1. Energy Changes in Chemical Reactions -- Chapter 17 1. First Law of Thermodynamics (Conservation of energy)  E = q + w where, q = heat absorbed by system w = work done on system SYSTEM Surroundings Heat in Heat out Work out Work in Q > 0 Q < 0 W < 0 W > 0 Sign Conventions: Q(+) --> System absorbs heat W (+) --> work done on system E is a “state function” -- independent of pathway, but, q and w are not! (they do depend on pathway) E = internal energy = sum of all potential and kinetic energy of the system:  E = E final - E initial

2. Entropy and Spontaneity Two factors affect any change or reaction (a) Enthalpy (H) (Recall that  H =  E + P  V) exothermic processes (negative  H) tend to be spontaneous, but not always (b) Entropy (S) – degree of disorder or randomness Change in entropy:  S = S final - S initial For a reaction:  S = S products - S reactants Positive  S means: an increase in disorder as reaction proceeds products more disordered (random) than reactants For a reaction, positive  S favors spontaneity (usually small)

3. Entropy and Second Law Entropy (S) increases with the number of energetically equivalent ways to arrange the components of a system to achieve a particular state. Trends: In general: S gas >> S liquid > S solid larger mass => more entropy more complex structure => more entropy more atoms in molecule => more entropy Second Law of Thermodynamics “…for any spontaneous process, the overall entropy of the universe increases…” A spontaneous process can have a negative  S for the system only if the surroundings have a larger positive  S.  S univ =  S sys +  S surr where  S surr = –  H sys T

4. Third Law Third Law of Thermodynamics “…the entropy of a pure crystalline substance equals zero at absolute zero…” S = 0 at T = 0 K (a) Standard Entropy (at 25 ºC) = Sº [Table 17.2, p789] (note the units!) Entropy change for a reaction: (e.g. J/mole K)  Sº =  Sº(products) -  Sº(reactants)

5. Gibbs Free Energy (b) Gibbs Free Energy = G Defined as: G = H - TS –A combination of enthalpy and entropy effects –Related to maximum useful work that system can do For a process (e.g. a reaction),  G =  H - T  S For any spontaneous change,  G is negative!!! (i.e. the free energy must decrease)

6. Standard Free Energy  Gº (at 1 atm) generally used to decide if a reaction is spontaneous (Yes, if negative) Three ways to obtain  Gº for a reaction (a) from  Hº and  Sº requires  Hº f and Sº data for all reactants and products  Gº =  Hº - T  Sº  Hº =   Hº f (products) -   Hº f (reactants)  Sº =  Sº(products) -  Sº(reactants)

7. Determining  Gº Three ways to obtain  Gº for a reaction…. (b) from standard “Free Energies of Formation”  Gº f  Gº =  Gº f (products) -  Gº f (reactants) where  Gº f is the free energy change for the formation of one mole of the compound from its elements, e.g.  Gº f for Al 2 (SO 3 ) 3(s) equals  Gº for the following reaction 2 Al (s) + 3 S (s) + 9/2 O 2(g) --> Al 2 (SO 3 ) 3(s)

8. Manipulating Chemical Equations to Get  Gº If reaction is reversed, change sign of  G°. If reaction is multiplied or divided by a factor, apply same factor to  G°.  G° for overall reaction = sum of  G° values for individual reactions. Problem Given the following thermochemical reactions: (eq 1) C 2 H 4(g) + 3 O 2(g) --> 2 CO 2(g) + 2 H 2 O (l)  G° = -1098.0 kJ/mol (eq 2) C 2 H 5 OH (l) + 3 O 2(g) --> 2 CO 2(g) + 3 H 2 O (l)  G° = -975.2kJ/mol Calculate  G° for the following reaction: C 2 H 4(g) + H 2 O (l) --> C 2 H 5 OH (l)

9. Example Problem, cont. Reverse 2nd reaction to put C 2 H 5 OH on product side then rewrite 1st equation and add them together. (eq 2) 2 CO 2(g) + 3 H 2 O (l) --> C 2 H 5 OH (l) + 3 O 2(g)  H° = + 975.2 kJ/mol (note the sign change!!!) (eq 1) C 2 H 4(g) + 3 O 2(g) --> 2 CO 2(g) + 2 H 2 O (l)  H° = -1098.0 kJ/mol Net: C 2 H 4(g) + H 2 O (l) --> C 2 H 5 OH (l) {note: 3 O 2, 2 CO 2, and 2 H 2 O cancel out}  H° =  H° 1 +  H° 2 = 975.2 + (-1098.0) = -122.8 kJ/mol

10. Effect of Temperature on  G (a) Effect of Temperature on  G  G depends on  H and  S:  G =  H – T  S but  H and  S are relatively independent of temperature, so  G at some temperature T can be estimated.  Gº T ≈  Hº 298 - T  Sº 298

11. Free Energy and Equilibrium (b) for a system at equilibrium: G products = G reactants and  G = 0 since  G =  H - T  S = 0  H = T  S or T =  H/  S Example Problem Given the following, determine the normal boiling point of mercury (Hg).  H vaporization of Hg = 60.7 kJ/mole entropies: liquid Hg: Sº = 76.1 J/mole K gaseous Hg: Sº = 175.0 J/mole K Equilibrium: Hg liq Hg gas  G = 0 so,  H - T  S = 0 or T =  H/  S T = [60.7 x 10 3 J/mole]/[(175.0 - 76.1) J/mol K] = 614 K = 341 ºC

12. Free Energy and Equilibrium Constant (c) Relationship between  Gº and Equilibrium Constant (K) For any chemical system:  G =  Gº + (RT) ln Q If  G is not zero, then the system is not at equilibrium. It will spontaneously shift toward the equilibrium state. At Equilibrium:  G = 0 and Q = K ∴  Gº = -RT ln K for gaseous reactions: K = K p for solution reactions: K = K c {units of  G must match those of R value}

13. Free Energy and Equilibrium (cont.)  Gº = – RT ln K K values can be determined from thermodynamic data ! when K > 1  Gº is negative ∴ spontaneous reactions have large K and negative  Gº values

14. Sample Problem 1. Consider the gas-phase reaction: N 2 O 5(g) + H 2 O (g) --> 2 HNO 3(g) and the following thermodynamic data. (a) Decide whether or not the above reaction is spontaneous at 25 ºC by calculating the value of the appropriate thermodynamic quantity. (b) Calculate the temperature (in ºC) at which the above reaction should have an equilibrium constant (K p ) equal to 1.00. Compound  Hº f (kJ/mole) Sº (J/mole K) N 2 O 5(g) 11.0356 H 2 O (g) - 242189 HNO 3(g) - 174156

15. Sample Problem 1. Consider the gas-phase reaction: N 2 O 5(g) + H 2 O (g) --> 2 HNO 3(g) and the following thermodynamic data. (a) Decide whether or not the above reaction is spontaneous at 25 ºC by calculating the value of the appropriate thermodynamic quantity. (b) Calculate the temperature (in ºC) at which the above reaction should have an equilibrium constant (K p ) equal to 1.00. (a)  G = – 48 kJ/mol, which is negative, so spontaneous. (b) 229 ºC Compound  Hº f (kJ/mole) Sº (J/mole K) N 2 O 5(g) 11.0356 H 2 O (g) - 242189 HNO 3(g) - 174156

16. Sample Problem For a certain reaction,  H ° = -95.2 kJ and  S ° = -157 J/K. Determine  S surr and  S univ (in J/K) for this reaction at 850 K. Based on these results, is the reaction spontaneous at this temperature?

17. Sample Problem For a certain reaction,  H ° = -95.2 kJ and  S ° = -157 J/K. Determine  S surr and  S univ (in J/K) for this reaction at 850 K. Based on these results, is the reaction spontaneous at this temperature? Answer:  S surr = +112 J/K  S univ = -45 J/K Not spontaneous, because  S univ would decrease, which is against the 2 nd Law of Thermodynamics.