1. Chemical Thermodynamics The chemistry that deals with the energy and entropy changes and the spontaneity of a chemical process.

2. First Law of Thermodynamics The change in the internal energy (  E) of a thermodynamic system is equal to the amount of heat energy (q) added to or lost by the system plus work done (w) on or by the system.  E = q + w For work that only involves gas expansion or compression, w = -p  V;

3. Values of Thermodynamic Functions FLoT:  E = q + w; –q is assigned a positive value if heat is absorbed, but a negative value if heat is lost by the system; –w is assigned a positive value if work is done on, but a negative value if work is done by the system. –For processes that do not involve phase changes, positive  E results in temperature increase.

4. Second Law of Thermodynamics Energy tends to flow from a high energy concentration to a dispersed energy state; Energy dispersion or diffusion is a spontaneous process. Dispersed or diffused energy is called entropy According to SLoT, a process/reaction is spontaneous if the entropy of the universe (system + surrounding) increases.

5. Third Law of Thermodynamics The entropy of a perfect crystalline substance is zero at absolute zero temperature (0.0 K) Is absolute zero temperature achievable?

6. What is Entropy? A thermodynamic (energy) function that describes the degree of randomness or probability of existence. As a state function – entropy change depends only on the initial and final states, but not on how the change occurs.

7. What is the significance of entropy? Nature spontaneously proceeds toward the state that has the highest probability of (energy) existence – highest entropy Entropy is used to predict whether a given process/reaction is thermodynamically possible;

8. Entropy: which are most probable?

9. Relative Entropy of Substances Entropy: –increases from solid to liquid to vapor/gas; –increases as temperature increases; –of gas increases as its volume increases at constant temperature; –increases when gases are mixed. –of elements increases down the group in the periodic table; –of compound increases as its structure becomes more complex.

10. Where do molecules have the higher entropy

11. Standard Entropy, S o The entropy of a substance in its most stable state at 1 atm and 25 o C. The entropy of an ionic species in 1 M solution at 25 o C.

12. Entropy and Second Law of Thermodynamics The second law of thermodynamics states that all spontaneous processes are accompanied by increase in the entropy of the universe. –Universe = System + Surrounding; –System: the process/reaction whose thermodynamic change is being studied; –Surrounding: the part of the universe that interacts with the system.

13. Conditions for Spontaneous Process Entropy change for a process:  S univ =  S sys +  S surr > 0,  process is spontaneous  S univ =  S sys +  S surr = 0,  process is at equilibrium If  S sys 0, and |  S surr | > |  S sys | If  S surr 0, and |  S sys | > |  S surr |

14. Gibb’s Free Energy For spontaneous reactions,  S univ =  S sys +  S surr > 0  S surr = -  H sys /T  S univ =  S sys -  H sys /T -T  S univ =  G sys =  H sys - T  S sys < 0  G sys is called Gibb’s free energy A new criteria for spontaneous process is  G sys < 0

15. What is Free Energy?

16. Free Energy?

17. Thermodynamic Free Energy It is the maximum amount of chemical energy derived from a spontaneous reaction that can be utilized to do work or to drive a nonspontaneous process. It is the minimum amount of energy that must be supplied to make a nonspontaneous reaction occur.

18. Effect of Temperature on  G and Spontaneity ——————————————————————————————————  H  S T  G Comments Examples —————————————————————————————————— - + high - spontaneous at 2H 2 O 2 (l)  2H 2 O (l) + O 2 (g) or low all temperature + + high - spontaneous at CaCO 3 (s)  CaO(s) + CO 2 (g) high temperature -- low - spontaneous at N 2 (g) + 3H 2 (g)  2NH 3 (g) low temperature + - high + nonspontaneous at 2H 2 O (l) + O 2 (g)  2H 2 O 2 (l) or low all temperature ________________________________________________________

19. Entropy Change in Chemical Reactions At constant temperature and pressure,  S o rxn =  n p S o products –  n r S o reactants In general,  S o rxn > 0 if  n p >  n r Example-1: C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (g), (  n p >  n r )  S o rxn = {(3 x S o CO 2 ) + (4 x S o H 2 O )} – {(S o C 3 H 8 ) + (5 x S o O 2 )} = {(3 x 214) + (4 x 189)}J/K – {270 + (5 x 205)}J/K = (642 + 756) J/K – (270 + 1025) J/K = 103 J/K

20. Entropy Change in Chemical Reactions  S o rxn < 0 if  n p <  n r Example-2: CO (g) + 2H 2 (g)  CH 3 OH (g), (  n p <  n r )   S o rxn = (S o CH 3 OH ) – {(S o CO ) + (2 x S o H 2 )} = 240 J/K – {198 J/K + (2 x 131 J/K) = 240 J/K – 460 J/K = -220 J/K

21. Effect of Temperature on  G o  G o =  H o - T  S o Example-1: For the reaction: N 2 (g) + 3H 2 (g)  2NH 3 (g),  H o = -92 kJ and  S o = -199 J/K = -0.199 kJ/K At 25 o C, T  S o = 298 K x (-0.199 J/K) = -59.3 kJ  G o =  H o - T  S o = -92 kJ – (-59.3 kJ) = -33 kJ;  reaction is spontaneous at 25 o C At 250 o C, T  S o = 523 K x (-0.199 J/K) = -104 kJ;  G o =  H o - T  S o = -92 kJ – (-104 kJ) = 12 kJ;  reaction is nonspontaneous at 250 o C

22. Effect of Temperature on  G o  G o =  H o – T  S o Example-2: For the reaction: CH 4 (g ) + H 2 O (g )  CO (g ) + 3H 2 (g ),  H o = 206 kJ and  S o = 216 J/K = 0.216 kJ/K At 25 o C, T  S o = 298 K x (0.216 J/K) = 64.4 kJ  G o =  H o - T  S o = 206 kJ – 64.4 kJ = 142 kJ;  reaction is nonspontaneous at 25 o C. At 1200 K, T  S o = 1200 K x (0.216 J/K) = 259 kJ;  G o =  H o - T  S o = 206 kJ – 259 kJ) = -53 kJ;  reaction is spontaneous at 1200 K

23.  G under Nonstandard Conditions Free energy change also depends on concentrations and partial pressures; Under nonstandard conditions (P i not 1 atm),  G =  G o + RTlnQ p, Consider the reaction: N 2 (g) + 3H 2 (g)  2NH 3 (g ), Q p = Under standard condition, P N 2 = P H 2 = P NH 3 = 1 atm, Q p = 1; lnQ p = 0, and  G =  G o

24.  G of reaction under nonstandard condition Consider the following reaction at 250 o C : N 2 (g) + 3H 2 (g)  2NH 3 (g), where, P N 2 = 5.0 atm, P H 2 = 15 atm, and P NH 3 = 5.0 atm Q p = 5 2 /(5 x 15 3 ) = 1.5 x 10 -3 lnQ p = ln(1.5 x 10 -3 ) = -6.5 Under this condition,  G =  G o + RTlnQ p ; (For this reaction at 250 o C, calculated  G o = 12 kJ)   G = 12 kJ + (0.008314 kJ/T x 523 K x (-6.5)) = 12 kJ – 28 kJ = -16 kJ  spontaneous reaction

25. Transition Temperature This is a temperature at which a reaction changes from being spontaneous to being nonspontaneous, and vice versa, when Q p or Q c equals 1 (standard condition) At transition temperature, T r,  G o =  H o – T r  S o = 0;  T r =  H o /  S o For reaction: N 2 (g) + 3H 2 (g)  2NH 3 (g), T r = -92 kJ/(-0.199 kJ/K) = 460 K = 190 o C Under standard pressure (1 atm), this reaction is spontaneous below 190 o C, but becomes nonspontaneous above this temperqature.

26. Transition Temperature For reaction: CH 4 (g ) + H 2 O (g )  CO (g ) + 3H 2 (g ),  H o = 206 kJ and  S o = 216 J/K = 0.216 kJ/K  G o =  H o + T r  S o = 0,  T r = 206 kJ/(0.216 kJ/K) = 954 K = 681 o C  Under standard pressure (1 atm), this reaction is not spontaneous below 681 o C, but becomes spontaneous above this temperature.  Reactions with both  H o and  S o < 0 favor low temperature;  Those with both  H o and  S o > 0 favor high temperature.

27. Free Energy and Equilibrium Constant For spontaneous reactions,  G decreases (becomes less negative) as the reaction proceeds towards equilibrium; At equilibrium,  G = 0;  G =  G o + RTlnK = 0  G o = -RTlnK lnK = -  G o /RT (  G o calculated at temperature T) Equilibrium constant, K = e -(  G o /RT)  Go 1; reaction favors products formation  Go > 0, K < 1; reaction favors reactants formation  Go = 0, K = 1; reaction favors neither reactants nor products

28. Calculating K from  G o Consider the reaction: N 2 (g) + 3H 2 (g)  2NH 3 (g), At 25 o C,  G o = -33 kJ lnK = -(-33 x 10 3 J/(298 K x 8.314 J/K.mol)) = 13 K = e 13 = 4.4 x 10 5 (reaction goes to completion) At 250 o C,  G o = 12 kJ; lnK = -(12 x 10 3 J/(523 K x 8.314 J/K.mol)) = -2.8 K = e -2.8 = 0.061 (very little product is formed)

29. Coupling Reactions A nonspontaneous reaction can be coupled to a spontaneous one to make it happen. Example: Fe 2 O 3 (s)  2Fe (s) + 3/2 O 2 (g) ;  G o = 740 kJ (eq-1) CO (g) + ½ O 2 (g)  CO 2 (g) ;  G o = -283 kJ 3CO (g) + 3/2 O 2 (g)  3CO 2 (g) ;  G o = -849 kJ (eq-2) Combining eq-1 and eq-2, Fe 2 O 3 (s) + 3CO (g)  2Fe (s) + 3CO 2 (g) ;  G o = -109 kJ

30. Coupling Reactions in Biological System The formation of ATP from ADP and H 2 PO 4 - is nonspontaneous, but it can be coupled to the hydrolysis of creatine-phosphate that has a negative  G o. ADP + H 2 PO 4 -  ATP + H 2 O;  G o = +30 kJ Creatine-phosphate  creatine + phosphate;  G o = -43 kJ Combining the two equations yields a spontaneous overall reaction: Creatine-phosphate + ADP  Creatine + ATP ;  G o = -13 kJ