The Gibbs Free Energy
Using the entropy criterion to decide whether a process can occur we have to calculate
the entropy change of the system and of the surroundings
We have concluded that,
∆S univ > 0
∆S univ = 0
∆S univ < 0
Spontaneous
Equilibrium
Impossible
Very impractical!
We need a state property to give a feasibility of the process without reference to
surroundings
Gibbs Free Energy (G): a state function (constant T and P)
G = H – TS (in Joules)
Since T is constant, ∆G system = ∆H system - T∆S system
Recall, ∆S surroundings =
− ∆H system
T
If the process is spontaneous: ∆S universe = ∆S system + ∆S surroundings > 0
Therefore, ( ∆S system +
− ∆H system
)>0
T
We multiply both sides of the inequality by T, and (T∆S system - ∆H system ) > 0 and
∆H system - T∆S system <0 ⇒ ∆G system = ∆H system - T∆S system <0 (at constant T and P)
• If the Gibbs free energy is forecast to decrease during a process at constant T and
P the process is spontaneous.
• If the Gibbs free energy is forecast to increase the process cannot occur under the
conditions specified (∆S univ < 0) but the reverse of the process under the specified
conditions is spontaneous.
• If the Gibbs free energy is forecast to stay constant, ∆S univ = 0 for the process and
no change in either direction can occur under the specified conditions and this is
synonymous to EQUILIBRIUM.
1
For a change at constant T and P:
∆G system < 0 Spontaneous
∆G system = 0 Equilibrium
∆G system >0 Non-spontaneous but reverse “is”
The Gibbs Energy and Phase Transitions
H2O(l) • H2O(s)
At the freezing point (273.15K) and 1 atm pressure, ∆H 273.15 = - 6007 J mol-1 and
∆S273.15
∆H 273.15 − 6007 Jmol −1
=
=
= −21.99 JK −1mol −1
T freeze
273.15K
∆G 273.15 = ∆H 273.15 - T∆S 273.15 = (-6007 J mol-1) – (273.15 K) (-21.99 J K-1 mol-1)
= 0 J mol-1 - No surprise, equilibrium!
What happens if water is cooled down to 263.15 K (- 10oC)?
We assume that both ∆H and ∆S do not change significantly with the drop in
temperature, and therefore,
∆G 263.15 = ∆H 263.15 - T∆S 263.15 = (-6007 J mol-1) – (263.15 K) (-21.99 J K-1 mol-1)
= - 220 J mol-1 - water freezes spontaneously (no external intervention)
What happens if water is heated up to 283.15 K (+ 10oC)?
Again we assume that both ∆H and ∆S do not change significantly with the rise in
temperature, and therefore,
∆G 283.15 = ∆H 283.15 - T∆S 283.15 = (-6007 J mol-1) – (283.15 K) (-21.99 J K-1 mol-1)
= + 220 J mol-1 - freezing of water is non-spontaneous
So, liquid water at atmospheric pressure never freezes when the temperature is greater
than 273.15 K. (Rather the reverse process occurs and solid water melts).
At 273.15 K the two curves cross, ∆G = 0 (ice
and water coexist)
At T < 273.15 K water spontaneously freezes to
ice
At T > 273.15 K the reverse process, the melting
of ice is spontaneous.
2
Example
CS2 is a liquid at room temperature. Calculate ∆G vap at 25oC given that ∆H vap = 27.66
kJ mol-1 and ∆S vap = 86.39 J K-1 mol-1. Calculate also the normal boiling point of CS2.
∆G vap= ∆H vap - T∆S vap = (27.66 kJ mol-1) – (298.15 K) (86.39 J K-1 mol-1) (1 kJ/1000J)
= 1.90 kJ mol-1
In order to calculate the normal boiling point of CS2 we consider ∆G vap = 0, because it
is associated with the process CS2 (l) • CS2 (g) (equilibrium)
∆G vap= ∆H vap - T∆S vap = 0 ⇒ Tb =
∆H vap
= 320.2 K
∆Svap
Trouton’s Rule
Most liquids have the approximately the same molar entropy of vaporization at their
normal boiling points
∆S vap = 88 ± 5 J K-1 mol-1
Trouton’s rule was derived because vaporization occurs within an increase in molar
volume that is very large and nearly the same from substance to substance. The volume
contribution to entropy is so large that we neglect any other.
∆Svap =
∆H vap
(get ∆H vap if we know Tbp)
Tbp
Trouton’s rule breaks when liquids have large amounts of order.
For water, ∆S vap = 109 J K-1 mol-1 - extensive H-bonding
Gibbs Free Energy and Chemical Reactions
∆G rxn = G products – G reactants A chemical rxn can occur under constant T and P only if it
loses Gibbs free energy to the surroundings
The negative of the Gibbs free energy (- ∆G rxn ) is a measure of the driving force of a
reaction under the conditions of constant T and P.
The units are J mol-1.
The “mol-1, per mole” refers to the reaction as written.
∆G rxn= ∆H rxn - T∆S rxn
3
Standard Gibbs Free Energies of Formation
∆G orxn = ∆H orxn - T∆S orxn
Reactants at standard states give products at standard states.
Standard Molar Gibbs Free Energies of Formation
The standard molar Gibbs free energy of formation is the change in Gibbs free energy
when 1 mol of substance forms in a standard state at a specified temperature from the
most stable forms of its constituent elements in standard states at the same temperature.
We use ∆G of = ∆H of - T∆S of to compute ∆G of of any substance at any temperature
Consider C (graphite) + O2 (g) → CO2 (g) at 25oC.
The ∆G of (CO2 (g)) equals ∆G orxn
We calculate ∆H of = ∆H of (CO2 (g)) – 0- 0 = - 393.5 kJ mol-1
∆S orxn = 1 x So (CO2 (g)) – [1 x So (O2 (g)) + 1 x So (C (graphite))] = 2.86 J K-1 mol-1
∆G of (CO2 (g)) = ∆H of (CO2 (g)) - T∆S of (CO2 (g)) = - 394. 36 kJ mol-1
For: aA + bB → cC + dD,
∆G orxn = [c ∆G of (C) + d ∆G of (D)] - [a ∆G of (A) + b ∆G of (B)]
In general,
∆G orxn = Σm∆G of (products) - Σn∆G of (reactants)
Example
Calculate ∆G orxn from enthalpy and entropy values or the reaction:
4KClO3(s)
∆
3KClO4(s) + KCl(s)
∆H orxn = [(3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol)] - (4 mol)(-397.7 kJ/mol)
= - 144 kJ
o
∆S rxn = [(3 mol) (151 J mol-1 K-1) + (1 mol) (82.6 J mol-1 K-1)] - (4 mol)(143.1 J mol-1
K-1) = - 36.8 J K-1
∆G orxn = ∆H orxn - T∆S orxn = (- 144 kJ) – (298.15K) (- 36.8 J K-1) ( 1 kJ/1000J) = - 133 kJ
∆G orxn <0 ∴ rxn is spontaneous!
4
We could the same calculation if we had in our disposal ∆G of values.
∆G orxn = [(3 mol)(-303.2 kJ/mol) + (1 mol)(-409.2 kJ/mol)] - (4 mol)(-296.3 kJ/mol)
= -134 kJ <0 ∴ rxn is spontaneous!
Effect of Temperature on ∆Go
∆G orxn = ∆H orxn - T∆S orxn
∆H orxn and ∆S orxn change
only slightly as the
temperature changes but
∆G orxn changes
considerably.
The spontaneity of a reaction at different temperatures
depends on the signs of ∆H orxn and ∆S orxn . When both
signs are the same, the temperature determines the
spontaneity of the reaction
We calculate ∆G orxn from ∆G of only at 298.15 K. At any other temperature
∆G orxn ≈ ∆H orxn - T∆S orxn . We assume that ∆H orxn and ∆S orxn are independent of
temperature in the temperature range we consider!!!!
In the yellow quarters there is a temperature T at which ∆G orxn = 0 and in this case∆G orxn
= ∆H
o
rxn
- T∆S
o
rxn
= 0 ⇒ ∆H =T∆S
o
rxn
o
rxn
o
∆H rxn
and T =
. At this T reactants and products
o
∆S rxn
5
are at their standard states (gases at partial pressure 1 bar and solutions at 1 M) This
temperature is usually called crossover temperature.
Example
Using Molecular Scenes to Examine the Signs of ∆H, ∆S, and ∆G
The following scenes represent a familiar phase change for water (blue spheres):
(a) What are the signs of ∆H and ∆S for this process? Explain.
The scenes represent condensation of a gas, ∆S < 0 (more order) and ∆H < 0
(exothermic)
(b) Is the process spontaneous at all T, no T, low T, or high T? Explain.
∆G = ∆H - T∆S. With a negative ∆S and negative ∆H, ∆G will be negative. Therefore,
the process is spontaneous at low T.
Example
Determining the Effect of Temperature on ∆G
A key step in the production of sulfuric acid is the oxidation of SO2 (g) to SO3 (g):
2SO2 (g) + O2 (g) → 2SO3 (g)
At 298 K, ∆G = -141.6 kJ; ∆H = -198.4 kJ; and ∆S = -187.9 J K-1
(a) Use the data to decide if this reaction is spontaneous at 25oC, and predict how ∆G
will change with increasing T.
The reaction is spontaneous at 25oC because ∆G is (-). Since ∆H is (-) but ∆S is also
(-), ∆G will become less negative, and the reaction less spontaneous as the temperature
increases.
(b) Assuming ∆H and ∆S are constant with T, is the reaction spontaneous at 900oC?
∆G= ∆H - T∆S ⇒ ∆G = -198.4 kJ – [(1173 K)(-187.9 J K-1)(1 kJ/1000 J) = 22.0 kJ
The reaction is not spontaneous at 900oC.
6
Example
At which temperature is the following process Br2 (l) → Br2 (g) spontaneous at 1 atm?
∆Ho = 31.0 kJ mol-1 and ∆So = 93.0 J K-1 mol-1
A bit deep thought
The vaporization process is spontaneous at all temperatures at which ∆Go < 0.
∆So favors vaporization and ∆Ho favors condensation (exothermic). These tendencies
balance at the boiling point.
∆H o
= 333 K
At the boiling point: Br2 (l) • Br2 (g) ∆G = 0 (equilibrium) and T =
∆S o
o
At T > 333 K, ∆So favors vaporization
At T < 333 K, ∆Ho favors an exothermic process
At T = 333 K, ∆Go = 0 – normal boiling point
Gibbs Free Energy and the Equilibrium Constant
∆G = ∆G orxn + RT ln Q Q is the reaction quotient. If all reactants and products are at
their standard states, then Q = 1, ln Q = 1 and ∆G = ∆G orxn
The reaction quotient gives the progress of the reaction. It varies from 0 (only pure
reactants) to infinity (only pure products) and it becomes equal to the equilibrium
constant, K at equilibrium. The relative values of Q and K establish the direction that
the reaction takes in coming to equilibrium.
Q < K reaction proceeds as written to the right
Q > K reaction proceeds as written to the left
Q = K equilibrium, ∆G = 0 ⇒ ∆G orxn = - RT ln K
∆G = ∆G orxn + RT ln Q = - RT ln K + RT ln Q = RT ln (Q/K)
7
Criteria for Spontaneity in a Chemical Reaction
Spontaneous Processes
∆S univ > 0
∆G rxn < 0
Q<K
Equilibrium Processes Non-spontaneous processes
∆S univ = 0
∆S univ < 0
∆G rxn = 0
∆G rxn > 0
Q=K
Q>K
Conditions
all
Const. T, P
Const. T, P
The relation between Gibbs free energy and the profess of reaction.
Keep in mind!
ln K = −
At equilibrium ∆G = 0 ⇒ ∆G orxn = - RT ln K
∆G
RT
o
rxn
and
K =e
o
− ∆G rxn
RT
Example
Consider N2 (g) + 3H2 (g) • 2NH3 (g). ∆Go = - 33.3 kJ mol-1 at 298.15 K. Predict the
direction in which the system will shift to reach equilibrium when
a) P of NH3 is 1 atm, P of N2 is 1. 47 atm and P of H2 is 1. 0 x 10-2 atm
We should calculate ∆G,
∆G = ∆G
o
rxn
−1
−1
+ RT ln Q = − 33.3kJ + (8.314 JK mol )(298.15K ) ln
−1
∆G = − 33.3kJ + ( 2.48kJmol ln
2
PNH
3
PH32 PN 2
12
= −33.3kJ + 33.3kJ = 0 EQUILIBRIUM
(1.47)(1.0 x10−2 ) 3
NO SHIFT OCCURS!
b) P of NH3 is 1.0 atm, P of N2 is 1.0 atm and P of H2 is 1. 0 atm
Q = 1 , ln 1 = 0, so ∆G=∆Go Rxn shifts to right to reach equilibrium
8
Example
Consider the following reaction:
-1
∆G (kJ mol )
o
f
CO (g) + 2H2 (g) → CH3OH (l)
-137
0
- 166
Calculate ∆G at 25oC when the partial pressure of CO is 5.0 atm and the partial pressure
of hydrogen is 3.0 atm.
∆G = ∆G orxn + RT ln Q
∆Go = - 166 – (-137) – 0 = - 29 kJ mol-1
1
1
Q=
=
= 2.2 x10 −12
2
2
PCO PH 2 (5.0)(3.0)
∆G = ∆G orxn + RT ln Q = (-2.9x104 J mol-1) + (8.314 J K-1 mol-1)(298.15K)ln(2.2x10-12)
∆G = - 38 kJ mol-1
We observe that ∆G is more negative than ∆Go. This implies that the reaction is more
spontaneous at reaction pressures greater than 1.0 atm.
The Temperature Dependence of Equilibrium Constants
o
o
o
∆Grxn
∆H rxn
∆S rxn
ln K = −
=−
+
RT
RT
R
o
To the extent that the temperature dependence of ∆H rxn and ∆S orxn can be neglected then
ln K is a linear function of 1/T.
A graph of ln K vs. 1/T is approximately a straight line with a slope of −
o
∆S rxn
intercept
R
o
o
∆H rxn
∆S rxn
+
ln K 2 = −
RT2
R
o
∆H rxn
and
RT
o
o
∆H rxn
∆S rxn
+
ln K1 = −
RT1
R
1 1
 − 
 T1 T2 
o
1 1
∆H rxn
K
 −  van’t Hoff Equation
ln 2 =
K1
R  T1 T2 
ln K 2 − ln K1 =
o
∆H rxn
R
9
The Variation of Vapor Pressure with Temperature
liquid (l) • vapor (g) Pvapor = K
Let’s write van’t Hoff equation for vaporization at two different temperatures T1 and T2
o
Pvap , 2 ∆H vap
1 1
K2
 −  We assume that ∆Ho and ∆So for vaporization are
ln
= ln
=
K1
Pvap ,1
R  T1 T2 
independent of temperature. Claussius-Clapeyron Equation
At the normal boiling point of a substance, Tb the vapor pressure is 1 atm. Taking T1 to
correspond to Tb and T2 some other temperature, T then
ln Pvapor ,T
o
1 1
∆H vap
 − 
=
R  Tb T 
10