1. Thermodynamics of Reactions I.Entropy and Chemical Reactions A.Example: N 2 (g) + 3H 2 (g) 2NH 3 (g) 1)System: positional probability a)4 reactant particles 2 product particles b)Fewer possible configurations,  S = - B.An increase in number of gas particles is entropically favored 1)4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) 2)Nine particles 10 particles,  S = + 3)Example: Predict sign of  S a)CaCO 3 (s) CaO(s) + CO 2 (g) b)2SO 2 (g) + O 2 (g) 2SO 3 (g)

2. C.Third Law of Thermodynamics = the entropy of a perfect crystal at 0K = 0 (there is only one possible configuration) 1)As temperature increases, random vibrations occur;  S = + 1)We can calculate S for any substance if we know how S depends on T [Standard S o values, Appendix IIB] 2)  S o reaction =  nS o prod -  nS o reactants 4)Example: Find  S o at 25 o C for 2NiS(s) + 3O 2 (g) 2SO 2 (g) + 2NiO(s) S o (J/K mol) 53 205 248 38  S o = 2(248) + 2(38) – 2(53) – 3(205) = -149 J/Kmol (3 gas particles 2 gas particles)

3. D.Factors Affecting Standard Entropies 1)Phase of Matter: as discussed earlier, S gas >> S liquid > S solid a)S o H 2 O(l) = 70.0 J/Kmol b)S o H 2 O(g) = 188.8 J/Kmol 2)Molar Mass: heavy elements have more entropy than light (in same state) Has to do with translation energy states 3)Allotropes: more rigid structures have less entropy 4)Molecular Complexity: more complex = more entropy a)S o Ar(g) = 154.8 (39.948 g/mol) b)S o NO(g) = 210.8 (30.006 g/mol)

4. II.Free Energy and Chemical Reactions A.Standard Free Energy Change =  G o = reactants/products at standard states 1)Gases at 1 atm, solution = 1 M, element at 25 o C and 1 atm has G o = 0 2)N 2 (g) + 3H 2 (g) 2NH 3 (g)  G o = -33.3 kJ  G o can’t be measured directly, it must be calculated from  H o and  S 4)Usefulness of  G o: The more negative  G o is, the more likely reaction is i.If  G o is negative, the reaction is spontaneous as written ii.If  G o is positive, the reaction is not spontaneous 5)Why use standard states? Because  G changes with P, T, concentration B.Calculating  G o :  G o =  H o - T  S o 1)Example: C(s) + O 2 (g) CO 2 (g) i.  H o = -393.5 kJ  S o = 3.05 J/K ii.  G o = (-3.935 x 10 5 J) – (298 K)(3.05 J/K) = -394.4 kJ 2)Example: Find  H o,  S o,  G o for 2SO 2 + O 2 2SO 3  H o (kJ/mol) -297 0 -396  S o (J/Kmol) 248 205 257 T = 298 K

5. 3)Use the  G o of known reactions to find  G o of unknown reactions a)2CO + O 2 2CO 2  G o = ? b)Known Reactions i)2CH 4 + 3O 2 2CO + 4H 2 O  G o = -1088 kJ/mol ii)CH 4 + 2O 2 CO 2 + 2H 2 O  G o = -801 kJ/mol c)If we reverse the first reaction and double the second… i)2CO + 4H 2 O 2CH 4 + 3O 2  G o = +1088 kJ/mol ii)2(CH 4 + 2O 2 CO 2 + 2H 2 O)  G o = -1602 kJ/mol iii)2CO + O 2 2CO 2  G o = -514 kJ/mol d)Example: Find  G o for C diamond C graphite Given that C d + O 2 CO 2  G o = -397 kJ/mol C g + O 2 CO 2  G o = -394 kJ/mol 4)The  G f o Method a)Free energy of formation  G f o is tabulated for many compounds b)We can sum these for reactants and products to find  G o

6. c)Example: 6C(s) + 6H 2 (g) + 3O 2 (g) C 6 H 12 O 6 (s) (glucose) d)  G f o = free energy change of formation of one mole of the product from its elements in their standard states (-911 kJ/mol for glucose) e)Example: 2CH 3 OH(g) + 3O 2 (g) 2CO 2 (g) + 4H 2 O(g)  G f o (kJ/mol) -163 0 -394 -229  G o =  nG f o prod -  nG f o reactants  G o = 2(-394) + 4(-229) - 2(-163) - 3(0) = -1378 kJ/mol  G and Pressure A.Pressure dependencies of the state functions 1)H is not pressure dependent 2)S is pressure dependent a)S(large volume) > S(small volume) b)S(low pressure) > S(high pressure) c)PV = nRT

7. 3)G must depend on pressure since G = H – TS G = G o + RTlnP Use R = 8.3145 J/Kmol B.Example: N 2 (g) + 3H 2 (g) 2NH 3 (g)  G = 2G(NH 3 ) – G(N 2 ) – 3G(H 2 )

8. = 0.0313 K = P H2O(g)

9. 2)Example: CO (g) + 2H 2(g) CH 3 OH (l) Find  G at 25 o C, P(CO) = 5 atm, P(H 2 ) = 3 atm a)From Appendix IIB, find  G o = -166 – (-137) – 0 = -29 kJ/mol b)Find RTlnQ = (8.3145)(298)ln(1/(5)(3) 2 ) = -9.4 kJ/mol c)  G =  G o + RTlnQ = -38 kJ/mol d)More spontaneous at these conditions than at standard conditions IV.Meaning of  G for Reaction at Equilibrium A.Even if  G is negative, the spontaneous reaction doesn’t necessarily go to completion 1)Phase changes always go to completion if spontaneous 2)Reactions often have a minimum G that is somewhere before completion of the reaction

10. 4) The equilibrium mixture of reactants and products might be more stable (lower  G) than the completely formed product alone CO (g) + 2H 2(g) CH 3 OH (l) B.Equilibrium 1)Kinetics: forward and reverse reaction rates are the same at equilibrium 2)Thermodynamics: the lowest free energy state is at equilibrium 3)A(g) B(g) a)G A = G A o + RTlnP A (this is decreasing as the reaction proceeds) b)G B = G B o + RTlnP B (this is increasing as the reaction proceeds) c)G = G A + G B (this is decreasing as the reaction proceeds) 4)At equilibrium G A = G B and we have a new P A E and P B E a)G is no longer decreasing, it is at its minimum point b)No further driving force for the reaction to proceed (  G = 0) Reaction StartsReaction Proceeds Equilibrium

11. 5)Example: A(g) B(g) a)P A E = 0.25(2 atm) = 0.5 atm b)P B E = 0.75(2 atm) = 1.5 atm c)K = P B /P A = 1.5/.5 = 3.0 6)The same Equilibrium Position would be reached from any initial condition where A + B = 2.0 atm 7)At Equilibrium: G prod = G react (  G = G prod – G react = 0) 1 mol A, 2 atm1 mol B, 2 atm1 mol A/B, 2 atm

12. 8)At Equilibrium:  G = 0 =  G o + RTlnK  G o = -RTlnK a)If  G o = 0, then K = 1 and we are at equilibrium b)If  G o 1 and the reaction proceeds forward c)If  G o > 0, then K < 1 and the reaction proceeds in reverse 9)Example: N 2 (g) + 3H 2 (g) 2NH 3 (g) a)Given  G o = -33.3 kJ/mol at 25 o C b)Predict the direction when P(NH 3 ) = 1 atm, P(N 2 ) = 1.47 atm and P(H 2 ) = 0.01 atm c)Predict direction when P(NH 3 ) = P(N 2 ) = P(H 2 ) = 1 atm 10) Example: 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) at 25 o C  H f o 0 0 -826 kJ/mol S o 27 205 90 J/Kmol Find K 11) Dependence of K on T 1/T lnK Slope = -  H/R Intercept =  S/R

14.  G and Work a)W max =  G All the free energy produced from a spontaneous reaction could be used to do work (Reversible Process only) b)For a non-spontaneous reaction,  G tells us how much work we would have to do on the system to get the reaction to occur c)W actual < W max We always lose some energy to heat in any process (Irreversible Processes) d)Theoretically, Reversible Process utilize all energy for work, Unfortunately, all real processes are Irreversible  G = -50.5 kJ  S = -80.8 J/K Irreversible (real) Process Increases S surr to make  S univ = +