THERMODYNAMICS – ENTROPY AND FREE ENERGY
Thermodynamics studies the energy of a system, how much work a system
could produce, and how to predict the spontaneous changes a system will
undergo
3A-1 (of 14)
1847 HERMANN VON HELMHOLTZ
Presented a mathematical argument for the Law of
Conservation of Energy
FIRST LAW OF THERMODYNAMICS – The change in energy of a system is
equal to heat that enters the system plus the work done on the system
ΔE = q + w
3A-2 (of 14)
ENTHALPY (H) – The total energy of a system plus the pressure-volume
product of a system
H = E + pV
ΔH = ΔE + ΔpV + pΔV
For processes at constant pressure
ΔH = ΔE + pΔV
From the first law
ΔE = q + w
ΔE = q + -pΔV
ΔE + pΔV = q
At constant pressure the enthalpy change of a system is equal to the
heat that enters the system
qp = ΔH
3A-3 (of 14)
1865 RUDOLF CLAUSIUS
Proposed that there is always some thermal energy not
available to be converted into work ( ΔT )
He called this ENTROPY
SECOND LAW OF THERMODYNAMICS – In any spontaneous process
there is always an increase in total entropy
3A-4 (of 14)
1890 LUDWIG BOLTZMANN
Proposed a statistical meaning for entropy
ENTROPY (S) – A measure of the number of arrangements available to a
system in a given state
3A-5 (of 14)
Boltzmann calculated the entropy of a system by
S = k log W
S = entropy
k = Boltzmann Constant (R/NA, or 1.38 x 10-23 J/moleculeK)
W = number of ways particles can be arranged in a given state while
keeping the total energy constant
3A-6 (of 14)
Calculate the entropies of the following samples of CO2 (g) and CO2 (s)
W = 20 x 19 = 380
W=2x1=2
(1.38 x 10-23 J/K) log 380
(1.38 x 10-23 J/K) log 2
= 3.56 x 10-23 J/K
= 4.15 x 10-24 J/K
The state of matter indicates the relative entropies of different substances
Ssolid < Sliquid < Sgas
3A-7 (of 14)
OTHER FACTORS IN PREDICTING S FOR CHEMICAL SUBSTANCES
After states, S is higher for substances with
1)
2)
3)
4)
larger masses
delocalized electron systems
weaker bonds
more internal complexity
Cl2 > Br2
Sgas is higher than Sliquid
Cl2 > F2
S is higher for larger masses
C3H8 > C2H6
S is higher for larger mass
C6H6 > C6H12
S is higher for delocalized pi systems
Be > Cdiamond
S is higher for delocalized metallic bonding
K > Cr
S is higher for weaker metallic bonding
CO > N2
S is higher for more internal complexity
3A-8 (of 14)
THIRD LAW OF THERMODYNAMICS – The entropy of a perfect crystal at
0 K is 0
At 0 K
C O
C O
C O
C O
C O
C O
C O
C O
C O
C O
C O
C O
C O
C O
C O
C O
C O
C O
C O
C O
C O
O C
C O
C O
C O
C O
C O
C O
C O
C O
C O
C O
W=1
W = 16
(1.38 x 10-23 J/K) log 1
(1.38 x 10-23 J/K) log 16
= 0 J/K
= 1.66 x 10-23 J/K
Because of the 3rd Law, all substances have entropy values > 0
3A-9 (of 14)
CALCULATING ΔS FOR CHEMICAL CHANGES
Third Law Entropies of substances in their standard states at 25ºC (Sº) are
found in the appendix of your textbook and on the class website
Thermodynamic data is tabulated under specified conditions called
STANDARD CONDITIONS
STANDARD STATE – Gaseous reactants and products are at 1 atm,
dissolved reactants and products are at 1 M, elemental reactants and
products are in their natural state at room temperature and pressure
Standard thermodynamic data is identified with a “º”, such as ΔHº
3A-10 (of 14)
Sº values are not ΔSºf values, they
are the absolute entropy (the innate
disorder) of each substance
All Sº values are positive
ΔSºreaction = Sºproducts - Sºreactants
Be (s) + O2 (g) + H2 (g) → Be(OH)2 (s)
3A-11 (of 14)
Calculate the standard entropy change for
2NiS (s) + 3O2 (g) → 2SO2 (g) + 2NiO (s)
Sº for NiS (s) = 53 J/molK
Sº for O2 (g) = 205 J/molK
Sº for SO2 (g) = 248 J/molK
Sº for NiO (s) = 38 J/molK
2 mol SO2 (248 J/molK SO2) + 2 mol NiO (38 J/molK NiO)
- 2 mol NiS (53 J/molK NiS)
- 3 mol O2 (205 J/molK O2)
= -149 J/K
This says the system is becoming more ordered
3A-12 (of 14)
PREDICTING ΔS FOR CHEMICAL CHANGES
ΔS is + for reactions that have
1) more gaseous products than reactants
2) solid reactants dissolving
3) dissolved gases coming out of solution
CaCO3 (s) → CaO (s) + CO2 (g)
0 gas molecules → 1 gas molecule
ΔS = +
2SO2 (g) + O2 (g) → 2SO3 (g)
3 gas molecules → 2 gas molecules
ΔS = –
Ag+ (aq) + Cl- (aq) → AgCl (s)
dissolved ions precipitate
ΔS = –
H+ (aq) + HCO3- (aq) → H2O (l) + CO2 (g)
gas evolved from solution
3A-13 (of 14)
ΔS = +
Predict the sign of the entropy change for each:
Dissolving of solid sugar
ΔS is positive
Deposition of iodine vapor
ΔS is negative
3A-14 (of 14)
SPONTANEITY
SPONTANEOUS PROCESS – One that proceeds without outside intervention
SECOND LAW OF THERMODYNAMICS – In any spontaneous process
there is always an increase in total entropy
ΔSsys + ΔSsurr
=
ΔSuniv
ΔSuniv > 0
3B-1 (of 18)
Energy + H2O (l) ⇆ H2O (g)
ΔSuniv = ?
To calculate ΔSuniv, one must know ΔSsys and ΔSsurr
(1) ΔSsys
ΔSsys = 1 mol (189 J/molK) – 1 mol(70. J/molK) = 119 J/K
= 0.119 kJ/K
3B-2 (of 18)
Energy + H2O (l) ⇆ H2O (g)
T = 298 K
(2) ΔSsurr
For a process at constant pressure
ΔSsurr = -ΔHsys
________
T
ΔHsys = 1 mol (-242 kJ/mol) – 1 mol(-286 kJ/mol) = 44 kJ
ΔSsurr = -44 kJ
________
298 K
3B-3 (of 18)
= -0.15 kJ/K
Energy + H2O (l) ⇆ H2O (g)
T = 298 K
ΔSuniv = ΔSsys + ΔSsurr
ΔSuniv = 0.119 kJ/K - 0.15 kJ/K = -0.03 kJ/K
∴ this process is nonspontaneous at 298 K
3B-4 (of 18)
Energy + H2O (l) ⇆ H2O (g)
T = 398 K
ΔHsys and ΔSsys are temperature independent
ΔSsys = 0.119 kJ/K
ΔSsurr = -44 kJ
________
= -0.11 kJ/K
398 K
ΔSuniv = 0.119 kJ/K - 0.11 kJ/K = 0.01 kJ/K
∴ this process is spontaneous at 398 K
3B-5 (of 18)
1873 JOSIAH WILLARD GIBBS
Proposed a way to determine the maximum amount of energy
that could be converted into work
Gibbs Free Energy (G) – The maximum energy of a system that can be
converted into work
3B-6 (of 18)
G = H - TS
ΔGsys = ΔHsys - TΔSsys
-ΔGsys = -ΔHsys + ΔSsys
________
________
T
T
ΔSsurr
-ΔGsys = ΔSsurr + ΔSsys
________
T
ΔSuniv
3B-7 (of 18)
-ΔGsys = ΔSuniv
________
T
ΔSuniv > 0 for a spontaneous process

ΔGsys < 0 for a spontaneous process
The change in Gibbs Free Energy (ΔG) indicates if a process is spontaneous
or nonspontaneous
3B-8 (of 18)
ΔG = ΔH - TΔS
Spontaneous processes are favored by a decrease in enthalpy and an
increase in entropy
3B-9 (of 18)
CONTRIBUTIONS OF ΔH AND ΔS TO SPONTANEITY
ΔG = ΔH - TΔS = (-)
(+) -- (+)
(-)
(-) =
(+)
ΔH
- exothermic
+ endothermic
+ endothermic
- exothermic
(+)
(?)
(-)
ΔS
+ more disorder
- more order
+ more disorder
- more order
ΔG
- spontaneous
+ nonspontaneous
? spontaneous at high T
? spontaneous at low T
While ΔH and ΔS are temperature independent, ΔG is temperature dependent
3B-10 (of 18)
For the reaction: H2O (s) ⇆ H2O (l)
ΔHºfus = 6,038 J
ΔSºfus = 22.1 J/K
Find ΔGº at -20.0ºC and tell if melting or freezing is spontaneous.
253ΔGº
= ΔHº - TΔSº
= 6,038 J - (253.2 K)(22.1 J/K) = 442 J
reaction is nonspontaneous  reverse reaction is spontaneous
H2O (l) freezes at -20.0ºC
3B-11 (of 18)
For the reaction: H2O (s) ⇆ H2O (l)
ΔHºfus = 6,038 J
ΔSºfus = 22.1 J/K
Find ΔGº at 0.0ºC.
273ΔGº
= ΔHº - TΔSº
= 6,038 J - (273.2 K)(22.1 J/K) = 0 J
The temperature of a phase change (like melting) is when the two phases
are in equilibrium
For any process at equilibrium, ΔG = 0
3B-12 (of 18)
For the reaction: H2O (l) ⇆ H2O (g)
ΔHºvap = 44,400 J
ΔSºvap = 119 J/K
Find the normal boiling point of water
At a phase change temperature there is an equilibrium between the 2 phases
ΔGº = ΔHº - TΔSº
0
= ΔHº - TΔSº
- ΔHº = - TΔSº
ΔHº = T = 44,400 J
_____
____________
ΔSº
119 J/K
3B-13 (of 18)
=
373 K
=
100. ºC
PREDICTING ΔG FOR STANDARD STATE CHEMICAL REACTIONS
1) ΔG = ΔH - TΔS or, in standard states, ΔGº = ΔHº - TΔSº
2) Standard Free Energies of Formation at 25ºC (ΔGºf) found in the
appendix of the textbook
ΔGºreaction = ΣΔGºf (products) - ΣΔGºf (reactants)
ΔGºf of elements in their standard states = 0
3B-14 (of 18)
Calculate the standard free energy change at 25°C for
2CH3OH (g) + 3O2 (g) → 2CO2 (g) + 4H2O (g)
298ΔGº
for CH3OH (g) = -163 kJ/mol
298ΔGº for O (g)
=
0 kJ/mol
f
2
f
2 mol CO2 (-394 kJ/mol CO2)
298ΔGº
for CO2 (g) = -394 kJ/mol
298ΔGº for H O (g) = -229 kJ/mol
f
2
f
+ 4 mol H2O (-229 kJ/mol H2O)
- 2 mol CH3OH (-163 kJ/mol CH3OH) - 3 mol O2 (0 kJ/mol O2)
= -1378 kJ
This says, when starting with a container having all of the reactants and
products in their standard states, the forward reaction is spontaneous
3B-15 (of 18)
Calculate ΔHº, ΔSº, and ΔGº at 25ºC for
2SO2 (g) + O2 (g) → 2SO3 (g)
SO2 (g)
O2 (g)
SO3 (g)
ΔHºf (kJ/mol)
-297
0
-396
Sº (J/molK)
248
205
257
ΔHº
2 mol SO3 (-396 kJ/mol SO3)
- 2 mol SO2 (-297 kJ/mol SO2) - 1 mol O2 (0 kJ/mol O2)
= -198 kJ ∴ reaction is exothermic
3B-16 (of 18)
Calculate ΔHº, ΔSº, and ΔGº at 25ºC for
2SO2 (g) + O2 (g) → 2SO3 (g)
SO2 (g)
O2 (g)
SO3 (g)
ΔHºf (kJ/mol)
-297
Sº (J/molK)
248
0
-396
205
257
ΔSº
2 mol SO3 (257 J/molK SO3)
- 2 mol SO2 (248 J/molK SO2) - 1 mol O2 (205 J/molK O2)
= -187 J/K ∴ reaction becomes more ordered
3B-17 (of 18)
Calculate ΔHº, ΔSº, and ΔGº at 25ºC for
2SO2 (g) + O2 (g) → 2SO3 (g)
SO2 (g)
O2 (g)
SO3 (g)
ΔHºf (kJ/mol)
-297
0
-396
Sº (J/molK)
248
205
257
ΔGº
ΔGº = ΔHº - TΔSº
= -198 kJ - (298.2 K)(-0.187 kJ/K)
= -142 kJ ∴ when all reactants and products are present and start in
their standard states, the forward reaction is spontaneous
3B-18 (of 18)
PREDICTING ΔG FOR NON-STANDARD STATE CHEMICAL REACTIONS
ΔGº predicts spontaneity when reactants and products are in standard
states
ΔG predicts spontaneity when reactants and products are not in
standard states
ΔG = ΔGº + RT ln Q
3C-1 (of 14)
When relating thermodynamic functions (ΔG) with Q, Q must be
expressed in ACTIVITIES instead of molarities or pressures
ACTIVITY (a) – The effective pressure or concentration of a substance
due to interactions between molecules, atoms, or ions
The ratio of the pressure of a gas to its standard state pressure,
or, the ratio of the concentration of a dissolved substance to its standard
state concentration
3C-2 (of 14)
Calculate the free energy change for
2CO (g) + O2 (g) → 2CO2 (g)
in a container that is 0.010 atm CO, 0.020 atm O2, and 3.0 atm CO2 at 25ºC.
The ΔGºf for CO (g) is -137 kJ/mol and the ΔGºf for CO2 (g) is -394 kJ/mol.
ΔG = ΔGº + RT ln Q
ΔGº = 2 mol CO2 (-394 kJ/mol CO2)
- 2 mol CO (-137 kJ/mol CO) - 1 mol O2 (0 kJ/mol O2)
= -514 kJ
Q = aCO22
__________
aCO2aO2
=
(3.0)2
_________________
= 4.5 x 106
(0.010)2(0.020)
ΔG = -514,000 J + (8.314 J/K)(298.2 K) ln 4.5 x 106 = -476,000 J
3C-3 (of 14)
For a reaction at equilibrium
ΔG = 0
ΔG = ΔGº + RT ln Q
0 = ΔGº + RT ln Q
0 = ΔGº + RT ln Keq
-RT ln Keq = ΔGº
ΔGº = -RT ln Keq
ΔGº is a valuable quantity because it is related to a reaction’s equilibrium
constant
3C-4 (of 14)
RELATIONSHIP BETWEEN Keq AND ΔGº
The magnitude of the Keq is indicated by the sign of ΔGº
ΔGº
Keq
1
>1
<1
ΔGº = -RT ln (1)
ΔGº = -RT ln (>1)
ΔGº = -RT ln (<1)
3C-5 (of 14)
0
+
∴ if ΔGº is 0 it means Keq = 1
∴ if ΔGº is - it means Keq > 1
∴ if ΔGº is + it means Keq < 1
HF (aq) + H2O (l) ⇆ H3O+ (aq) + F- (aq)
The Keq (Ka) for the above reaction is 7.2 x 10-4 at 25ºC. Calculate ΔGº.
ΔGº = -RT ln Keq
ΔGº = -(8.314 J/K)(298.2 K) ln (7.2 x 10-4)
ΔGº = 18,000 J
3C-6 (of 14)
An equilibrium mixture at 25ºC has pNO = 0.125 atm, pN2O = 3.50 atm, and
pO2 = 1.74 atm. Find Kp and ΔGº for the reaction
4NO (g) ⇆ 2N2O (g) + O2 (g)
Kp = pN2O2pO2
___________
=
pNO4
(3.50 atm)2(1.74 atm)
__________________________
(0.125 atm)4
ΔGº = -RT ln Keq
ΔGº = -(8.314 J/K)(298.2 K) ln (8.731 x 104)
ΔGº = -28,200 J
3C-7 (of 14)
=
8.731 x 104
Find ΔGº and Keq at 25ºC for the reaction
N2O4 (g) ⇆ 2NO2 (g)
If the ΔGºf for N2O4 (g) is 98 kJ/mol and the ΔGºf for NO2 (g) is 52 kJ/mol.
ΔGº = 2 mol NO2 (52 kJ/mol NO2) - 1 mol N2O4 (98 kJ/mol N2O4) = 6 kJ
ΔGº = -RT ln Keq
ΔGº = ln Keq
_____
-RT
e-ΔGº/RT
= Keq
Keq = e-(6,000 J)/(8.314 J/K)(298.2 K) = 0.09
3C-8 (of 14)
Find ΔGº and Ka at 25ºC for the reaction
HNO2 (aq) + H2O (l) ⇆ H3O+ (aq) + NO2- (aq)
If the ΔGºf for HNO2 (aq) is -54 kJ/mol, the ΔGºf for H2O (l) is -237 kJ/mol, the
ΔGºf for H3O+ (aq) is -237 kJ/mol, and the ΔGºf for NO2- (aq) is -35 kJ/mol.
ΔGº =
1 mol H3O+ (-237 kJ/mol H3O+) + 1 mol NO2- (-35 kJ/mol NO2-)
- 1 mol HNO2 (-54 kJ/mol HNO2) - 1 mol H2O (-237 kJ/mol H2O)
= 19 kJ
e-ΔGº/RT
= Keq
= e-(19,000 J)/(8.314 J/K)(298.2 K) = 4.7 x 10-4
How would you find ΔGº and Ka at 35ºC for the reaction?
308ΔGº
=
298ΔHº
308
e ΔGº/R(308 K) = Keq
3C-9 (of 14)
+ (308 K)298ΔSº
Solid NH4Cl is placed in a 5.00 L flask at 25ºC and at equilibrium the total
pressure is 0.250 atm. Find Kp and ΔGº for the reaction
NH4Cl (s) ⇆ NH3 (g) + HCl (g)
NH4Cl (s)
Initial atm’s
Change in atm’s
Equilibrium atm’s
Kp = pNH3pHCl
⇆
NH3 (g)
+
HCl (g)
0
+x
0
+x
x
x
ptotal
=
pNH3
+
pHCl
= x2
0.250 atm
=
x
+
x
= (0.125)2 = 0.0156
0.125 atm
=
x
3C-10 (of 14)
Solid NH4Cl is placed in a 5.00 L flask at 25ºC and at equilibrium the total
pressure is 0.250 atm. Find Kp and ΔGº for the reaction
NH4Cl (s) ⇆ NH3 (g) + HCl (g)
ΔGº = -RT ln Keq
= -(8.314 J/K)(298.2 K) ln 0.0156 = 10,300 J
3C-11 (of 14)
1.511 g HCl (g) are placed in a 3.00 L container and partially decomposes.
The equilibrium pressure of HCl (g) is 0.310 atm at 25ºC. Find Kp and ΔGº for
2HCl (g) ⇆ H2 (g) + Cl2 (g)
1.511 g HCl x
1 mol HCl
________________
= 0.041443 mol HCl
36.46 g HCl
pV = nRT
p = nRT = (0.041443 mol)(0.08206 Latm/molK)(298.2 K) = 0.3380 atm
_____
V
3C-12 (of 14)
_________________________________________________________
(3.00 L)
1.511 g HCl (g) are placed in a 3.00 L container and partially decomposes.
The equilibrium pressure of HCl (g) is 0.310 atm at 25ºC. Find Kp and ΔGº for
2HCl (g) ⇆ H2 (g) + Cl2 (g)
2HCl (g)
Initial atm’s
Change in atm’s
Equilibrium atm’s
⇆
H2 (g)
+
0
+x
0.338
- 2x
0.338 - 2x
x
0.338 – 2x = 0.310
x = 0.014
Kp = pH2 pCl2
_________
pHCl2
3C-13 (of 14)
= (0.014)2
__________
(0.310)2
=
2.0 x 10-3
Cl2 (g)
0
+x
x
1.511 g HCl (g) are placed in a 3.00 L container and partially decomposes.
The equilibrium pressure of HCl (g) is 0.310 atm at 25ºC. Find Kp and ΔGº for
2HCl (g) ⇆ H2 (g) + Cl2 (g)
ΔGº = -RT ln Keq
= -(8.314 J/K)(298.2 K) ln 2.04 x 10-3 = 15,000 J
3C-14 (of 14)