Gibbs Free energy and Helmholtz free energy

Learning objectivesAfter reviewing this presentation

learnerwill be able to • Explain entropy and enthalpy• Describe Gibb’s free energy • Derive a relation for Helmholtz free

energy.

Entropy

Entropy, S: Measure of dispersal or disorder.       Can be measured with a calorimeter. Assumes in a perfect crystal at absolute zero, no

disorder and S = 0.       If temperature change is very small, can

calculate entropy change, S = q/T (heat absorbed / T at which change occurs)

      Sum of S can give total entropy at any desired temperature.

 

Entropy Examples (positive S) Boiling water Melting ice Preparing solutions CaCO3 (s) CaO (s) + CO2 (g)

Entropy Examples (negative S) Molecules of gas collecting Liquid converting to solid at room

temp 2 CO (g) + O2 (g) 2 CO2 (g) Ag+ (aq) + Cl-(aq) AgCl (s)

Entropy Generalizations

Sgas > S liquid > Ssolid

Entropies of more complex molecules are larger than those of simpler molecules (Spropane > Sethane>Smethane)

Entropies of ionic solids are higher when attraction between ions are weaker.

      Entropy usually increases when a pure liquid or solid dissolves in a solvent.

   Entropy increases when a dissolved gas escapes from a solution

Laws of Thermodynamics

First law: Total energy of the universe is a constant.

Second law: Total entropy of the universe is always increasing.

Third law: Entropy of a pure, perfectly formed crystalline substance at absolute zero = 0.

Calculating So system

So system = So (products) - So (reactants)

So surroundings = q surroundings / T

= - Hsystem / T

Calculating So universe

So universe = So surroundings + So

system

So universe =- Hsystem / T + So system

 

• Enthalpy, H: Heat transferred between the system and surroundings carried out under constant pressure.

• Enthalpy is a state function.• If the process occurs at constant pressure,

EnthalpyEnthalpy

PVEH

VPE

PVEH

• Since we know that

• We can write

• When H is positive, the system gains heat from the surroundings.

• When H is negative, the surroundings gain heat from the system.

EnthalpyEnthalpy

VPw

P

P

P

q

VPVPq

VPwq

VPEH

)(

Gibbs Free Energy

Gibbs free energy is a measure of chemical energy.

All chemical systems tend naturally toward states of

minimum Gibbs free energy

G = H - TSWhere:

G = Gibbs Free Energy

H = Enthalpy (heat content)

T = Temperature in Kelvins

S = Entropy (can think of as randomness)

Gibbs Free Energy G is a measure of the

maximum magnitude of the net useful work that can be obtained from a reaction.

Gibbs Free Energy

Gsystem = - T Suniverse

= Hsystem - TSsystem

Gosystem = Ho

system - T Sosystem

  Go

rxn = Horxn - T So

rxn

Gibbs Free Energy

Gosystem or Go

rxn If negative, then product-favoured. If positive, then reactant-favoured.

Go reaction = Gfo (products) - Gf

o (reactants)

Thermodynamics and KIf not at standard conditions,G = Go + RT ln Q  (Equilibrium is characterized by the inability to do

work.)At equilibrium, Q = K and G = O

 Therefore, substituting into previous equation gives

0 = Go + RT ln K and Go = - RT ln K (can use Kp or Kc)

 

Thermodynamics and K       Understand relationship

between Go, K, and product-favoured reactions

Go<0 K>1 Product-favoured Go=0 K=1 Equilibrium Go>0 K<1 Reactant-favoured

The Helmholtz free energy is a thermodynamic potential that measures the “useful” work obtainable from a closed thermodynamic system at a constant temperature and volume. 

Helmholtz Free Energy

The Helmholtz energy is defined as: A= U - TSwhereA  is the Helmholtz free energy (SI: joules, CGS: ergs),U  is the internal energy of the system (SI: joules, CGS: ergs),T  is the absolute temperature (Kelvins),S  is the entropy (SI: joules per Kelvin, CGS: ergs per kelvin).

Helmholtz Free EnergyFrom the first law of thermodynamics dU = δQ - δW,where U is the internal energy,  δQ is the energy added by heating and δW is the work done by the system. From the second law of thermodynamics, for a reversible process we may say that δQ = TdS. Also, in case of a reversible change, the work done can be expressed as δW = pdV dU = TdS - pdVApplying the product rule for differentiation to d(TS) = TdS + SdT, we have: dU = d(TS) – SdT – pdVd(U-TS) = – SdT – pdV,and The definition of A = U - TS enables to rewrite this as: dA = – SdT – pdV