Spontaneity, Entropy and Free Energy

1. Spontaneity, Entropyand Free Energy

2. Spontaneous Processes and Entropy • First Law • “Energy can neither be created nor destroyed" • The energy of the universe is constant • Spontaneous Processes • Processes that occur without outside intervention • Spontaneous processes may be fast or slow • Many forms of combustion are fast • Conversion of diamond to graphite is slow

3. Entropy (S) • A measure of the randomness or disorder • The driving force for a spontaneous process is an increase in the entropy of the universe • Entropy is a thermodynamic function describing the number of arrangements that are available to a system • Nature proceeds toward the states that have the highest probabilities of existing

4. Positional Entropy • The probability of occurrence of a particular state depends on the number of ways (microstates) in which that arrangement can be achieved • Ssolid < Sliquid << Sgas

5. Example • Which has higher positional S? • Solid CO2 or gaseous CO2? • N2 gas at 1 atm or N2 gas at 0.01 atm? • Predict the sign of DS • Solid sugar is added to water • + because larger volume • Iodine vapor condenses on cold surface to form crystals • - because gs, less volume

6. Second Law of Thermodynamics • "In any spontaneous process there is always an increase in the entropy of the universe" • "The entropy of the universe is increasing" • For a given change to be spontaneous, Suniverse must be positive • Suniv = Ssys + Ssurr

7. 2nd Law of Thermodynamics • In any spontaneous reactions, there is always an increase in entropy of the universe • ∆Suniv = ∆Ssys + ∆Ssurr • If ∆Ssys is negative, it can still be spont. as long as the ∆Ssurr is larger and positive • ∆Suniv > 0 : spontaneous • ∆Suniv = 0 : no tendency, at equilibrium • ∆Suniv < 0 : not spontaneous

8. Ch. 16: Spontaneity, Entropy, and Free Energy 16.3 The Effect of Temperature on Spontaneity

9. H2O(l)  H2O(g) • ∆Ssys has + sign b/c of the increase in # of positions • ∆Ssurr is determined mostly by heat flow • Vaporization is endothermic so it removes heat from the surroundings • So decreases random motion of surroundings • Negative ∆Ssurr

10. Temperature’s Effects • If the ∆Ssys and ∆Ssurr have different signs, the temperature determines the ∆Suniv • For vaporization of water • Above 100°C, ∆Suniv is positive • Below 100°C, ∆Suniv is negative • Impact of the transfer of heat will be greater at lower temperatures

11. Calculating Entropy Change in a Reaction • Entropy is an extensive property (a function of the number of moles) • Generally, the more complex the molecule, the higher the standard entropy value

12. Gibb’s Free Energy D Suniv= D Ssystem + D Ssur. Entropy of something will increase as it gets more heat. The particles will move more. The effect this heat has on the particles is dependant upon the temp. A small amount of heat at a high temp will not make much of an impact, while a small amount of heat at a small temp. will.

13. This leads to this relationship. S= D H T

14. Thus the D Ssurr. Can be expressed as - DHsys T The sign should be opposite that of the systems, so it is negative because from the systems perspective, the surrounding have an opposite sign.

15. So D Suniv= D Ssys + D Ssur Can be changed to D Suniv= D Ssys + -DHsys T

16. Multiply the whole thing be T and this gives: T D Suniv= T D Ssys + - DHsys Multiple by –1 and you get :

17. T D Suniv= -T D Ssys + DHsys • Rearranging this gives: • T D Suniv= DHsys -T D Ssys

18. DG is what we call the relationship -T D Suniv So: D G= DHsys - T D Ssys

19. Standard Free Energy Change • G0 is the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states • G0 cannot be measured directly • The more negative the value for G0, the farther to the right the reaction will proceed

20. Calculating Free Energy Method #1 For reactions at constant temperature: G0 = H0 - TS0

21. H, S, G and Spontaneity G = H - TS H is enthalpy, T is Kelvin temperature

22. Calculating Free Energy Method #2 An adaptation of Hess's Law: Cdiamond(s) + O2(g)  CO2(g) G0 = -397 kJ Cgraphite(s) + O2(g)  CO2(g) G0 = -394 kJ CO2(g)  Cgraphite(s) + O2(g) G0 = +394 kJ Cdiamond(s)  Cgraphite(s) G0 = -3 kJ

23. Calculating Free Energy Method #3 Using standard free energy of formation (Gf0): Gf0 of an element in its standard state is zero

24. Sample problem using free energy to find the boiling point of a substance. At what temperature is the following process spontaneous at 1 atm? Br2 (l)  Br2(g) DH= 31.0 Kj/mol & DSo = 93.0 j/k.mol G0 = H0 - TS0 make G0 = 0, so 0 =H0 - TS0 0 = 31000 j/mol - T (93.0 j/k.mol), solve for T…. T = 333 K is boiling point

25. The Dependence of Free Energy on Pressure • Enthalpy, H, is not pressure dependent • Entropy, S • entropy depends on volume, so it also depends on pressure • Slarge volume > Ssmall volume • Slow pressure > Shigh pressure