a system,
converted to heat, q
used to do work, w
Energy conservation requires that the energy change in the
system = heat exchanged + work done on the system.
 E = q + w (E = internal energy change)
 E = H – PV (at const. P, qp = H, enthalpy
change)
• State functions (H, P, V). q and w are not.
internal energy change (E) independent of how done 4
Enthalpy, H
•
•
•
•
•
•
related to (includes) the internal energy
H generally kJ/mol
stronger bonds = more stable molecules
if products more stable than reactants, energy
released; exothermic
 H = negative
if reactants more stable than products, energy
absorbed; endothermic
 H = positive
The enthalpy is favorable for exothermic reactions
and unfavorable for endothermic reactions.
• Hess’ Law: H°rxn = S(Hf°prod) - S(Hf°react)
5
Thermodynamics and Spontaneity
• thermodynamics predicts whether a process will
proceed (occur) under the given conditions
spontaneous process
nonspontaneous process does not occur under
specific conditions.
• spontaneity is determined by comparing the free
energy (G) of the system before the reaction with the free
energy of the system after reaction.
if the system after reaction has less free energy than
before the reaction, the reaction is thermodynamically
favorable.
• spontaneity ≠ fast or slow (rate); this is kinetics
6
Spontaneous
ice melts @ 25oC
water freezes @ -10 oC
ball rolls downhill
Nonspontaneous
water freezes @ 25oC
ice melts @ -10oC
ball rolls uphill
2Na(s) + 2H2O(l) 
H2(g) + 2NaOH(aq) 
H2(g) + 2NaOH(aq)
2Na(s) + 2H2O(l)
7
Diamond → Graphite
kinetics:
how fast
Spontaneity:
direction & extent
kinetics
Graphite is thermodynamically more stable than
diamond, so the conversion of diamond into
graphite is spontaneous – but it’s kinetically too
slow (inert) it will never happen in many, many
generations.
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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.
bond energy = amount needed to break a bond.
 H
• The entropy factors relate to the
randomness/orderliness of a system
 S
• The enthalpy factor is generally more important
than the entropy factor
9
Substance
Hf°
kJ/mol
Substance
Hf°
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, S
Entropy is usually described as a measure of the
randomness or disorder; the greater the disorder of a
system, the greater its S.
The greater the order  the smaller its S.
• Entropy is a thermodynamic function that
increases as the number of energetically equivalent
ways of arranging the components increases.
 S generally in J/K (joules/K)
• S = k lnW
k = Boltzmann Constant (R/NA) = 1.38  10-23 J/K
W is the number of energetically equivalent ways,
11
(microstates). It is unitless.
Entropy & Microstates,
W
S = k ln W
Energetically Equivalent
States for the Expansion of
a Gas (4 gas molecules) 1 microstate
S = k ln Wf - k ln Wi
 S  k ln
1 microstate
Wf
Wi
if Wf > Wi , S > 0 &
entropy increases. 6 microstates
(most probable distribution)
12
Changes in Entropy, S
• entropy change is favorable when the result is a
more random system (State C: higher entropy).
 S is positive (S > 0)
Some changes that increase the entropy are:
rxns where products are in a more disordered state.
(solid > liquid > gas) less order
(solid< liquid < gas) larger S (disorder)
reactions which have larger numbers of product
molecules than reactant molecules.
increase in temperature (more movement)
solids dissociating into ions upon dissolving
13
Changes in Entropy in a System
(melting)
Particles fixed in space
Particles can occupy many positions
14
Changes in Entropy in a System
(vaporization)
Particles occupy more space
(larger volume)
15
Changes in Entropy in a System
(solution process)
Structure of solute and solvent disrupted
(also more solute particles)
16
Predict entropy change for a process/reaction
For which process/reaction is S negative?
Freezing ethanol  entropy dec
Mixing CCl4 with C6H6
Condensing bromine vapor  entropy dec
2O3(g)  3O2(g)
4Fe(s) + 3O2(g)  2Fe2O3(s)  entropy dec
2H2O2(aq)  2H2O(l) + O2(g)
2Li(s) + 2H2O(l)  2LiOH(aq) + H2(g)
2NH3(g)  N2(g) + 3H2(g)
17
The 2nd Law of Thermodynamics
• The entropy of the universe increases
in a spontaneous process.
Suniverse = Ssystem + Ssurroundings > 0
Suniverse = Ssystem + Ssurroundings = 0 (equilibrium)
If Ssystem >> 0, Ssurroundings < 0 for Suniverse > 0!
If Ssystem < 0, Ssurroundings >> 0 for Suniverse > 0!
• the increase in Ssurroundings often comes from the heat
released in an exothermic reaction, Hsystem < 0.
18
The 3rd Law of Thermodynamics
-allows determination of entropy of substances.
(W = 1, there is only one way to
arrange the particles to form a perfect
crystal)
S = k ln W = k ln 1 = 0
the 3rd Law states that for a perfect
crystal at absolute zero, the absolute
entropy = 0 J/mol∙K
S = Sf – Si; where Si = 0 @ 0 K
the absolute entropy of a substance
is always (+) positive at the new T
S = k ln W
19
Standard Entropies
• S°
• 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
Values can be used to calculate the standard
entropy change for a reaction, Sorxn (= Sosys20 )
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
Trends: Standard Entropies
Molar Mass
• For monatomic species, the
larger the molar mass, the
larger the entropy
• available energy states more
closely spaced, allowing
more dispersal of energy
through the states
22
Trends: Standard Entropies
States
• the standard entropy of a substance in the
gas phase is greater than the standard
entropy of the same substance in the solid or
liquid phase at a particular temperature
Substance
S°,
(J/mol∙K)
H2O (l)
70.0
H2O (g)
188.8
23
Trends: Standard Entropies
Allotropes
-different forms of an element
• the more highly
ordered form has
the smaller entropy
24
Trends: Standard Entropies
Molecular Complexity (inc # of atoms)
Molar
S°,
Substance
Mass (J/mol∙K)
Ar (g)
39.948
154.8
larger, more complex
molecules generally have NO (g)
larger entropy
• more available energy
states, allowing more
dispersal of energy
through the states
30.006
210.8
25
Trends: 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
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Q. Arrange the following in order of increasing entropy
@ 25oC! (lowest to highest)
Ne(g), SO2(g), Na(s), NaCl(s) and H2(g)
Na(s) < NaCl(s) < H2(g) < Ne(g) < SO2(g)
Q . Which has the larger entropy in each pair?
a) Li(s) or Li(l)
b) C2H5OH(l) or CH3OCH3 (l)
c) Ar(g) or Xe(g)
d) O2(g) or O3(g)
27
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
Calculate S for the reaction
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
Find: S, J/K
Concept Plan:


Relationships:


 S  S n pS
 S  S n pS
Solution:
S
SNH3, SO2, SNO, SH2O,
 [ 4 (S


products
NO( g ) )
 [ 4 ( 210 . 8
 178 . 8
J
K
 6 (S

products
  S n S
  S n S
r

H 2 O( g ) )]
)  6 (188 . 8
J
K
r

reactants


reactants
 [ 4 (S


NH
)]  [ 4 (192 . 8
3(g)
J
K
)  5 (S

O 2 ( g ) )]
)  5 ( 205 . 2
J
K
)]
J
K
Check: S is +, as you would expect for a reaction with
more gas product molecules than reactant molecules
Calculating ∆So for a Reaction
∆So = S So (products) - S So (reactants)
Consider 2 H2(g) + O2(g)  2 H2O(liq) @ 25oC
∆So = 2 So (H2O) - [2 So (H2) + So (O2)]
∆So = 2 mol (69.9 J/K•mol) [2 mol (130.6 J/K•mol) +
1 mol (205.0 J/K•mol)]
∆So = -326.4 J/K
Note that there is a decrease in S because 3 mol of
gas give 2 mol of liquid.
Temperature Dependence of Ssurroundings
Hsystem < 0 (exothermic), it adds heat to the
surroundings, increasing the entropy of the
surroundings (Ssurroundings > 0 )
Hsystem > 0 (endothermic), it takes heat from the
surroundings, decreasing the entropy of the
surroundings (Ssurroundings < 0 )
ΔS s urroundin gs   ΔH s y s t em
ΔS s urroundi n gs 
ΔS s urroundin gs 
 ΔH s y s t em
T
1
T
The more negative Hsyst
and the lower the
temperature the higher
(more positive) Ssurr
30
Calculating ∆So for the surroundings
2 H2(g) + O2(g)  2 H2O(liq) @ 25 oC
∆Sosystem = -326.4 J/K
ΔS s urroundi n gs 
 ΔH s y s t em
T
Can calculate ∆Hosystem = ∆Horxn = -571.7 kJ
(also from tabulated data)
S
o
surroundin gs
=
- (-571.7 kJ)(1000 J/kJ)
298 K
∆Sosurroundings = +1917 J/K
Suniverse = Ssystem + Ssurroundings
Given Sosurr , Sosys and T,
determine Souniv and predict
if the reaction will be
spontaneous.
2 H2(g) + O2(g)  2 H2O(liq) @ 25oC
∆Sosystem
= -326.4 J/K
∆Sosurroundings = +1917 J/K
∆Souniverse = +1591 J/K
The entropy of the
universe is increasing,
so the reaction is
spontaneous ( productfavored). (see slide # 18)
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
ΔS s urr 
Relationships:
Solution:
S
T, H
 S surr 
 ΔH s y s
  H sys
T
 S surr  6 . 86
T

   2044 kJ 
298 K
kJ
K
  6.86  10
3 J
K
Check: combustion is largely exothermic, so the entropy of
the surroundings should increase (inc in # gas mol)
Spontaneous or Not?
Suniverse = Ssystem – Hsystem/T
Hsystem
Ssystem
Spontaneous?
Exothermic
Hsys < 0
Less order
Ssys > 0
Spontaneous under all
conditions; Suniv > 0
Exothermic
Hsys < 0
More order
Ssys < 0
Favorable at low T
Endothermic
Hsys > 0
Less order
Ssys > 0
Favorable at high T
Endothermic
Hsys > 0
More order
Ssys < 0
Not spontaneous
under any conditions
Suniv < 0
originally: Suniverse = Ssystem + Ssurroundings
but
ΔS s urr 

ΔH s y s
T
Without doing any calculations, determine the sign of Ssys and
Ssurr for each reaction. Predict under what temperatures (all T,
low T, or high T) the reaction will be spontaneous.
Suniverse = Ssystem + (-Hsys/T)
Ssurr = (-Hsys/T)
2CO(g) + O2(g)  2CO2(g)
Hrxn = -566.0 kJ
Ssystem = (-); 3 mol gas form 2 mol gas
Ssurr = (+); spontaneous @ low T
2NO2(g)  O2(g) + 2NO(g)
Hrxn = +113.1 kJ
Ssystem = (+); 2 mol gas form 3 mol gas
Ssurr = (-); spontaneous @ high T
35
Without doing any calculations, determine the sign of Ssys and 
Ssurr for each reaction. Predict under what temperatures (all T,
low T, or high T) the reaction will be spontaneous.
Suniverse = Ssystem + (-Hsys/T)
2H2(g) + O2(g)  2H2O(g)
Hrxn = -483.6 kJ
Ssystem = (-); 3 mol gas form 2 mol gas
Ssurr = (+) ; spontaneous @ low T
CO2(g)  C(s) + O2(g)
Hrxn = +393.5 kJ
Ssystem = (-); complicated gas forms a solid & gas
Ssurr = (-); nonspontaneous @ all T
Ssurr = (-Hsys/T)
36
At what temperature is the change in entropy for the
reaction equal to the change in entropy for the
surroundings, if Horxn = -127 kJ and Sorxn = 314 J/K.
Plan: set Sorxn = Sosurr and solve for T;
convert kJ to J
ΔS
o
s urr

 ΔH
ΔS
o
 ΔH
T
o
sys
sys
o
sys
 ΔS
 T
o
sys
rxn implies system!!!
 (  127 , 000 J )
T
314 J/K
Ans: T = +404 K
37
Gibbs Free Energy, G = H -TS
∆Suniv = ∆Ssurr + ∆Ssys
ΔS univ =
 ΔH s y s
T
+ Δ S sys
Multiply through by -T
-T∆Suniv = ∆Hsys - T∆Ssys
J. Willard Gibbs
1839-1903
-T∆Suniv = change in Gibbs free energy for
the system = ∆Gsystem
hence, ∆G = ∆H -T∆S
Under standard conditions —
∆Gosys = ∆Hosys - T∆Sosys
∆G = ∆H - T∆S
Gibbs free energy change = total energy
change for system
- energy lost in disordering the system
If the reaction is
• exothermic (negative ∆H)
• and entropy increases (positive ∆So)
• then ∆G must be NEGATIVE
• the reaction is spontaneous (and productfavored) at ALL temperatures.
∆G = ∆H - T∆S
• G will be positive (∆G > 0) when H is
positive (endothermic) and S is negative (more
ordered). So the change in free energy will be
positive at all temperatures.
• The reaction will therefore be nonspontaneous
at ALL temperatures
• When G = 0 the reaction is at equilibrium
40
G = H – TS
Spontaneous or Not?
A decrease in Gibbs free energy (G < 0) corresponds
to a spontaneous process
An increase in Gibbs free energy (G > 0) corresponds
to a nonspontaneous process
41
Calculating ∆Go : ∆Go = ∆H -T∆So
Combustion of acetylene @ 25 oC
C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(g)
Use enthalpies of formation to calculate
∆Horxn = -1238 kJ
Use standard molar entropies to calculate
∆Sorxn = -97.4 J/K or -0.0974 kJ/K
∆Gorxn = -1238 kJ - (298 K)(-0.0974 J/K)
= -1209 kJ
(spontaneous)
Reaction is product-favored in spite of negative
∆Sorxn.
Reaction is “enthalpy driven”
Calculating ∆Go : ∆Go = ∆H -T∆So
Is the dissolution of ammonium nitrate productfavored?
If so, is it enthalpy- or entropy-driven?
NH4NO3(s) + heat  NH4NO3(aq)
From tables of thermodynamic data we find
∆Horxn = +25.7 kJ
(endothermic)
∆Sorxn = +108.7 J/K or +0.1087 kJ/K
(disorder)
∆Gorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K)
= -6.7 kJ
(spontaneous)
Reaction is product-favored in spite of positive ∆Horxn.
Reaction is “entropy driven”
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:
Relationships:
Solution:
H = +95.7 kJ, S = 142.2 J/K, T = 298 K
G, kJ
T, H, S
G
G  H  TS
G  H  TS
  95.7  10 J    298 K  142 . 2 
3
J
K
  5 . 33  10 J
Answer: Since G is +, the reaction is not spontaneous at
this temperature. To make it spontaneous, we need
to increase the temperature.
4
The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has
H = +95.7 kJ and S = +142.2 J/K.
Calculate the minimum T @ which it will be spontaneous.
Given:
Find:
Concept Plan:
H = +95.7 kJ, S = 142.2 J/K, G < 0
T, K
G, H, S
G  H  TS
Relationships:
Solution:
T
 G   H  T  S   0
3
 95.7  10 J   T  142 .2   0
 95.7  10 J   T  142 .2 
3
J
K
3
 95.7  10 J   T
 142 .2 
J
K
J
K
673 K  T
Answer: The temperature must be higher than 673K for the
reaction to be spontaneous (i.e. 674 K)
Gibbs Free Energy, G
Calculating ∆Go (two ways)
a)
Determine ∆Horxn and ∆Sorxn and use
Gibbs equation (at various temps).
b)
free
energies of formation, ∆Gfo
Use tabulated values of
@ 25oC
∆Gorxn = S ∆Gfo (products) - S ∆Gfo (reactants)
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
47
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)
Calculate G at 25C for the reaction
CH4(g) + 8 O2(g)  CO2(g) + 2 H2O(g) + 4O3(g)
Given: standard free energies of formation from Appendix IIB
Find: G, kJ
Concept Plan:
Gf of prod & react
Relationships:
G  Sn p G



f products
G
  S n  G
r

f reactants

Solution:

G  Sn pG

 [(  G

CO
f

products
f
)  2(G
2
  S n  G

H O
f 2
r
)  (G

reactants
f

O
f 3

)]  [(  G

CH
f
)  8( G
4

O
f 2
)]
 [(  394 . 4 kJ )  2 (  228 . 6 kJ )  4 (  163 . 2 kJ) ]  [(  50 . 5 kJ )  8 ( 0 . 0 kJ )]
  148 . 3 kJ
(spontaneous)
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




G  H  TS
Relationships:

Solution:
(PRACTICE PROBLEM)
G  H  TS




  98.9  10 J  398 K   94 . 0
3
J
K

3
  61 . 5  10 J   61 . 5 kJ
Answer: Since G is -ve, the rxn is spontaneous at this
temperature, but less spontaneous than at 25C
(-127 kJ)
∆G, ∆G˚, and Keq
• ∆G is the change in free energy at non•
•
•
•
•
standard conditions.
∆G is related to ∆G˚
∆G = ∆G˚ + RT ln Q
where Q = reaction quotient
When Q < K or Q > K, the reaction may be
spontaneous or nonspontaneous.
When Q = K reaction is at equilibrium
When ∆G = 0 reaction is at equilibrium
• Therefore, ∆G˚ = - RT ln K
Thermodynamics and Keq

FACT: ∆Gorxn is the change in free energy when
pure reactants convert COMPLETELY to pure
products, both at standard conditions.

FACT: Product-favored systems have
Keq > 1 (∆G˚rxn < 0).

Therefore, both ∆G˚rxn and Keq are related
to reaction favorability.
Summary: ∆G˚ = - RT ln K
Thermodynamics and Keq
∆Gorxn = - RT lnK
Calculate K for the reaction @ 25 oC
N2O4  2 NO2
∆Gorxn = +4.8 kJ
∆Gorxn = +4800 J = - (8.31 J/K)(298 K) ln K
ln K =
-
4800 J
(8.31 J/K)(298
K)
=
- 1.94
K = e–1.94 = 0.14 (reactant favored)
When ∆Gorxn > 0 (nonspontaneous),
then K < 1!!
• Estimate the equilibrium constant and position of
equilibrium for the following reaction at 427°C
N2(g) + 3 H2(g) ⇄ 2 NH3(g)
NH3
N2
H° = [ 2(-46.19)] − [ 0
H2
+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 (nonspontaneous)
G° = -RT lnK
+46400 J = -(8.314 J/K)(700 K) lnK
ln K = -7.97
K = e−7.97 = 3.45  10−4 small!!!
since K is << 1, the position of equilibrium favors reactants
• 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)
PNH32
(2.0 atm)2
-7
Q = P 1  P 3 = (33.0 atm)1 (99.0)3 = 1.2  10
N2
H2
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 (nonspontaneous)
G = G° + RTlnQ
spontaneous
G = +46400 J + (8.314 J/K)(700 K)(ln 1.2  10-7)
G = 46400 J − 92700 J = -46300 J = −46 kJ  −G°
54
Q. Rank the following in order of increasing molar
entropy (So) @ 25oC!
a) Cl2(g), I2(g), Br2(g), and F2(g)
F2(g) < Cl2(g) < Br2(g) < I2(g)
b) H2O(g), H2O2 (g), H2S(g)
H2O(g) < H2S(g) < H2O2(g)
55
Use only the information provided here to determine the value
of Ssurr @ 355 K. Predict whether this reaction shown will
be spontaneous @ this temperature, if Hrxn = -114 kJ.
2N O (g)  O 2 (g)  2N O 2 (g)
A) Ssurr = +114 kJ/K, reaction is not spontaneous
B) Ssurr = +321 J/K, reaction is spontaneous
C) Ssurr = +114 kJ/K, reaction is spontaneous
D) Ssurr = -355 J/K, reaction is not spontaneous
E) Ssurr = +321 J/K, it is not possible to predict
the spontaneity of this reaction without more information.
56
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
and
G° = H − T S°
H − TS° = − RTlnK,
by rearranging
RTlnK = −H − TS°,
and dividing by R T
Temperature Dependence of K
RTlnK
−H
TS°
———— = ———— + ————
RT
RT
RT


 H rxn  1   S rxn
ln K  
 
R T
R
• 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