1. Chapter 16 Spontaneity, Entropy and Free Energy

2. 16.1 Spontaneity and Entropy  A reaction that will occur without outside intervention. We can’t determine how fast.  We need both thermodynamics and kinetics to describe a reaction completely.  Thermodynamics compares initial and final states but not how long it will take.  A reaction that will occur without outside intervention. We can’t determine how fast.  We need both thermodynamics and kinetics to describe a reaction completely.  Thermodynamics compares initial and final states but not how long it will take. l Kinetics describes the pathway between. (YDVD)

3. Thermodynamics  1st Law - the energy of the universe is constant. (Cannot create or destroy)  Keeping track of thermodynamics doesn’t correctly predict spontaneity.  Entropy (S) is disorder or randomness.  2nd Law - the entropy of the universe increases.  1st Law - the energy of the universe is constant. (Cannot create or destroy)  Keeping track of thermodynamics doesn’t correctly predict spontaneity.  Entropy (S) is disorder or randomness.  2nd Law - the entropy of the universe increases.

4. Entropy  The driving force for a spontaneous process is an increase in the entropy of the universe.  Defined in terms of probability.  Substances take the arrangement that is most likely.  The most likely is the most random.  The driving force for a spontaneous process is an increase in the entropy of the universe.  Defined in terms of probability.  Substances take the arrangement that is most likely.  The most likely is the most random.

5. Think about this.  It would seem like reactions with a negative  H value would proceed spontaneously.  But NH 3 = O 2  NO + H 2 O will not occur spontaneously even though  H = -909kJ/mol  It would also seem like reactions with a positive  H value would not occur spontaneously.  But KCl(s)  K + (aq) + Cl - (aq) will occur spontaneously even though  H =+19.2 kJ/mol  Activation energy is involved in both reactions but there must be something else involved.  Entropy is that driving force!  It would seem like reactions with a negative  H value would proceed spontaneously.  But NH 3 = O 2  NO + H 2 O will not occur spontaneously even though  H = -909kJ/mol  It would also seem like reactions with a positive  H value would not occur spontaneously.  But KCl(s)  K + (aq) + Cl - (aq) will occur spontaneously even though  H =+19.2 kJ/mol  Activation energy is involved in both reactions but there must be something else involved.  Entropy is that driving force!

6. Positional Entropy  A gas expands into a vacuum because the expanded state has the highest positional probability of states available to the system.  S solid < S liquid << S gas  There are many more ways for the molecules to be arranged as a liquid than a solid.  Gases have a huge number of positions possible.  A gas expands into a vacuum because the expanded state has the highest positional probability of states available to the system.  S solid < S liquid << S gas  There are many more ways for the molecules to be arranged as a liquid than a solid.  Gases have a huge number of positions possible.

7. Practice For each of the following pairs, choose the substance with the higher positional entropy (per mole) at a given temperature. solid CO 2 and gaseous CO 2 N 2 gas at 1 atm and N 2 gas at 1.0 x 10 -2 atm Predict the sign of the entropy change for each of the following processes. Solid sugar is added to water to form a solution. Iodine vapor condenses on a cold surface to form crystals. For each of the following pairs, choose the substance with the higher positional entropy (per mole) at a given temperature. solid CO 2 and gaseous CO 2 N 2 gas at 1 atm and N 2 gas at 1.0 x 10 -2 atm Predict the sign of the entropy change for each of the following processes. Solid sugar is added to water to form a solution. Iodine vapor condenses on a cold surface to form crystals.

8. Proof?  The change in entropy (  S) in a reaction can be found by subtracting the absolute values of the entropy of the reactants from the absolute entropies of the products   S reaction =  S products -  S reactants  For the reaction below, the absolute entropies of the chemicals involved are 193 J/K mol forNH 3 (g), 192 J/K mol-1 for N 2 (g) and 131 J/K mol-1 for H 2 (g).  N 2 (g) + 3H 2 (g)  2NH 3 (g)   S reaction = (2 x 193) – (192 + (3 x 131)) = -199 J/K  This makes sense since the negative value implies that the system has become more ordered  (four gas molecules are converted to two gas molecules).  The change in entropy (  S) in a reaction can be found by subtracting the absolute values of the entropy of the reactants from the absolute entropies of the products   S reaction =  S products -  S reactants  For the reaction below, the absolute entropies of the chemicals involved are 193 J/K mol forNH 3 (g), 192 J/K mol-1 for N 2 (g) and 131 J/K mol-1 for H 2 (g).  N 2 (g) + 3H 2 (g)  2NH 3 (g)   S reaction = (2 x 193) – (192 + (3 x 131)) = -199 J/K  This makes sense since the negative value implies that the system has become more ordered  (four gas molecules are converted to two gas molecules).

9. 16.2 Entropy  Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate.  2nd Law - in any spontaneous process there is always an increase in the entropy of the universe.   S univ =  S sys +  S surr  If  S univ is positive the process is spontaneous.  If  S univ is negative the process is spontaneous in the opposite direction.  If  S univ is 0 the system is at equilibrium.  Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate.  2nd Law - in any spontaneous process there is always an increase in the entropy of the universe.   S univ =  S sys +  S surr  If  S univ is positive the process is spontaneous.  If  S univ is negative the process is spontaneous in the opposite direction.  If  S univ is 0 the system is at equilibrium.

10.  For exothermic processes  S surr is positive.  For endothermic processes  S surr is negative.  Consider this process H 2 O(l)  H 2 O(g)   S sys is positive   S surr is negative   S univ depends on temperature.  For exothermic processes  S surr is positive.  For endothermic processes  S surr is negative.  Consider this process H 2 O(l)  H 2 O(g)   S sys is positive   S surr is negative   S univ depends on temperature.

11. 16.3 Temperature and Spontaneity  Entropy changes in the surroundings are determined by the heat flow.  An exothermic process is favored because by giving up heat the entropy of the surroundings increases.  The size of  S surr depends on temperature.   S surr = -  H/T  Entropy changes in the surroundings are determined by the heat flow.  An exothermic process is favored because by giving up heat the entropy of the surroundings increases.  The size of  S surr depends on temperature.   S surr = -  H/T

12. Practice  In the metallurgy of antimony, the pure metal is recovered by two different reactions. Ex16.4  Sb 2 S 3 (s) + 3Fe(s) 2Sb(s) +3FeS(s)  H = -125 kJ  Sb 4 O 6 (s) + 6C(s) 4Sb(s) + 6CO(g)  H = 778 kJ  Calculate the  S surr for each of these reactions at 25°C and 1 atm.  In the metallurgy of antimony, the pure metal is recovered by two different reactions. Ex16.4  Sb 2 S 3 (s) + 3Fe(s) 2Sb(s) +3FeS(s)  H = -125 kJ  Sb 4 O 6 (s) + 6C(s) 4Sb(s) + 6CO(g)  H = 778 kJ  Calculate the  S surr for each of these reactions at 25°C and 1 atm.

13. 16.4 Gibb's Free Energy   G=  H-T  S at constant temperature  Dividing by -T and substituting -  H/T for  S surr we’ve shown that:  -  G/T =  S univ (at constant T and P)  This is very important! It means if the process is carried out at constant T and P, the process will be spontaneous only if  G is negative.  -  G means +  S univ   G=  H-T  S at constant temperature  Dividing by -T and substituting -  H/T for  S surr we’ve shown that:  -  G/T =  S univ (at constant T and P)  This is very important! It means if the process is carried out at constant T and P, the process will be spontaneous only if  G is negative.  -  G means +  S univ

14. Let’s Check  For the reaction H 2 O(s)  H 2 O(l)   Sº = 22.1 J/K mol  Hº =6030 J/mol  Calculate  G at 10ºC and -10ºC  Look at the equation  G=  H-T  S  Spontaneity can be predicted from the sign of  H and  S.  For the reaction H 2 O(s)  H 2 O(l)   Sº = 22.1 J/K mol  Hº =6030 J/mol  Calculate  G at 10ºC and -10ºC  Look at the equation  G=  H-T  S  Spontaneity can be predicted from the sign of  H and  S.

15.  G=  H-T  S HH SS Spontaneous? +- At all Temperatures ++ At high temperatures, “entropy driven” -- At low temperatures, “enthalpy driven” +- Not at any temperature, Reverse is spontaneous

16.  G and Spontaneity  At what temperature is the following process spontaneous at 1 atm?  Br 2 (l)Br2(g)   kJ/mol  S° = 93.0 J/Kmol  Hint: Use  G=  H-T  S and set  G to 0  At what temperature is the following process spontaneous at 1 atm?  Br 2 (l)Br2(g)   kJ/mol  S° = 93.0 J/Kmol  Hint: Use  G=  H-T  S and set  G to 0

17. 16.5 Entropy Changes in Chemical Reactions Reactions involving gaseous molecules:  The change in positional entropy is dominated by the relative number of molecules of gaseous reactants and products.  2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O   S increases because there are more molecules on the product side than the reactant side. Reactions involving gaseous molecules:  The change in positional entropy is dominated by the relative number of molecules of gaseous reactants and products.  2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O   S increases because there are more molecules on the product side than the reactant side.

18. Practice Predict the sign of  S° for each. Ex16.6  The thermal decomposition of solid calcium carbonate:  CaCO 3 (s) CaO(s) + CO 2 (g)  The oxidation of SO 2 in air:  2SO 2 (g) + O 2 (g) 2SO 3 (g) Predict the sign of  S° for each. Ex16.6  The thermal decomposition of solid calcium carbonate:  CaCO 3 (s) CaO(s) + CO 2 (g)  The oxidation of SO 2 in air:  2SO 2 (g) + O 2 (g) 2SO 3 (g)

19. Predict the sign of ΔS ｡ for each of the following changes. (a)Na(s) + 1/2 Cl 2 (g) NaCl(s) (b) N 2 (g) + 3 H 2 (g) 2 NH 3 (g) (c) NaCl(s) Na + (aq) + Cl - (aq) (d) NaCl(s) NaCl(l)

20. Third Law of Thermodynamics  The entropy of a pure crystal at 0 K is 0.  No disorder. This gives us a starting point.  All others must be > 0.  Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed.   Products -  reactants to find  Sº (a state function).  More complex molecules higher Sº.  The entropy of a pure crystal at 0 K is 0.  No disorder. This gives us a starting point.  All others must be > 0.  Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed.   Products -  reactants to find  Sº (a state function).  More complex molecules higher Sº.

21. Practice 1.Predict then calculate  S° at 25°C for the reaction. (use the  S° values in appendix A21) Ex 16.7 2NiS(s) + 3O 2 (g) 2SO 2 (g) + 2NiO(s) 2.Predict then calculate  S° for the reduction of aluminum oxide by hydrogen gas. Ex 16.8 Al 2 O 3 (s) + 3H 2 (g) 2Al(s) + 3H 2 O(g) 1.Predict then calculate  S° at 25°C for the reaction. (use the  S° values in appendix A21) Ex 16.7 2NiS(s) + 3O 2 (g) 2SO 2 (g) + 2NiO(s) 2.Predict then calculate  S° for the reduction of aluminum oxide by hydrogen gas. Ex 16.8 Al 2 O 3 (s) + 3H 2 (g) 2Al(s) + 3H 2 O(g)

22. Practice 1.For the reaction #35 C 2 H 2 (s) + 4F 2 (g) 2CF 4 (g) + H 2 (g)  S° is equal to - 358J/K. Use this value and data from the appendix to calculate the S° for CF 4 (g) Consider the reaction #39 1.2O(g)O 2 (g) a  Predict the signs of  H and  S. b. Would the reaction be more spontaneous at high or low temperatures? 1.For the reaction #35 C 2 H 2 (s) + 4F 2 (g) 2CF 4 (g) + H 2 (g)  S° is equal to - 358J/K. Use this value and data from the appendix to calculate the S° for CF 4 (g) Consider the reaction #39 1.2O(g)O 2 (g) a  Predict the signs of  H and  S. b. Would the reaction be more spontaneous at high or low temperatures?

23. From data in Appendix 4, calculate ΔH°, ΔS°, and ΔG° for the following reaction at 25 ｡ C. #41 CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g)

24. 16.6 Free Energy in Reactions   Gº = standard free energy change.  Free energy change that will occur if reactants in their standard state turn to products in their standard state.  Can’t be measured directly, can be calculated from other measurements.  Method #1  Gº=  Hº-T  Sº  Method #2 Use Hess’s Law with known reactions.   Gº = standard free energy change.  Free energy change that will occur if reactants in their standard state turn to products in their standard state.  Can’t be measured directly, can be calculated from other measurements.  Method #1  Gº=  Hº-T  Sº  Method #2 Use Hess’s Law with known reactions.

25. Free Energy in Reactions  There are tables of  Gº f.(A21)  Method #3  G  =  n p  G f  (products)  n r  G f  (reactants)  Products - reactants because it is a state function.  The more negative G, the further to the right the reaction will proceed to reach equilibrium.  The standard free energy of formation for any element in its standard state is 0.  Remember - Spontaneity tells us nothing about rate.  There are tables of  Gº f.(A21)  Method #3  G  =  n p  G f  (products)  n r  G f  (reactants)  Products - reactants because it is a state function.  The more negative G, the further to the right the reaction will proceed to reach equilibrium.  The standard free energy of formation for any element in its standard state is 0.  Remember - Spontaneity tells us nothing about rate.

26. Practice 1.Consider the reaction: 2SO 2 (g) + O 2 (g) 2SO 3 (g) Carried out at 25°C and 1 atm. Calculate  H° &  S° using the data in A21. Find  G° using  G°=  H°-T  S° 2.Using the following data (at 25°C) C diamond (s) + O 2 (g) CO 2 (g)  G° = -397 kJ C graphite (s) + O 2 (g) CO 2 (g)  G° = -394 kJ calculate  G° for the reaction: C diamond (s) C graphite (s) 3. Calculate  G° for the reaction using data on A21 2CH 3 OH(g) + 3O 2 (g) 2CO 2 (g) + 4H 2 O(g) 1.Consider the reaction: 2SO 2 (g) + O 2 (g) 2SO 3 (g) Carried out at 25°C and 1 atm. Calculate  H° &  S° using the data in A21. Find  G° using  G°=  H°-T  S° 2.Using the following data (at 25°C) C diamond (s) + O 2 (g) CO 2 (g)  G° = -397 kJ C graphite (s) + O 2 (g) CO 2 (g)  G° = -394 kJ calculate  G° for the reaction: C diamond (s) C graphite (s) 3. Calculate  G° for the reaction using data on A21 2CH 3 OH(g) + 3O 2 (g) 2CO 2 (g) + 4H 2 O(g)

27. 16.7 Free energy and Pressure   G =  Gº +RTln(Q) where Q is the reaction quotients (P of the products /P of the reactants).  CO(g) + 2H 2 (g)  CH 3 OH(l)  Would the reaction be spontaneous at 25ºC with the H 2 pressure of 5.0 atm and the CO pressure of 3.0 atm?   Gº f CH 3 OH(l) = -166 kJ   Gº f CO(g) = -137 kJ  Gº f H 2 (g) = 0 kJ   G =  Gº +RTln(Q) where Q is the reaction quotients (P of the products /P of the reactants).  CO(g) + 2H 2 (g)  CH 3 OH(l)  Would the reaction be spontaneous at 25ºC with the H 2 pressure of 5.0 atm and the CO pressure of 3.0 atm?   Gº f CH 3 OH(l) = -166 kJ   Gº f CO(g) = -137 kJ  Gº f H 2 (g) = 0 kJ

28. 16.8 Free Energy and Equilibrium   G tells us spontaneity at current conditions. When will it stop?  It will go to the lowest possible free energy which may be an equilibrium.  At equilibrium  G = 0, Q = K   Gº = -RTlnK from [  G =  Gº + RTlnK]   G tells us spontaneity at current conditions. When will it stop?  It will go to the lowest possible free energy which may be an equilibrium.  At equilibrium  G = 0, Q = K   Gº = -RTlnK from [  G =  Gº + RTlnK]  Gº = 0 < 0 > 0 K = 1 > 1 < 1

29. Temperature Dependence of K   Gº= -RTlnK =  Hº - T  Sº  ln(K) =  Hº/R(1/T)+  Sº/R  A straight line of lnK vs 1/T   Gº= -RTlnK =  Hº - T  Sº  ln(K) =  Hº/R(1/T)+  Sº/R  A straight line of lnK vs 1/T

30. 16.9 Free Energy and Work  “Free energy” is energy “free” to do work.  The maximum amount of work possible at a given temperature and pressure.  Never really achieved because some of the free energy is changed to heat during a change, so it can’t be used to do work.  “Free energy” is energy “free” to do work.  The maximum amount of work possible at a given temperature and pressure.  Never really achieved because some of the free energy is changed to heat during a change, so it can’t be used to do work.