8-3 Gibbs Free Energy

G=TSuniv

Story of Free Energy

Suniv 0

Suniv

Suniv S + Ssurr

Suniv = S qrev/T

Suniv = S H/T

suniv G

Suniv = S H/T 0

T Suniv = TS H 0

T Suniv = H TS 0

G = T Suniv

G = H TS 0

SunivG

Suniv

Suniv=0

Suniv 0Suniv 0

Equilibrium

G=0

G=TSuniv

G

Definition of Free Energy

Gibbs Free Energy G is defined as G = H TS

G = H S T T S

dG = dH S dT T dS

G

G = H S T T Swhere T=0

G = H T S

G = H T S 0 G=0

G

Example: How does H influence G?

0K

0K

G = H T S 0

Example: How does S influence G?

Can a non-spontaneous process with negative S become

spontaneous if the temperature increase?

Can a non-spontaneous process with positive S become

spontaneous if the temperature increase?

G = H T S 0

and Nonexpansion Work , when 0

, when 0 and 0

For a reversible process

, ,

,

,

is the Maxima of Nonexpansion Work ,

One important equation in the derivation of Clausius Inequality

If the nonexpansion work delivered by the system is

,

,

Reaction Free Energy

or

or

or

of

of

or

om

om

om

or

STHG

GnGnG

G

GnGnG

Standard Gibbs Energy of Formation

Gf

The standard reaction free energy per mole for the

formation of a compound from its elements in their

most stable form

The standard free energies of the formation of elements

in their most stable are zero, e.g. Gf(H2,g)=0

Examples of Gf at 25

Use G = nGf()- nGf() to calculate

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

Appendix 2 Thermodynamic Data at 1 Bar and 25

Example

Example

Use G = nGf()- nGf() to calculate

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

Ans:

G= [4Gf (NO) + 6Gf (H2O) ] [4Gf (NH3) + 5Gf (O2) ]

G= [4 (87.6) + 6(-228.6) ] [4 (-16.4) + 5 (0) ] = -955.4 kJ/mol

Example: H and S are independent of T

Here is the reaction 2Fe2O3(s) + 3C(s) 4Fe(s) + 3CO2(g). Assume that H and

S are independent of temperature.

Calculate G at 25.

Calculate G at 1000.

At what temperature does G become zero?

Ans:

H= [3Hf (CO2)] [2Hf (Fe2O3 )] = +467.9 kJ/mol

S= [3Sm(CO2)+ 4Sm(Fe )+] [2Sm(Fe2O3 ) + 3Sm(C ) ] = +558.4 J/K/mol

T = H / S = 838 K

Appendix 2 Thermodynamic Data at 1 Bar and 25

2Fe2O3(s)+3C(s)4Fe(s)+3CO2(g).

Example: H and S are independent of T

Here is the reaction 2Fe2O3(s) + 3C(s) 4Fe(s) + 3CO2(g). Assume that H and S are

independent of temperature.

Calculate G at 25.

Calculate G at 1000.

At what temperature does G become zero?

Ans:

H= [3Hf (CO2)] [2Hf (Fe2O3)] = +467.9 kJ/mol

S= [3S(CO2)+ 4S(Fe)+] [2S(Fe2O3) + 3S(C) ] = +558.13 J/K/mol

G(25)= H-(25+273) S= 467.9 -298 558.13/1000 = 301.6 kJ

G(1000)= H-(1000+273) S= 467.9 -1273 558.13/1000 = -242.6 kJ

T = H / S = 838.34 K

of a mixing process

ln

ln

ln ln

AccordingtoAvogadroslaw,wecanderive

of a mixing process

Foranidealsolution, 0

For an ideal solution

Foranidealsolution, 0

an ideal solution

Example

Calculate the Gibbs free energy and entropy due to mixing 2.5

moles of Ar with 3.5 mole of O2, both at 1 bar and 25. Assume

ideal gas behavior.

G is Coupled to Help Spontaneous Rxns

G is Coupled in Biological System

C6H12O6 + 6O2(g) 6CO2(g) + 6H2O(l)

Gr = -2879kJ