1. Spontaneity, Entropy and 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 S solid < S liquid << S gas

5. Example Which has higher positional S? –Solid CO 2 or gaseous CO 2 ? –N 2 gas at 1 atm or N 2 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. What about Chemical Changes? We have been working with only physical changes so far… Compare using # of independent states N 2 (g) + 3H 2 (g)  2NH 3 (g) Entropy decreases (-∆S) Compare using # molecules in higher entropy states 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) Entropy increases (+∆S)

7. Example Predict the sign of ∆S for the following reactions: CaCO 3 (s)  CaO(s) + CO 2 (g) –Increasing S, +∆S 2SO 2 (g) + O 2 (g)  2SO 3 (g) –Decreasing S, -∆S

8. 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,  S universe must be positive  S univ =  S sys +  S surr

9.  H,  S,  G and Spontaneity Value of  H Value of T  S Value of  G Spontaneity Negative PositiveNegativeSpontaneous Positive NegativePositiveNonspontaneous Negative ???Spontaneous if the absolute value of  H is greater than the absolute value of T  S (low temperature) Positive ???Spontaneous if the absolute value of T  S is greater than the absolute value of  H (high temperature)  G =  H - T  S H is enthalpy, T is Kelvin temperature

10. 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

11. Example Find the ∆S° at 25°C for: 2NiS(s) + 3O 2 (g)  2SO 2 (g) + 2NiO(s)

12. Standard Free Energy Change   G 0 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   G 0 cannot be measured directly  The more negative the value for  G 0, the farther to the right the reaction will proceed in order to achieve equilibrium  Equilibrium is the lowest possible free energy position for a reaction

13. For reactions at constant temperature:  G 0 =  H 0 - T  S 0 H 2 O(s)  H 2 O(l) where ∆H°=6.03x103 J/mol, ∆S°=22.1 J/mol.K Since reaction’s values are in opposition, the T determines spontaneity. At -10°C, is it spontaneous? Calculating Free Energy Method #1

14. Example NO! the reverse is spontaneous Is it spontaneous at 0°C? NO! it is at equilibrium so no shift

15. Example Is it spontaneous at +10°C? YES! When H and S are in opposition, the spontaneity depends on T.

16. Calculating Free Energy: Method #2 An adaptation of Hess's Law: C diamond (s) + O 2 (g)  CO 2 (g)  G 0 = -397 kJ C graphite (s) + O 2 (g)  CO 2 (g)  G 0 = -394 kJ CO 2 (g)  C graphite (s) + O 2 (g)  G 0 = +394 kJ C diamond (s)  C graphite (s)  G 0 = C diamond (s) + O 2 (g)  CO 2 (g)  G 0 = -397 kJ -3 kJ

17. Calculating Free Energy Method #3 Using standard free energy of formation (  G f 0 ):  G f 0 of an element in its standard state is zero

18. 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 S large volume > S small volume S low pressure > S high pressure

19. Free Energy and Equilibrium  Equilibrium point occurs at the lowest value of free energy available to the reaction system  At equilibrium,  G = 0 and Q = K G0G0 K  G 0 = 0K = 1  G 0 < 0K > 1  G 0 > 0K < 1

20. Temperature Dependence of K So, ln(K)  1/T

21. Free Energy and Work  The maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy  The amount of work obtained is always less than the maximum  Henry Bent's First Two Laws of Thermodynamics  First law: You can't win, you can only break even  Second law: You can't break even