1. Do Now: In the last quarter we studied physical and chemical changes… how fast they occur, how far does it go to completion, how changes in conc. and temperature effect the reaction. In this chapter we will study why the change takes place….

2. Ponder … A gas fills its container uniformly. It never spontaneously collects at one end of the container. Sodium explosivley reacts with water producing NaOH and hydrogen gas, but H 2 + NaOH -> no reaction A lump of sugar spontaneously dissolves in a cup of tea but dissolved sugar does not spontaneously reappear in the original form. Iron exposed to water and oxygen forms rust but rust does not spontaneously change to iron. All are spontaneous in one direction are nonspontaneous in the reverse direction

3. Apply thermodynamics To help us predict in which way a reaction will occur spontaneously. Spontaneous – occurs without a continuous input of energy, proceeds without any outside intervention. Can be fast or slow, does not give indication of speed of reaction… what does?

4. Which tends to be spontaneous ? What makes these occurrences spontaneous? Enthalpy (  H) Entropy (  S) - energy flow as heat (const. P)  H > 0 endothermic  H < 0 exothermic - measure of disorder or randomness

5. Entropy S (J/K) - describes the number of ways the particles in a system can be arranged in a given state More arrangements = greater entropy Increase in temperature, increases entropy (more kinetic energy)  S = S final – S initial  S > 0 represents increased randomness or disorder

6. Patterns of Entropy Change For the same or similar substances : S solid < S liquid < S gas solid liquid vapor

7. Patterns of Entropy Change Solution formation typically leads to increased entropy solute solvent solution particles more disordered

8. Determine the sign of  S for the following H 2 O (l) -> H 2 O (g) NaCl (s) -> NaCl (aq) 2A(g) + B(g) -> 2C (g) Which has higher entropy in each pair? 1mol SO 2 (g) or 1mol SO 3 (g) 1mol KBr or 1 mol KCl Seawater at 2 o C or 23 o C Remember… as the system becomes more disordered there are more possible arrangements… the entropy increases … therefore  S would be positive

9. Laws of Thermodynamics 1 st - Energy is neither created nor destroyed 2 nd - The entropy of the universe (system + surroundings) increases during any spontaneous process.  Suniv =  Ssys +  Ssurr > 0 if spontaneous Depends on direction of heat flow

10. Will reaction be spontaneous if?  H 0 exothermic reaction that becomes more disordered… Always!  H > 0,  S < 0 endothermic reaction that becomes more ordered… Never! (but reverse reaction will be) What if the reaction is endothermic, but entropy increases ??? What if the reaction is exothermic, but entropy decreases ??? We need to examine these scenarios more closely

11. So spontaneous if  Suniv =  Ssys +  Ssurr > 0 If system undergoes exothermic reaction H sys 0 If system undergoes endothermic reaction H sys >0, heat is transferred from surroundings…  S surr <0 Ssurr and Hsys have opposite relationship +/- When a reaction takes place, will  Ssurr change more when existing temperature is high or low? Low… so  S surr = -  H sys /T (T in Kelvin) When temp is high, a release in energy from the system to surrounding, won’t effect the entropy of the surroundings as significantly as if the temp was low and energy was released to the surrounding. So Ssurr and T have inverse relationship S  1/T.

12.  Suniv =  Ssys +  Ssurr Becomes  Suniv =  Ssys  Hsys/T Multiply by -T  Suniv = -T  Ssys +  Hsys Gibbs Free Energy,  G =  Suniv =  Hsys -T  Ssys

13. Gibb’s Free Energy  G =  H – T  S Helps us determine if the reaction will proceed spontaneously If  G < 0 (negative) Forward reaction is spontaneous! If  G > 0 (positive) Forward reaction is not spontaneous (but reverse reaction is!) If  G = 0 At equilibrium, and  Ssys =  Hsys/T Term measures useful work obtainable from the system,  G = w

14. Gibbs Free Energy and Temperature  G =  H - T  S HH SS GG Spontaneous? < 0>0 <0 > 0 < 0Yes, at any temperature > 0Not spontaneous at any T Spontaneous at low temperatures Spontaneous at high temperatures ? ?

15. Try This! For a particular reaction,  H rxn = 53 kJ and  S rxn = 115 J/K. Is this process spontaneous a)at 25°C? b)at 250°C? c)At what temperature will this reaction be at equilibrium?

16.  G o =  H o – T  S o  G is a state function (independent of path), so we can use Hess’s Law to determine  G To compare  G (or  H or  S) of different reactions we compare at standard state symbolized by o. Standard state = 1 atm, 25 o C

17. For a given reaction, we found the  H at standard state using  Hº reaction =  H f º (products) -  H f º (reactants) … and a Thermodynamic Properties Table (such as the one in your text book Appendix A19) – We did this in Ch 6- the Thermochemistry chapter

18. We assigned  H f (Heat of formation) = 0 for elements in their most stable state at 1 atm. We will do the same for  G f.  G f (Gibb’s Free Energy of formation) = 0 for elements in their most stable state. What about S f ? Take a look at the Thermodynamic Property Table (appendix A-19 in the back of your book) Sº and  Gº can also be found in the same manner.

19. S f (Entropy of formation) Will have a value > 0 EXCEPT for a perfect crystal at absolute zero, S = 0 (3 rd Law of Thermodynamics)crystalabsolute zero For any other structure or temperature there are different possibilities for arrangement, hence entropy, S>0

20. Determine the entropy change for a phase transition What is the standard molar entropy of vaporization of water at 100.0ºC, given that the standard molar enthalpy of vaporization is 40.7 kJ/mol? Remember this change, although physical, is an example of an equilibrium situation. How are the terms, G,H, and S, related at equilibrium?

21. The reaction is neither spontaneous nor nonspontaneous. The forward and reverse reactions show an equal tendency to occur. At equilibrium, both forward and reverse reactions are occurring, so  G = 0. If  G = 0  H = T  S  S =  H/T = 40.7 kJ/mol / 373 K = 0.109 J/molK

22. Ok… so we should be able to Predict sign of entropy for reaction Compare entropies of substances Relate/solve for H, S, T, and G and determine if a reaction is spontaneous using  G =  H – T  S Find  G o,  H o,and  S o using Thermodynamic Properties (table in back of book) or Hess’s Law and describe equilibrium conditions at standard state Try #27,29,31,33,35,39,51,53,55 in your text book. Check answers. Fix any errors.

23. Now we will use thermodynamic properties to describe equilibrium for a variety of conditions, not just standard state. What if pressure is higher or lower than standard state, 1 atm? It won’t effect the  H, but will it effect  S? As pressure increases, molecules are closer, how will this effect entropy? S high pressure <S low pressure (closer will have fewer arrangements) Can’t use  G o (only good for 1 atm), would need to use  G. G = G o + RTln(P) (this R = 8.314 J/mol K)

24. Consider: 2NO (g) + Cl 2(g) -> 2NOCl (g)  G o = -41.0 kJ/mol So the reaction is spontaneous at standard state(1 atm and 25 o C) because  G<0 What if the reaction occurs at 25 o C, but the following initial pressures of substances were added to the vessel.  G =  n p  G prod -  n r  G react  G = 2G NOCl – 2G NO -G Cl2 Where G NOCl = G o NOCl + RT ln (P NOCl ) G Cl2 = G o Cl2 + RT ln (P Cl2 ) G NO = G o NO + RT ln (P NO ) so…  G = 2[G o NOCl + RT ln (P NOCl )] –[G o Cl2 + RT ln (P Cl2 )] -2[ G o NO + RT ln (P NO )] P NO P Cl2 P NOCl 1 x 10 -5 atm 1 x 10 -2 atm

25. From the last slide…  G = 2[G o NOCl +RTln(P NOCl )] –[G o Cl2 +RTln(P Cl2 )]-2[G o NO +RTln(P NO )] Rearrange the equation a bit…  G=2G o NOCl -G o Cl2 -2G o NO + 2RTln(P NOCl ) –RTln(P Cl2 ) -2RT ln (P NO )  G =  G o reaction +RT [2ln (P NOCl ) – ln (P Cl2 ) - ln (P NO )]  G =  G o reaction +RTln[P 2 NOCl /(P Cl2 P 2 NO )]  G =  G o reaction +RTln Q We can use this to find  G and hence spontaneity for any conditions, not just standard state

26. Lastly, we can relate Gibb’s Free Energy to the equilibrium constant, K Remember, at equilibrium both forward and reverse reactions are occurring, thus  G = 0 and at equilibrium so Q = K Using  G =  G o reaction +RTln Q  =  G o reaction +RTlnK  G o reaction = -RTlnK

27. Try #57,59,61,63,65,67, 74, 76, 97, 98 in your text book. Check answers. I’ll post answers to even problems on my website. Fix any errors. You will hand in the entire assignment on Monday and we can go over any problems that you were not sure of. Tuesday we have our quarterly, Wednesday we are starting Ch. 17 We will have a short assessment on Chapter 16 on Friday 4/8. No makeup. Reference sheet provides the following…  G o reaction = -RTlnK  Hº reaction =  H f º (products) -  H f º (reactants)  Sº reaction =  S f º (products) -  S f º (reactants)  Gº reaction =  G f º (products) -  G f º (reactants)  G o =  H o – T  S o and the value of R = 8.314 J/mol K