Gibbs Energy

To check spontaneity of a process we require knowledge of system

as well as of surrounding.

It is desirable and also more useful to define a parameter with the

help of which we can check spontaneity by considering system only.

A new mathematical term Gibbs energy was introduced to explain

criteria of spontaneity in terms of system only.

The Gibbs energy of a system is defined as the maximum amount of

energy available to a system during a process that can be converted
into useful work.
Relationship between ∆G(P,T) and ∆Suniverse

ΔS univ = ΔS sys + ΔSsurr

−q sys −q sys
ΔS univ = ΔS sys + As, ΔS surr =
T T

For a process at constant pressure, qp = H

ΔHsys
ΔS univ = ΔS sys −
T

TΔS univ = TΔSsys − ΔH sys

−TΔS univ = (ΔH sys − TΔSsys൯

ΔGsys (P,T) = −TΔSuniv ΔG sys = ΔH sys − TΔSsys

Gibbs function 'G' is defined as,

G = H – TS.

Since 'G' is a combination of state variables, it is a state function.

Gibb's energy is defined at constant pressure and temperature to

predict spontaneity of a process considering system only.
Criteria of Spontaneity

GsysP = –TStotal

(i) If, Stotal = +ve  ΔGsys (P.T) = –ve

 spontaneous process

(ii) If, Stotal = –ve  ΔGsys (P.T) = +ve

 non-spontaneous process

(iii) If, Stotal = 0  ΔGsys (P.T) =0

 process is at equilibrium
 Spontaneity of Reaction

Prediction of spontaneity on basis of sign of ∆H, ∆S

(G = H – TS)system

H S G Description
Reaction is spontaneous at all
– + –
temperature

+ – + Reaction is non-spontaneous at
all temperature
– Reaction is spontaneous at low
– –
temperature
Reaction is spontaneous at high
+ + –
temperature
 Note

The temperature at which equilibrium is attained,

G = H – TS

At equilibrium  G = 0 ,

ΔH
Teq =
ΔS
Standard State

Standard state of a substance is its most stable state at a pressure

of 1 bar and at a specified temperature. (If temperature is not
specified then it is conventionally taken to be 298 K).

For a substance/ions in solution, standard state is 1M concentration.

∆Go is change in Gibbs energy at standard state while ΔG is change

in Gibbs energy.
Standard Gibbs free energy of formation (∆fG°)

It is the change in Gibbs energy when 1 mole of a substance is

formed from its elements in standard state.

Conventionally, the Gibbs energy of formation of elements in their

standard state is taken to be zero.

EXAMPLE

fGo (C, graphite) = 0

fGo [H2(g)] = 0
fGo [Br2(l)] = 0

fGo [I2(s)] = 0
 Gibbs Free Energy
change calculation

Chemical Phase Expansion /

Reaction Change Compression
1. For chemical reaction at constant P and T

(a) If H and S are given

rG = rH – TrS

rH = ∑ f H (products) – ∑ f H(reactants)

rS = ∑ Sm (products) – ∑ Sm (reactants)

(b) If standard Gibbs energy of formation(∆G∘f ) is provided

∆fG° = ∑ ∆fG° (products) – ∑ ∆fG° (reactants)

Note

The Gibbs energy of formation of elements in their standard state is

taken to be zero.
EXAMPLE
Calculate the G° at 300 K for the reaction;
Br2 (l) + Cl2(g)   2BrCl(g).
For the reaction H° = 29.3 kJ mole-1 and the standard entropies of
Br2 (l), Cl2(g) & BrCl(g) at the 300 K are 150, 220, 240 J mol–1K–1
respectively.

SOLUTION

Ans : - 3.7 kJ
EXAMPLE
Show that the reaction CO(g) + 1 O 2(g)   CO2(g) at 300 K is
2
spontaneous and exothermic, when the standard entropy is – 0.094 kJ
mole–1 K–1. The standard Gibbs free energies of formation for CO2 and
CO are –394.4 and –137.2 kJmole–1, respectively.

SOLUTION

Ans : – 285.4 kJmole–1

EXAMPLE
Which of the following conditions regarding a chemical process ensures
its spontaneity at all temperature?
(A) H > 0, S < 0 (B) H < 0, S > 0
(C) H < 0, S < 0 (D) H > 0, S < 0

SOLUTION

Ans : (B)
EXAMPLE
For the reaction at 25°C, X 2 O4 (l)   2XO2(g), H =2.1 kcal and
S = 20 calK–1. The reaction wouldbe
(A) Spontaneous (B) Non-spontaneous
(C) At equilibrium (D) Unpredictable

SOLUTION

Ans : (A)
EXAMPLE
For the reaction at 298K, 2A+B   C
H = 100 kcal and S = 0.050 kcal K–1. If H and S are assumed to be
constant over the temperature range, above what temperature will the
reaction become spontaneous?
(A) 1000 K (B) 1500 K
(C) 2000 K (D) 2500 K

SOLUTION

Ans : (C)
EXAMPLE
Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 JK—1 mol—1,
respectively. For reaction 1X2 + 3Y2 → XY3 ; H = — 30kJ to be at
2 2
equilibrium the temperature will be
(A) 500 K (B) 750 K
(C) 1000 K (D) 1250 K

SOLUTION

Ans : (B)
 2. For substance under going expansion or compression

Consider a pure substance having a Gibbs function as follows :

G = H – TS

= (U + PV) – TS

Let, the pressure and temperature are varied by infinitesimally small

value :
dG = dU + PdV + VdP – TdS – SdT
dG = (dq + w) + PdV + VdP – TdS – SdT
As, w = – PdV and dq = TdS [Assume reversible process.]

dG = VdP – SdT
 As dG = VdP – SdT
(a) If pressure is varied keeping temperature constant -

dG = VdP

Ideal Gas liquid/solid

nRT (volume remains almost constant)
dG = dP
P P2
P2
ΔG = න VdP
nRT
ΔG = න dP P1
P
P1
P2 Any change in pressure results in
ΔG = nRT ln very small change in volume.
P1
V2 ΔG = V(P2–P1)
Or, ΔG = −nRT ln
V1
(b) If temperature is changed keeping pressure constant -

dG = –SdT

T2

ΔG = න SdT
T1

Entropy as a function of temperature will be given for problem solving.

EXAMPLE
Calculate G (in bar-L) when pressure over 1 mole of H2O is change from
1 bar to 10 bar.

(i) At 4°C (ii) At 127°C

SOLUTION

Ans. (i) 0.162 bar–L

(ii) 400 R ln 10
EXAMPLE
Calculate G (in bar-L) when a definite mass of a monoatomic ideal gas
at 1 bar & 27oC is expanded adiabatically against vacuum from 10 L to
20 L (ln 2 = 0.7).

SOLUTION

Ans : –7 bar-L
3. For substance under going phase change

All phase changes are considered to occur at constant pressure and

temperature.

All phase change at it’s boiling point, melting point or sublimation

point are reversible in nature.

Hence phase changes at its boiling point, melting point or sublimation

point will have

(G)(P,T) = 0
EXAMPLE
Calculate Gm for the following process:
(i) H2O (l, 100oC, 1 atm)    H2O (g, 100oC, 1 atm)
(ii) H2O (l, 100oC, 1 atm)    H2O (g, 100oC, 0.5
atm)

SOLUTION

Ans : (i) G = 0
(ii) G = –2149.5 J
EXAMPLE
5 mole H2O (l) at 373K and 1 atm is converted into H2O (g) at 373K and
5 atm. Calculate G for this process. [Given : R = 2Cal/K-mol, ln 5 = 1.6]

SOLUTION

Ans : 5.968 kCal

EXAMPLE
A liquid freezes into a solid (H = –1000 J/mole) at 200 K and 1 atm,
(it's normal melting point).
(i) What is the S value at 200 K ?
(ii) Will the freezing be spontaneous at 150 K and 1 atm ?
(iii) What is the value of G at 250 K and 1 atm ?
(And it is assumed that S and H do not depend on temperature)

SOLUTION

Ans. (i) S = – 5 J/mole.K

(ii) Spontaneous
iii G = + 250 J/mole
Gibbs free Energy change (ΔG) and non-pv work

By first law of thermodynamics

U = q + w
U = q + (wPV + wnon PV)
U = q – PV – wnon PV Q S = q/T
U = TS – PV – wnon PV
(U + PV) – TS = – wnon PV
H – TS = – wnon PV
wnon PV = wuseful = – (G) (P,T)
(G)(P,T) = – w non PV

The decrease in Gibb’s energy of system is equal to the useful work

or non-PV work or electrical work.
EXAMPLE
For reaction :
2A(s) + B(g)  3C(l), standard entropy change of reaction is –2 kJ
/ mol-K and standard enthalpy of combustion of A(s), B(g) and C(l) are
–100, –60, –285 kJ/mol respectively, then find maximum useful work
that can be obtained at 27°C and 1 bar pressure from the reaction (in
kJ/mol) .

SOLUTION

Ans : 1195
Relationship between Standard Gibb's Energy Change (G°) and Equilibrium
Constant (K) :-

For a reaction aA + bB   cC + dD

rG = ∆rG° + RT l n Q

At equilibrium G = 0 and Q = Keq

So, G° + RT ln K = 0

∴ ΔG° = – RT lnKeq

G° = –2.303RT logKeq

Q is reaction quotient which contains concentration/partial pressure

at any instant.
Keq is equilibrium constant which contains concentration/partial
pressure at equilibrium (All pressure terms should be kept in bar).
EXAMPLE
For reaction
N2(g) + 3H2(g)   2NH3(g)
At equilibrium partial pressure of N2, H2 and NH3 are 2 bar, 1 bar and
10 2 bar respectively. At 500 K calculate
i rG° in kcal.
(ii) If at any instant partial pressure of N2, H2 and NH3 are 2 bar, 5
bar and 20 bar respectively. Calculate ΔrG (ln 1.6 = 0.5).

SOLUTION

Ans. (i) –4.6 kcal/mole

(ii) –4.1 kcal/mole
EXAMPLE
The equilibrium constant of the reaction 2C3H6(g) ⇌ C2H4(g) + C4H8(g) is

found to fit the expression lnK = – 1.04 – 1088𝐾

𝑇
Calculate the standard reaction enthalpy and entropy at 400 K.

SOLUTION

Ans. H° = 9.04 kJ/mole

S° = –8.64 J/moleK
EXAMPLE
-D Glucose undergoes mutarotation to -D-Glucose in aqueous solution.
If at 300K there is 60% conversion. Calculate G° of the reaction. (ln 2
= 0.7, ln 3 = 1.1) -D-Glucose ⇌ -D-Glucose

SOLUTION

Ans : –997.68 J/mol