1. Gibbs Free energy and Helmholtz free energy

2. Learning objectives After reviewing this presentation learner will be able to Explain entropy and enthalpy Describe Gibb’s free energy Derive a relation for Helmholtz free energy.

3. 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.

4. Entropy Examples (positive  S) Boiling water Melting ice Preparing solutions CaCO 3 (s)  CaO (s) + CO 2 (g)

5. Entropy Examples (negative  S) Molecules of gas collecting Liquid converting to solid at room temp  2 CO (g) + O 2 (g)  2 CO 2 (g)  Ag + (aq) + Cl - (aq)  AgCl (s)

6. Entropy Generalizations S gas > S liquid > S solid Entropies of more complex molecules are larger than those of simpler molecules (S propane > S ethane >S methane ) 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

7. 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.

8. Calculating  S o system  S o system = S o (products) - S o (reactants)  S o surroundings = q surroundings / T = -  H system / T

9. Calculating  S o universe  S o universe =  S o surroundings +  S o system  S o universe =-  H system / T +  S o system

10. 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, Enthalpy

11. 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. Enthalpy

12. 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 - TS Where: G = Gibbs Free Energy H = Enthalpy (heat content) T = Temperature in Kelvins S = Entropy (can think of as randomness)

13. Gibbs Free Energy  G is a measure of the maximum magnitude of the net useful work that can be obtained from a reaction.

14. Gibbs Free Energy  G system = - T  S universe =  H system - T  S system  G o system =  H o system - T  S o system  G o rxn =  H o rxn - T  S o rxn

15. Gibbs Free Energy  G o system or  G o rxn If negative, then product-favoured. If positive, then reactant-favoured.  G o reaction = G f o (products) - G f o (reactants)

16. Thermodynamics and K If not at standard conditions,  G =  G o + 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 =  G o + RT ln K and  G o = - RT ln K(can use Kp or Kc)

17. Thermodynamics and K  Understand relationship between  G o, K, and product-favoured reactions  G o 1Product-favoured  G o =0 K=1 Equilibrium  G o >0 K<1Reactant-favoured

18. 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 - TS where A is the Helmholtz free energy (SI: joules, CGS: ergs),SIjoulesergs U is the internal energy of the system (SI: joules, CGS: ergs),internal energy T is the absolute temperature (Kelvins), S is the entropy (SI: joules per Kelvin, CGS: ergs per kelvin).entropy

19. Helmholtz Free Energy From the first law of thermodynamics 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.second law of thermodynamicsreversible process Also, in case of a reversible change, the work done can be expressed as δW = pdV dU = TdS - pdV Applying the product rule for differentiation to d(TS) = TdS + SdT, we have: dU = d(TS) – SdT – pdV d(U-TS) = – SdT – pdV,and The definition of A = U - TS enables to rewrite this as: dA = – SdT – pdV