Spontaneity, entropy and free energy

1. Topic 6: Part II Thermochemistry Spontaneity, entropy and free energy

2. Spontaneous • A reaction that will occur without outside intervention. • We can’t determine how fast. • We need both thermodynamics and kinetics to describe a reaction completely. • Thermodynamics compares initial and final states. • Kinetics describes pathway between.

3. Thermodynamics • 1st Law of Thermodynamics - the energy of the universe is constant (conservation of energy). • This keeps track of thermodynamics but doesn’t correctly predict spontaneity. • We need a new concept called entropy. • Entropy (S) is a measure of disorder or randomness • 2nd Law of thermodynamics - the entropy of the universe increases. • Natural progression order → disorder

4. Entropy • Defined in terms of probability. • Substances take the arrangement that is most likely. • The most likely is the most random. • We often calculate the number of arrangements for a system. • The greater the number of possible arrangements the greater the entropy.

5. One Molecule • 2 possible arrangements • 50 % chance of finding the left empty (1/2)

6. 2 Molecules • 4 possible arrangements • 25% chance of finding the left empty (1/4) • 50 % chance of them being evenly dispersed (2/4)

7. 4 Molecules • 12 possible arrangements • 8% chance of finding the left empty (1/12) • 50 % chance of them being evenly dispersed (6/12)

8. Gases • Gases completely fill their chamber because there are many more ways to do that than to leave half empty. • Ssolid <Sliquid <<Sgas • there are many more ways for the molecules to be arranged as a liquid than a solid. • Gases have a huge number of positions possible.

9. 16.1 Positional Entropy For each of the following pairs, choose the substance with the higher positional entropy (per mole) at a given temperature. • Solid CO2 and gaseous CO2 • 1 mole of N2 gas at 1atm and N2 gas at 1.0x10-2 atm

10. 16.2 Predicting Entropy Changes • Predict the sign of the entropy for each of the following processes. • Solid sugar is added to water to form a solution • Iodine vapor condenses on a cold surface to form crystals

11. Entropy • Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate. • 2nd Law • DSuniv = DSsys + DSsurr • If DSuniv is positive the process is spontaneous. • If DSuniv is negative the process is spontaneous in the opposite direction.

12. For exothermic processes DSsurr is positive. • For endothermic processes DSsurr is negative. • Consider this process H2O(l)® H2O(g) • DSsys is positive • DSsurr is negative • DSuniv depends on temperature.

13. Temperature and Spontaneity • Entropy changes in the surroundings are determined by the heat flow. • An exothermic process is favored because by giving up heat the entropy of the surroundings increases. • The size of DSsurr depends on temperature • DSsurr = -DH/T

14. Determining ∆Ssurr • Iron is used to reduce antimony in sulfide ores: • Sb2S3(s) + 3Fe(s)→ 2Sb(s) + 3FeS(s) ∆H = -125kJ • Carbon is used as the reducing agent for oxide ores: • Sb4O6(s) + 6C (s) → 4Sb (s) + 6CO (g) ∆H = 778kJ • Calculate ∆Ssurr for each of these reactions at 25oC and 1atm.

15. - + ? DSsurr DSuniv Spontaneous? DSsys - - - No, Reverse + + + Yes + - ? Yes, if ∆Ssys > ∆Ssurr Yes, if ∆Ssys <∆Ssurr

16. Gibb's Free Energy • G=H-TS • Never used this way. • DG=DH-TDS at constant temperature • Divide by -T • -DG/T = -DH/T+DS • -DG/T = DSsurr + DS • -DG/T = DSuniv • If DG is negative at constant T and P, the Process is spontaneous.

17. Let’s Check • For the reaction H2O(s) ® H2O(l) • DSº = 22.1 J/K mol DHº =6030 J/mol • Calculate DG at 10ºC and -10ºC • Look at the equation DG=DH-TDS • Spontaneity can be predicted from the sign of DH and DS.

18. Spontaneous? DS DH DG=DH-TDS + - At all Temperatures At high temperatures, “entropy driven” + + At low temperatures, “enthalpy driven” - - Not at any temperature, Reverse is spontaneous - +

19. Free Energy and Spontaneity • At what temperature is the following process spontaneous at 1atm? • Br2(l)→ Br2(g) • ∆Ho = 31.0 kJ/mol and • ∆So = 93.0 J/K∙mol • What is the normal boiling point of liquid Br2?

20. Entropy Changes in Chemical Reactions • In chemical reactions at constant T and P • Entropy changes in surroundings ∆Ssurr are determined by heat flow that occurs as the reaction takes place • Entropy change in the system ∆Ssyst are determined by the positional probability • When a reaction involves gaseous molecules the change in positional entropy is dominated by the relative numbers of molecules of reactants and products.

21. Predicting the sign of ∆So • Predict the sign of ∆So for each of the following reactions. • A. The thermal decomposition of solid calcium carbonate: • CaCO3(s) → CaO (s) + CO2(g) • B. The oxidation of SO2 in air: • 2SO2(g) + O2(g) → 2SO3(g)

22. Third Law of Thermodynamics • The entropy of a pure crystal at 0 K is zero. • Gives us a starting point. • All others must be>0. • Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed. • Products - reactants to find DSº (a state function). • More complex molecules higher Sº.

23. Calculating ∆So • Calculate ∆So at 25oC for the reaction • 2NiS (s) + 3O2 (g) → 2SO2 (g) + 2NiO (s) Given the following standard entropy values: SO2 (g) So = 248 J/K∙mol NiO (s) So = 38 J/K∙mol O2 (g) So = 205 J/K∙mol NiS (s) So = 53 J/K∙mol

24. Free Energy in Chemical Reactions • DGº = standard free energy change. • Free energy change that will occur if reactants in their standard state turn to products in their standard state. • Can’t be measured directly, can be calculated from other measurements. • DGº=DHº-TDSº • Use Hess’s Law with known reactions.

25. Free Energy in Reactions • There are tables of DGºf . • Products-reactants because it is a state function. • The standard free energy of formation for any element in its standard state is 0. • Remember- Spontaneity tells us nothing about rate. • Systems at constant temperature and pressure will proceed spontaneously in the direction that lowers free energy.

26. Calculating ∆Ho, ∆So, and ∆Go • Consider the reaction • 2SO2(g) + O2(g)→ 2SO3(g) Carried out at 25oC and 1atm. Calculate ∆Ho, ∆So, ∆Go using data in the table.

27. Free energy and Pressure • DG = DGº + RT ln(Q) • where Q is the reaction quotient • P of the products R = 8.3145 J/K · mol P of the reactants Q is Keq before the reaction starts (initial pressures) Enthalpy is not pressure dependent Entropy is pressure dependent it is greater in a larger volume S large volume > S small volume

28. Free energy and Pressure • CO(g) + 2H2(g) ® CH3OH(l) • Would the reaction above be spontaneous at 25ºC with the H2 pressure of 5.0 atm and the CO pressure of 3.0 atm? DGºf CH3OH(l) = -166 kJ DGºf CO(g) = -137 kJ DGºf H2(g) = 0 kJ

29. The Meaning of ∆G for a Reaction • DG tells us spontaneity at current conditions. When will it stop? • It will go to the lowest possible free energy which may be at equilibrium not necessarily to completion. • At equilibrium DG = 0, Q = K • DGº = -RT lnK

30. Free Energy and Equilibrium • Equilibrium Point – occurs at the lowest value of free energy available to the reaction system • From earlier: ∆G = ∆Go + RT ln(K) • At equilibrium ∆G = 0 • So: ∆Go = -RT ln(K)

31. Free energy And Work • Free energy is that energy free to do work. • The maximum amount of work possible at a given temperature and pressure. • Wmax = ∆G • Never really achieved because some of the free energy is changed to heat during a change, so it can’t be used to do work.