1. System strives for minimum Free Energy

2. aA + bB cC + dD G0G0 rxn d  G 0 (D) f c  G 0 (C) f = [+] - b  G 0 (B) f a  G 0 (A) f [+] G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - The standard free-energy of reaction (  G 0 ) is the free- energy change for a reaction when it occurs under standard- state conditions. rxn Standard free energy of formation (  G 0 ) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f  G 0 of any element in its stable form is zero. f

3. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - What is the standard free-energy change for the following reaction at 25 0 C? G0G0 rxn 6  G 0 (H 2 O) f 12  G 0 (CO 2 ) f = [+] - 2  G 0 (C 6 H 6 ) f [] G0G0 rxn = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ Is the reaction spontaneous at 25 0 C?  G 0 = -6405 kJ < 0 spontaneous 18.4

4.  G =  H - T  S  G gives direction of reaction If A  B  G < 0 then B  A  G > 0 A  B  G = 0  G does not indicate the rate of the process.

5. CaCO 3 (s) CaO (s) + CO 2 (g)  H 0 = 177.8 kJ  S 0 = 160.5 J/K  G 0 =  H 0 – T  S 0  G 0 = 0 =  H sys -T  S sys  H sys /  S sys = T 177.8 kJ / 0.1605 kJ/K = T T = 835 0 C At 25 0 C,  G 0 = 130.0 kJ 18.4 Temperature and Spontaneity of Chemical Reactions When does reaction switch from being spontaneous to non- spontaneous? Watch units !!!

6. Gibbs Free Energy and Phase Transitions H 2 O (l) H 2 O (g)  G 0 = 0=  H 0 – T  S 0  S = T HH = 40.79 kJ 373 K = 109 J/K 18.4

7. Gibbs Free Energy and Chemical Equilibrium  G =  G 0 + RT lnQ R is the gas constant (8.314 J/K mol) T is the absolute temperature (K) Q is the reaction quotient = [Products] 0 or P° Products 18.4 Can correct Gibbs Free Energy change for non-standard state [Reactants] 0 P° Reactants

8. N 2 (g) + 3H 2 (g) 2 NH 3 (g) What is the standard free-energy change for the following reaction at 25 0 C? G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - G0G0 rxn  G 0 (N 2 ) f 2  G 0 (NH 3 ) f = - - 3  G 0 (H 2 ) f G0G0 rxn = [ 2(–16.66 kJ/mole) – 0 – 0 = - 33.32 kJ What is the standard free-energy change for the following mixture of gases at 25 0 C: 0.10 atm N 2, 0.10 atm H 2 and 0.10 atm NH 3 ?  G =  G 0 + RT lnQ =  G 0 + RT ln P NH3 2 P N2 P H2 3 = -33.32 kJ + (0.008314 kJ /molK)(298K) ln 0.10 2 (0.10)(0.10) 3 = -33.32 kJ + 11.4 kJ = -21.92 kJ

9. Gibbs Free Energy and Chemical Equilibrium  G =  G 0 + RT lnQ R is the gas constant (8.314 J/K mol) T is the absolute temperature (K) Q is the reaction quotient At Equilibrium  G = 0 Q = K 0 =  G 0 + RT lnK  G 0 =  RT lnK 18.4

10. N 2 (g) + 3H 2 (g) 2 NH 3 (g) What is K p for the following reaction at 25 0 C? K p = e -  G 0  RT K p = e - ( -33.32 kJ)/(0.008314 kJ /molK)(298K) K p = e 13.4 K p = 6.6 x 10 5

11.  G 0 < 0  G 0 > 0 18.4

12.  G 0 =  RT lnK 18.4

13.  G 0 =  RT lnK NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) K b = 1.8 x 10 -5 Consider the reaction: What is  G 0 for the reaction at 25 0 C?  G 0 =  RT lnK = -(0.008314 kJ /molK)(298K) ln(1.8 x 10 -5 )  G 0 = 27.08 kJ /mol K < 1, reinforces that this is a weak base and left side is favored.

14.  RT lnK =  G 0 =  H 0 – T  S 0 How does K change with Temperature? At equilibrium: RT lnK +  H 0 = T  S 0 R lnK +  H 0 /T =  S 0 constant Compare 2 Temperatures: R lnK 1 +  H 0 /T 1 = R lnK 2 +  H 0 /T 2 R(lnK 2 - lnK 1 ) =  H 0 /T 1 -  H 0 /T 2 ln(K 2 / K 1 ) =  H 0 /R (1/T 1 - 1/T 2 ) Similar to Clausius-Clapeyron Equation for vapor pressure and rate constant change equation

15. H 2 (g) + I 2 (g) 2 HI (g) What is K p for the following reaction at 25 0 C? K p = e -  G 0  RT - ( 2.6 kJ)/(0.008314 kJ /molK)(298K) -1.05 K p = 0.350  G 0 = 2.6 kJ = e What is K p at 50 0 C? H0H0 rxn =2(25.9 kJ/mole) – 0 – 0 = 51.8 kJ ln(K 2 / K 1 ) =  H 0 /R (1/T 1 + 1/T 2 ) ln(K 2 / 0.350 = (51.8 kJ )/(0.008314 kJ /molK)(1/298K + 1/323K) lnK 2 - ln0.350 = (6.23 x 10 3 )(2.60 x 10 -4 ) lnK 2 = 1.62 - 1.05 = 0.60 K 2 = 1.82

16. 2 NO (g) + O 2 (g) 2 NO 2 (g) Calculate  G 0 for the following reaction at 25 0 C? Given:  G 0 =  H 0 – T  S 0 S0S0 rxn = 2 x S 0 (NO 2 ) – [2 x S 0 (NO) – S 0 (O 2 )] H0H0 rxn = 2 x  H 0 (NO 2 ) – [2 x  H 0 (NO) +  H 0 (O 2 )] f fff = 2(33.84 kJ) - 2(90.37 kJ) - 0 = -113.06 kJ = 2(240.45J/molK) - 2(210.62 J/molK) - 205.0 j/molK = -145.34 J/K

17.  G 0 =  H 0 – T  S 0 G0G0 rxn = 2 x  G 0 (NO 2 ) – [2 x  G 0 (NO) +  G 0 (O 2 )] fff = 2(51.84 kJ/mol) - 2(86.71 kJ/mol) - 0 = 103.68 kJ - 173.42 kJ = -69.74 kJ = -113.06 kJ - (298 K)(- 0.14534 kJ/K = -113.06 kJ + 43.31 kJ = -69.75 kJ Can also calculate with  G 0 ’s from tabulated data f Get same answer either way  G 0,  H 0 and  S 0 are all state functions