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Chapter 15: Principles of Chemical Equilibrium

Chapter 15: Principles of Chemical Equilibrium. Contents. 15-1 Dynamic Equilibrium 15-2 The Equilibrium Constant Expression 15-3 Relationships Involving Equilibrium Constants 15-4 The Magnitude of an Equilibrium Constant

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Chapter 15: Principles of Chemical Equilibrium

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  1. Chapter 15: Principles of Chemical Equilibrium

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

  3. H2O NaCl(s) NaCl(aq) CO(g) + 2 H2(g) CH3OH(g) 15-1 Dynamic Equilibrium • Equilibrium – two opposing processes taking place at equal rates. H2O(l) H2O(g) I2(H2O) I2(CCl4)

  4. Dynamic Equilibrium

  5. 15-2 The Equilibrium Constant Expression • Methanol synthesis is a reversible reaction. k1 CO(g) + 2 H2(g) CH3OH(g) k-1 CH3OH(g) CO(g) + 2 H2(g) k1 CO(g) + 2 H2(g) CH3OH(g) k-1

  6. Three Approaches to the Equilibrium

  7. Three Approaches to Equilibrium

  8. Three Approaches to Equilibrium k1 CO(g) + 2 H2(g) CH3OH(g) k-1

  9. k1 CO(g) + 2 H2(g) CH3OH(g) k-1 k1 [CH3OH] = k-1 [CO][H2]2 The Equilibrium Constant Expression k1 Forward: CO(g) + 2 H2(g) → CH3OH(g) Rfwrd = k1[CO][H2]2 k-1 Reverse: CH3OH(g) → CO(g) + 2 H2(g) Rrvrs = k-1[CH3OH] At Equilibrium: Rfwrd = Rrvrs k1[CO][H2]2 = k-1[CH3OH] = Kc

  10. General Expressions a A + b B …. → g G + h H …. [G]g[H]h …. Equilibrium constant = Kc= [A]m[B]n ….

  11. 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. [N2O] = N2(g) + ½O2 N2O(g) Kc(2)= 2.710-18 [N2][O2]½ [NO]2 N2(g) + O2 2 NO(g) Kc(3)= 4.710-31 = [N2][O2] [NO]2 [NO]2 [N2][O2]½ 1 Kc= = Kc(3) = [N2O][O2]½ [N2][O2] [N2O] Kc(2) Combining Equilibrium Constant Expressions N2O(g) + ½O2 2 NO(g) Kc= ? = 1.710-13

  13. KP = Kc(RT)Δn :

  14. C(s) + H2O(g) CO(g) + H2(g) [CO][H2] Kc = [H2O]2 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).

  15. Worked Examples Follow:

  16. 15-1 Practice Example B

  17. CRS Questions Follow:

  18. 3 2 H2 moles of substance 1 CO CH4 = H2O 0 time Which of the following statements is correct? • 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.

  19. 3 2 H2 moles of substance 1 CO CH4 = H2O 0 time Which of the following statements is correct? • 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.

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