1. Slide 1 of 33 Chapter 15: Principles of Chemical Equilibrium

2. Slide 2 of 33 Contents 15-1Dynamic Equilibrium 15-2The Equilibrium Constant Expression 15-3Relationships Involving Equilibrium Constants 15-4The Magnitude of an Equilibrium Constant 15-5The Reaction Quotient, Q: Predicting the Direction of a Net Change 15-6Altering Equilibrium Conditions: Le Châtelier’s Principle 15-7Equilibrium Calculations: Some Illustrative Examples

3. Slide 3 of 33 15-1 Dynamic Equilibrium  Equilibrium – two opposing processes taking place at equal rates. H 2 O(l) H 2 O(g) I 2 (H 2 O) I 2 (CCl 4 ) NaCl(s) NaCl(aq) H2OH2O CO(g) + 2 H 2 (g) CH 3 OH(g)

4. Slide 4 of 33 Dynamic Equilibrium

5. Slide 5 of 33 15-2 The Equilibrium Constant Expression  Methanol synthesis is a reversible reaction. CO(g) + 2 H 2 (g) CH 3 OH(g) k1k1 k -1 CH 3 OH(g) CO(g) + 2 H 2 (g) CO(g) + 2 H 2 (g) CH 3 OH(g) k1k1 k -1

6. Slide 6 of 33 Three Approaches to the Equilibrium

7. Slide 7 of 33 Three Approaches to Equilibrium

8. Slide 8 of 33 Three Approaches to Equilibrium CO(g) + 2 H 2 (g) CH 3 OH(g) k1k1 k -1

9. Slide 9 of 33 The Equilibrium Constant Expression Forward:CO(g) + 2 H 2 (g) → CH 3 OH(g) Reverse:CH 3 OH(g) → CO(g) + 2 H 2 (g) At Equilibrium: R fwrd = k 1 [CO][H 2 ] 2 R rvrs = k -1 [CH 3 OH] R fwrd = R rvrs k 1 [CO][H 2 ] 2 = k -1 [CH 3 OH] [CH 3 OH] [CO][H 2 ] 2 = k1k1 k -1 = K c CO(g) + 2 H 2 (g) CH 3 OH(g) k 1 k -1 k1k1

10. Slide 10 of 33 General Expressions a A + b B …. → g G + h H …. Equilibrium constant = K c = [A] m [B] n …. [G] g [H] h ….

11. Slide 11 of 33 15-3 Relationships Involving the Equilibrium Constant  Reversing an equation causes inversion of K.  Multiplying by coefficients by a common factor raises the equilibrium constant to the corresponding power.  Dividing the coefficients by a common factor causes the equilibrium constant to be taken to that root.

12. Slide 12 of 33 Combining Equilibrium Constant Expressions N 2 O(g) + ½O 2 2 NO(g)K c = ? Kc=Kc= [N 2 O][O 2 ] ½ [NO] 2 = [N 2 ][O 2 ] ½ [N 2 O][N 2 ][O 2 ] [NO] 2 K c(2) 1 K c(3) = = 1.7  10 -13 [N 2 ][O 2 ] [NO] 2 = [N 2 ][O 2 ] ½ [N 2 O] = N 2 (g) + ½O 2 N 2 O(g)K c(2) = 2.7  10 -18 N 2 (g) + O 2 2 NO(g)K c(3) = 4.7  10 -31

13. Slide 13 of 33 K P = K c (RT) Δn :

14. Slide 14 of 33 Pure Liquids and Solids  Equilibrium constant expressions do not contain concentration terms for solid or liquid phases of a single component (that is, pure solids or liquids). K c = [H 2 O] 2 [CO][H 2 ] C(s) + H 2 O(g) CO(g) + H 2 (g)

15. Slide 15 of 33  Worked Examples Follow:

16. Slide 16 of 33 15-1 Practice Example B

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19. Slide 19 of 33  CRS Questions Follow:

20. Slide 20 of 33 H2H2 CO CH 4 = H 2 O time moles of substance 0 1 2 3 Which of the following statements is correct? 1.At equilibrium the reaction stops. 2. At equilibrium the rate constants for the forward and reverse reactions are equal. 3. At equilibrium the rates of the forward and reverse reactions are equal. 4. At equilibrium the rates of the forward and reverse reactions are zero.

21. Slide 21 of 33 H2H2 CO CH 4 = H 2 O time moles of substance 0 1 2 3 Which of the following statements is correct? 1.At equilibrium the reaction stops. 2. At equilibrium the rate constants for the forward and reverse reactions are equal. 3. At equilibrium the rates of the forward and reverse reactions are equal. 4. At equilibrium the rates of the forward and reverse reactions are zero.