Chapter 17 Lesson 2 Free Energy and Thermodynamics

Gibbs Free Energy, G = H -TS Multiply through by -T -T∆S univ = ∆H sys - T∆S sys -T∆S univ = change in Gibbs free energy for the system = ∆G system hence, ∆ G = ∆ H -T ∆ S hence, ∆ G = ∆ H -T ∆ S Under standard conditions — ∆G o sys = ∆H o sys - T∆S o sys ∆S univ = ∆S surr + ∆S sys J. Willard Gibbs

∆G = ∆H - T∆S Gibbs free energy change = total energy change for system - energy lost in disordering the system If the reaction is exothermic (negative ∆H) and entropy increases (positive ∆S o ) then ∆ G must be NEGATIVEthen ∆ G must be NEGATIVE the reaction is spontaneous (and product- favored) at ALL temperatures.

4 ∆G > 0)  G will be positive ( ∆G > 0) when  H is positive (endothermic) and  S is negative (more ordered). So the change in free energy will be positive at all temperatures. The reaction will therefore be nonspontaneous at ALL temperatures W hen  G = 0 the reaction is at equilibrium ∆G = ∆H - T∆S

5  G =  H – T  S Spontaneous or Not? A decrease in Gibbs free energy (  G < 0) corresponds to a spontaneous process An increase in Gibbs free energy (  G > 0) corresponds to a nonspontaneous process

Calculating ∆G o ∆G o ∆∆S o Calculating ∆G o : ∆G o = ∆H -T∆S o Combustion of 25 o C C 2 H 2 (g) + 5/2 O 2 (g)  2 CO 2 (g) + H 2 O(g) Use enthalpies of formation to calculate ∆H o rxn = kJ Use standard molar entropies to calculate ∆S o rxn = J/K or kJ/K ∆G o rxn = kJ - (298 K)( J/K) = kJ (spontaneous) Reaction is product-favored in spite of negative ∆S o rxn. Reaction is “enthalpy driven”

From tables of thermodynamic data we find ∆H o rxn = kJ (endothermic) ∆H o rxn = kJ (endothermic) ∆S o rxn = J/K or kJ/K (disorder) ∆G o rxn = kJ - (298 K)( kJ/K) = -6.7 kJ (spontaneous) = -6.7 kJ (spontaneous) Reaction is product-favored in spite of positive ∆H o rxn. Reaction is “entropy driven” NH 4 NO 3 (s) + heat  NH 4 NO 3 (aq) Is the dissolution of ammonium nitrate product- favored? If so, is it enthalpy- or entropy-driven? Calculating ∆G o ∆G o ∆∆S o Calculating ∆G o : ∆G o = ∆H -T∆S o

The reaction CCl 4(g)  C (s, graphite) + 2 Cl 2(g) has  H = kJ and  S = J/K at 25°C. Calculate  G and determine if it is spontaneous. Since  G is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature.  H = kJ,  S = J/K, T = 298 K  G, kJ Answer: Solution: Concept Plan: Relationships: Given: Find: GGT,  H,  S

The reaction CCl 4(g)  C (s, graphite) + 2 Cl 2(g) has  H = kJ and  S = J/K. Calculate the minimum which it will be spontaneous. The temperature must be higher than 673K for the reaction to be spontaneous (i.e. 674 K)  H = kJ,  S = J/K,  G < 0  Answer: Solution: Concept Plan: Relationships: Given: Find: T  G,  H,  S

Gibbs Free Energy, G Calculating ∆G o Calculating ∆G o (two ways) a) Determine ∆H o rxn and ∆S o rxn and use Gibbs equation (at various temps). b) b) Use tabulated values of free energies of formation, ∆G f 25 o C ∆G o rxn =  ∆G f o (products) -  ∆G f o (reactants)

11

Calculate  G  at 25  C for the reaction CH 4(g) + 8 O 2(g)  CO 2(g) + 2 H 2 O (g) + 4O 3(g) standard free energies of formation from Appendix IIB  G , kJ Solution: Concept Plan: Relationships: Given: Find: GG  G  f of prod & react Substance  G  f, kJ/mol CH 4 (g)-50.5 O2(g)O2(g)0.0 CO 2 (g) H 2 O(g) O3(g)O3(g)163.2 (spontaneous)

The reaction SO 2(g) + ½ O 2(g)  SO 3(g) has  H  = kJ and  S  = J/K at 25°C. Calculate  G  at 125  C and determine if it is spontaneous. Since  G is -ve, the rxn is spontaneous at this temperature, but less spontaneous than at 25  C (-127 kJ)  H  = kJ,  S  = J/K, T = 398 K  G , kJ Answer: Solution: Concept Plan: Relationships: Given: Find: GG T,  H ,  S  (PRACTICE PROBLEM)

Copyright McGraw-Hill 2009 Free Energy and Equilibrium  G =  G° + RT ln Q  G = non-standard free energy  G° = standard free energy (from tables) R = J/K·mole T = temp in K Q = reaction quotient

∆G, ∆G˚, and K eq ∆G is the change in free energy at non- standard conditions. ∆G is related to ∆G˚ ∆G = ∆G˚ + RT ln Q where Q = reaction quotient When Q K, the reaction may be spontaneous or nonspontaneous. When Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln K

 FACT: ∆G o rxn is the change in free energy when pure reactants convert COMPLETELY to pure products, both at standard conditions. ∆G˚ rxn 1 (∆G˚ rxn < 0).  Therefore, both ∆G˚ rxn and K eq are related to reaction favorability. Thermodynamics and K eq Summary: Summary: ∆G˚ = - RT ln K

Calculate K for the 25 o C N 2 O 4  2 NO 2 ∆G o rxn = +4.8 kJ ∆G o rxn = J = - (8.31 J/K)(298 K) ln K ∆G o rxn = - RT lnK Thermodynamics and K eq K = e –1.94 = 0.14 (reactant favored) When ∆G o rxn > 0 (nonspontaneous), then K 0 (nonspontaneous), then K < 1!!

Estimate the equilibrium constant and position of equilibrium for the following reaction at 427°C N 2 (g) + 3 H 2 (g) ⇄ 2 NH 3 (g)  G° = -RT lnK J = -(8.314 J/K)(700 K) lnK ln K = K = e − 7.97 = 3.45  10 − 4 small!!! since K is << 1, the position of equilibrium favors reactants        H° = [ 2(-46.19)] − [ 0 +3( 0)] = kJ = J  S° = [2 (192.5)] − [(191.50) + 3(130.58)] = J/K  G° = J - (700 K)( J/K)  G° = J (nonspontaneous)

Copyright McGraw-Hill 2009 Relationship Between  G° and K

20 Calculate  G at 427°C for the reaction below if the P N2 = 33.0 atm, P H2 = 99.0 atm, and P NH3 = 2.0 atm N 2 (g) + 3 H 2 (g) ⇄ 2 NH 3 (g) Q = P NH3 2 P N2 1  P H2 3 (2.0 atm) 2 (33.0 atm) 1 (99.0) 3 = = 1.2   G =  G° + RTlnQ spontaneous  G = J + (8.314 J/K)(700 K)(ln 1.2  )  H° = [ 2(-46.19)] - [0 +3( 0)] = kJ = J  S° = [2 (192.5)] - [(191.50) + 3(130.58)] = J/K  G° = J - (700 K)( J/K)  G° = J (nonspontaneous)  G = J − J = J = − 46 kJ  −  G°

21 Q. Rank the following in order of increasing molar entropy (S o 25 o C! a) Cl 2 (g), I 2 (g), Br 2 (g), and F 2 (g) b) H 2 O(g), H 2 O 2 (g), H 2 S(g) F 2 (g) < Cl 2 (g) < Br 2 (g) < I 2 (g) H 2 O(g) < H 2 S(g) < H 2 O 2 (g)

 G under Nonstandard Conditions   G =  G  only when the reactants and products are in their standard states  there normal state at that temperature  partial pressure of gas = 1 atm  concentration = 1 M  under nonstandard conditions,  G =  G  + RTlnQ  Q is the reaction quotient  at equilibrium  G = 0   G  = − RTlnK and  G° =  H  − T  S°  H  − T  S° = − RTlnK, by rearranging RTlnK = −  H  − T  S°, and dividing by R T

Temperature Dependence of K for an exothermic reaction, increasing the temperature decreases the value of the equilibrium constant for an endothermic reaction, increasing the temperature increases the value of the equilibrium constant RTlnK −  H  T  S° ———— = ———— + ———— RT RT R T

24 High Entropy Low Entropy Chemistry In Action: The Thermodynamics of a Rubber Band

Copyright McGraw-Hill Thermodynamics in Living Systems Coupled reactions Thermodynamically favorable reactions drives an unfavorable one Enzymes facilitate many nonspontaneous reactions

26 The Structure of ATP and ADP in Ionized Forms ATP  ADP  G° = -31 kJ/mol

Copyright McGraw-Hill 2009 ATP-ADP Interconversions C 6 H 12 O 6 oxidation:  G° =  2880 kJ/mol ATP  ADP  G° = -31 kJ/mol Synthesis of proteins: (first step) alanine + glycine  alanylglycine  G° = 29 kJ/mol Protein synthesis is now favored.