1. Chemistry 101 : Chap. 19 Chemical Thermodynamics (1) Spontaneous Processes (2) Entropy and The Second Law of Thermodynamics (3) Molecular Interpretation of Entropy (4) Entropy Change in Chemical Reactions (5) Gibbs Free Energy (6) Free Energy and Temperature

2. Spontaneous Processes Which process looks “normal” at 25 o C? Ice [H 2 O(s)]Liquid Water [H 2 O(l)]

3. Spontaneous Processes A spontaneous process is one that proceeds on its own without any outside assistance. 4 Fe (s) + 3 O 2 (g)  2 Fe 2 O 3 (s) rust Spontaneous chemical process

4. Spontaneous Processes  At a given temperature and pressure, processes are spontaneous only in one direction H 2 O (l)  H 2 O (s) spontaneous at -10 o C H 2 O (l)  H 2 O (s) spontaneous at 25 o C Why some processes are spontaneous and while others are not? Are all spontaneous processes exothermic? No

5. Spontaneous Processes Any spontaneous process is irreversible. Left: Vacuum, Right: Full of gas Gas will be evenly distributed You will not see the reverse process in billion years!

6. Spontaneous Processes Initial State 4 Possible Configurations Number of ways to arrange two spheres All in the left = 1, One in the left and one in the right = 2, All in the right = 1 Each configuration is equally possible However, the top two configurations are identical

7. Number of ways to arrange 4 identical spheres (N L =4, N R = 0) = 1, (N L =3, N R =1) = 4, (N L =2, N R =2) = 6, (N L =1,N R =3) = 4, (N L =0,N R =4) = 1 Spontaneous Processes  If we have four identical spheres.. -- There are 16 configurations -- 5 distinguishable arrangements What happens if we have 1000 identical spheres? Virtually, zero probability to have all in one side The even distribution, (N L = 500, N R =500), has close to 10 300 configurations, but completely uneven distribution (N L =1000, N R =0 or N L =0, N R =1000) has just 2 configuration.

8. Spontaneous Processes Spontaneous processes proceed towards a more probable state Generally, the more probable state is associated with more disorder S = k log W S = Entropy, k= Boltzman Constant W = Number of configurations associated with a given arrangement ( In general, number of microstates ) Ludwig Eduard Boltzman (1844-1906)

9. Entropy  Entropy is a state function  The change in entropy for any process is  S = S final - S initial What is the sign of  S for the following processes at 25 o C ? H 2 O (s) → H 2 O (l) CO 2 (g) → CO 2 (s)  S = +  S = -

10. Second Law of Thermodynamics The total entropy of the universe increases in any spontaneous process  S o universe =  S o system +  S o surroundings > 0 Rudolf Clausius (1822-1888) “I propose to name the quantity S the entropy of the system, after the Greek word [trope], the transformation. I have deliberately chosen the word entropy to be as similar as possible to the word energy.” (1865)

11. Second Law of Thermodynamics  Example : The enthalpy of fusion for H 2 O is  H fusion = 6.01kJ/mol. What is the entropy change when a mole of ice melts in the palm of your hand at 310K?

12. Entropy S solid < S liquid < S gas gas liquidsolid Of all phases, gases have the highest entropy In gas phase, molecules are more randomly distributed  Entropy of a substance in different phase

13. Entropy Larger Molecules generally have a larger entropy Larger molecules have more internal motion S small < S medium < S large  Entropies of molecules with different sizes or complexity

14. Entropy  Usually, dissolving a solid or liquid will increase the entropy dissolves higher entropy more disordered arrangement lower entropy

15. Entropy  Dissolving gas in liquid decreases the entropy dissolves overall more disordered arrangement: higher entropy lower entropy

16. Entropy  Decreasing the number of molecules usually decrease the entropy 2 NO(g) + O 2 (g)  2NO 2 (g) Higher Entropy Lower Entropy

17. Entropy  Example : Predict whether  S is positive or negative for each of the following processes under constant temperature H 2 O (l)  H 2 O (g) Ag + (aq) + Cl - (aq)  AgCl (s) 4Fe (s) + 3O 2 (g)  2Fe 2 O 3 (s) Ba(OH) 2 (s)  BaO(s) + H 2 O (g)

18. Standard Molar Entropy  Absolute entropy (S) : This is based on the reference point of zero entropy for perfect crystalline solids at 0K

19. Standard Molar Entropy  Standard Molar Entropy (S o ) : The molar entropy value of substance in their standard state (25 o C, 1 atm) } size of molecules increases } S gas > S liquid } dissolving a gas in a liquid is accompanied by a lowering of the entropy } dissolving a liquid in another liquid is accompanied by an increase in entropy O 2 (g)205 Standard molar entropies of elements are not zero

20. Standard Molar Entropy  Example : For each of the following pairs, which substance has a higher molar entropy at 25 o C ? HCl (l) HCl (s) C 2 H 2 (g) C 2 H 6 (g) Li (s) Cs (s) Pb 2+ (aq) Pb (s) O 2 (g) O 2 (aq)HCl (l) HBr (l) CH 3 OH (l) CH 3 OH (aq)N 2 (l) N 2 (g)

21. Standard Molar Entropy  Entropy change in chemical reactions : If you know the standard molar entropies of reactants and products, you can calculate  S o for a reaction:  S o rxn = Σ n S o (products) – Σ m S o (reactants) stoichiometric coefficients for products stoichiometric coefficients for reactants NOTE : Compare this equation with the standard enthalpy change formula.

22. Standard Molar Entropy  Example : What is  S o for the following reaction? Do you expect  S o to be positive or negative? C 2 H 4 (g) + H 2 (g) → C 2 H 6 (g) substance S o (J/K-mol) H 2 (g)130.6 C 2 H 4 (g)219.4 C 2 H 6 (g)229.5 S o for elements are NOT zero

23. Standard Molar Entropy  Example : What is  S o for the following reaction? Do you expect  S o to be positive or negative? CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O (g) substance S o (J/K-mol) O 2 (g)205.0 CH 4 (g)186.3 CO 2 (g)213.6 H 2 O (g) 188.3

24. Gibbs Free Energy  Recall : We need to know both  S sys and  S surr to predict the direction of changes (second law of thermodynamics) Can we predict the direction of change based on the system properties only? However, in general,  S surr is not easy to evaluate. … YES! An important key to solve this problem is to realize that there is a close relation between  H (heat) and  S.

25. Gibbs Free Energy  Gibbs Free Energy (G = H - TS):  G =  H - T  S (under constant Temperature and Pressure) Josiah W. Gibbs (1839-1903) Why “free” energy?  G for a given spontaneous process is the maximum amount of “available” work that a system can do on its surrounding at constant temperature and pressure.

26. Gibbs Free Energy   G and the direction of spontaneity G = H - TS  G =  H - T  S (at constant temperature, pressure)  S surr = q surr /T = -  H sys /T ( at constant temperature, pressure)  S univ =  S sys +  S surr =  S sys -  H sys /T Therefore,  H sys - T  S sys = -T  S univ =  G  G < 0 : spontaneous in forward direction  G > 0 : non-spontaneous in forward direction  G = 0 : System is in equilibrium

27. Gibbs Free Energy  Why Gibbs free energy is useful in chemistry?  Most chemical reactions take place under constant temperature and pressure conditions.   G relates to the system alone and allows us to avoid examination of  S surr for the prediction of spontaneity.  Free energy of a chemical system decreases until it reaches a minimum value, where a state of equilibrium exists.

28. Gibbs Free Energy  Example : The reaction of sodium metal with water takes place according to the following equation. 2 Na (s) + 2 H 2 O (l) → 2 NaOH (aq) + H 2 (g) Is the reaction spontaneous? What is the sign of  G? What is the sign of  ? What is the sign of  S?

29. Standard Free Energy Change  Standard free energy of formation (  G f o ) : Free energy change for the formation of 1 mol of compound from elements with all substances in their standard states.  standard state = pure compound at 25 o C and 1 atm.  For solution, standard state is 1M solution   G f o for pure elements are zero.  G is a state function !  Standard free energy change (  G o )  G o = Σ n  G f o (products) – Σ m  G f o (reactants)

30. Standard Free Energy Change  Example : What is the  G o value of the following reaction? 2 C diamond (s) + O 2 (g) → 2 CO (g) substance  H f o (kJ/mol)  G f o (kJ/mol) S o (J/K-mol) O 2 0 0205.0 C (diamond, s) 1.88 2.84 2.43 C (graphite, s) 0 0 5.69 CO 2 (g)-393.5-394.4213.6

31. Standard Free Energy Change  Example : What is  G o for the following reaction at 25 o C using the values of  H o and S o provided? CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O (g) substance  H f o (kJ/mol) S o (J/K-mol) O 2 (g) 0 205.0 CH 4 (g) -74.80 186.3 CO 2 (g)-393.5 213.6 H 2 O (g) -241.8 188.3

32. Standard Free Energy Change  Example : What is  G o for the following reaction at 25 o C using the values of  G o provided? CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O (g) substance  G f o (kJ/mol) CH 4 (g) -50.8 CO 2 (g) -394.4 H 2 O (g) -228.57

33. Free Energy and Temperature  G =  H - T  S  H  S + - GG - + - - sign depends on T ! low T =>  G is negative high T =>  G is positve + sign depends on T ! low T =>  G is positive high T =>  G is negative +

34. Free Energy and Temperature  G =  H - T  S H 2 O (s) → H 2 O (l) spontaneous at 298 K (25 o C)  H > 0,  S > 0 Which is larger,  H or T  S ? T  S >  H (at 298 K) If you decrease the temperature, the difference between T  S and  H becomes smaller and, eventually, T  S <  H. At that point, the process becomes no longer spontaneous (  G > 0) What happens when  G = 0?  G = 0 at melting point (equilibrium) and  G < 0

35. Free Energy and Temperature At the normal melting point, the Gibbs free energies of the solid and liquid phase of any substance are equal: H 2 O (s) → H 2 O (l) at 0 o C  G o = 0 H 2 O (l) → H 2 O (g) at 100 o C  G o = 0 At the normal boiling point, the Gibbs free energies of the liquid and gas phase of any substance are equal: C 2 H 5 OH (s) C 2 H 5 OH(l) at -117 o C  G o = 0 C 2 H 5 OH (l) C 2 H 5 OH(g) at 79 o C  G o = 0

36. Free Energy and Temperature  Example : Determine the temperature at which the following reaction will become spontaneous in the opposite direction. Assume that  H and  S do not change with temperature. CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O (g)

37. Free Energy and Temperature  Example : Use the enthalpy of fusion (  H o fus ) and enthalpy of vaporization (  H o vap ) to calculate the values of  S o fus and  S o vap for 1 mol of water. (  H o fus = 6.02 kJ/mol,  H o vap = 40.7 kJ.mol)