1. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chemistry FIFTH EDITION by Steven S. Zumdahl University of Illinois

2. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 2 Chemistry FIFTH EDITION Chapter 16 Spontaneity, Entropy, and Free Energy

3. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 3 Section 16.5 ENTROPY CHANGES in CHEMICAL REACTIONS Entropy changes in surroundings are determined by the heat flow that occurs as the reaction takes place. Entropy changes in system (reactants & products) are determined by positional probability.

4. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 4  S system When a reaction involves gaseous molecules, the change in positional entropy is dominated by the relative numbers of molecules of gaseous reactants and products. If the # of molecules increases,  S system will be positive for the reaction. Homework: # 33 - 35

5. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 5 The Third Law of Thermodynamics... the entropy of a perfect crystal at 0 K is zero. Because S is explicitly known (= 0) at 0 K, S values at other temps can be calculated. Read pages 763-764. Perfect crystal is unattainable ideal!

6. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 6 Figure 16.5Entropy Perfect crystal at 0 K. Temp. above 0 K S = 0 S > 0 STANDARD ENTROPY VALUES, S°: APPENDIX 4 ON PAGE A19 Increasing Temp.  random vibrational motion increases;  Disorder within the crystal increases.

7. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 7  S ° reaction Entropy change for a chemical reaction can be calculated.  S ° reaction =  n p S ° products -  n r S ° reactants Entropy is a state function. It is not pathway- dependent. Entropy is an extensive property.

8. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 8 Figure 16.6 H 2 O Molecule Water has higher entropy than simple diatomic molecule like H 2 with fewer possible motions. The more complex the molecule, the higher the standard entropy value. Go to Sample Exercise 16.8 On pages 809 – 810. Let’s do: # 37, 39, 41, 43

9. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 9 Section 16.6 STANDARD FREE ENERGY CHANGE  G ° ( for chemical rxns.) THE CHANGE IN FREE ENERGY THAT WILL OCCUR IF THE REACTANTS IN THEIR STANDARD STATES ARE CONVERTED TO THE PRODUCTS IN THEIR STANDARD STATES.

10. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 10  G° cannot be measured directly. We have no instrument that measures free energy! We calculate it from other measured values. The more negative the  G , the further the reaction will go to the right to reach equilibrium.

11. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 11 The value of  G  tells us nothing about the rate of reaction, only its eventual equilibrium position. For accurate comparison of reaction tendencies, use  G  (same pressure or concentration conditions.)

12. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 12 THREE WAYS TO CALCULATE  G °  G ° =  H   T  S ° Recall  H ° reaction =  n p  H ° f products -  n r  H ° f reactants and  S ° reaction =  n p S ° products -  n r S ° reactants Let’s do # 45

13. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 13 2)Use Procedure similar to Hess’s Law for  H ° i.e., manipulating known equations (  G is a state function.)

14. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 14 3) Using Standard Free Energy of Formation  G  f = change in free energy that accompanies the formation of 1 mole of that substance from its constituent elements with all reactants and products in their standard state. Example: 6 C (s) + 6 H 2 (g) + 3 O 2 (g)  C 6 H 12 O 6 (s)

15. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 15 Free Energy Change and Chemical Reactions  G  = standard free energy change that occurs if reactants in their standard state are converted to products in their standard state.  G  =  n p  G f  (products)   n r  G f  (reactants) Std. Free energy of formation of an element in its std. state is ZERO. Let’s do: #49 & 52