3. Wednesday, March 19 th : “A” Day Thursday, March 20 th : “B” Day Agenda  Homework questions/problems?  Section 14.1 Quiz  Begin Section 14.2: “Systems at Equilibrium”  Homework: Practice pg. 504: #1, 2 Practice pg. 506: #1, 2 Concept Review: “Systems at Equilibrium”: #1-14 We will finish section 14.2 next time…

4. Homework Questions/Problems?  Section 14.1 review Pg. 501: #1-5

5. Quiz 14.1: “Reversible Reactions and Equilibrium”  You may use your book and your notes to complete the quiz… Good Luck! (Pick up the notes for today when you turn in your quiz)

6. The Equilibrium Constant, K eq  There is a mathematical relationship between product and reactant concentrations at equilibrium.  Equilibrium constant, K eq : a number that relates the concentrations of starting materials and products of a reversible chemical reaction to one another at a given temperature.

7. The Equilibrium Constant, K eq  Equilibrium constants: do NOT have any units depend on temperature apply only to systems in equilibrium must be found experimentally or from tables

8. ** Rules for determining K eq ** 1.Write a balanced chemical equation. Make sure that the reaction is at equilibrium before you write a chemical equation. 2.Write an equilibrium expression. To write the expression, put the product concentrations in the numerator and the reactant concentrations in the denominator. The concentration of any solid or a pure liquid that takes part in the reaction is left out. For a reaction occurring in aqueous solution, water is omitted.

9. ** Rules for determining K eq ** 3.Complete the equilibrium expression Raise each substance’s concentration to the power equal to the substance’s coefficient in the balanced chemical equation.

10. The Equilibrium Constant, K eq  Limestone caverns form as rainwater, slightly acidified by H 3 O +, dissolves calcium carbonate.  The reverse reaction also takes place, depositing calcium carbonate and forming stalactites and stalagmites. CaCO 3 (s) + 2 H 3 O + (aq) Ca 2+ (aq) + CO 2 (g) + 3 H 2 O(l)  When the rates of the forward and reverse reactions become equal, the reaction reaches chemical equilibrium.

11. The Equilibrium Constant, K eq  Write the equilibrium constant expression for limestone reacting with acidified water at 25°C according to the balanced equation: CaCO 3 (s) + 2 H 3 O + (aq) Ca 2+ (aq) + CO 2 (g) + 3 H 2 O(l) (left out) (left out) K eq = [Ca 2+ ] [CO 2 ] [H 3 O + ] 2  K eq for this reaction at 25˚C is 1.4 X 10 -9  The [ ] means “concentration” in M, moles/liter.  Equilibrium constants do not have any units and apply only to systems at equilibrium.

12. Calculating K eq from Concentrations of Reactants and Products ( Sample Problem)  Calculate K eq for the following reaction: Br 2 (g) 2 Br (g) At equilibrium [Br 2 ] = 0.99 M, [Br] = 0.020 M  Write Equilibrium expression: [Br] 2 [Br 2 ]  Plug in values = (0.020) 2 and Solve… 0.99 K eq = 4.0 X 10 -4

13. Calculating K eq from Concentrations of Reactants and Products Sample Problem A, pg. 504  An aqueous solution of carbonic acid reacts to reach equilibrium as described below: : The solution contains the following concentrations: - carbonic acid, 3.3 × 10 −2 mol/L - bicarbonate ion, 1.19 × 10 −4 mol/L - hydronium ion, 1.19 × 10 −4 mol/L  Determine the K eq

14. Sample Problem A, pg. 504, cont. 3.3 X 10 -2 M 1.19 X 10 -4 M 1.19 X 10 -4 M K eq = [HCO 3 - ] [H 3 O + ] [H 2 CO 3 ]  Plug in values: K eq = (1.19 X 10 -4 ) (1.19 X 10 -4 ) (3.3 X 10 -2 ) = 1.42 X 10 -8 3.3 X 10 -2 K eq = 4.3 X 10 -7

15. ** K eq Shows if the Reaction is Favorable **  If K eq is small, reaction favors reactants. This is called an “unfavorable” reaction.  If K eq is large, reaction favors products. This is called a “favorable” reaction.  If K eq =1, products and reactants are equal.

16. K eq Shows if the Reaction is Favorable  The synthesis of ammonia is very favorable at 25°C and has a large K eq value.

17. K eq Shows if the Reaction is Favorable  However, the reaction of oxygen and nitrogen to give nitrogen monoxide is not favorable at 25°C.  It’s a good thing that this reaction is not favorable or we would have trouble breathing! K eq = [NO] 2 = 4.5 X 10 -31 at 25˚C [N 2 ] [O 2 ]

18. Calculating Concentrations of Products and Reactants from K eq ; Sample Problem B, pg. 506  K eq for the equilibrium below is 1.8 × 10 −5 at a temperature of 25°C. Calculate [NH 4 + ] when [NH 3 ] = 6.82 × 10 −3  Write equilibrium expression: K eq = [NH 4 + ] [OH - ] [NH 3 ]  NH 4 + and OH - ions are produced in equal numbers, so [NH 4 + ] = [OH - ] = X 1.8 X 10 -5 = ___X 2 ___ 6.82 X 10 -3 [NH 4 + ] = 3.5 X 10 -4

19.  Find [H 2 ] equilibrium at 700K when [CH 3 OH] = 0.25, [CO] = 0.0098, K eq = 290 CO (g) + 2 H 2 (g) CH 3 OH (g)  Write equilibrium expression: K eq = [CH 3 OH] [CO] [H 2 ] 2  Plug in values: 290 = 0.25__ (.0098) [H 2 ] 2  Cross multiply: 2.8 [H 2 ] 2 = 0.25 Divide each side by 2.8: [H 2 ] 2 =.089 [H 2 ] = 0.30 Calculating Concentrations of Products and Reactants from K eq ; Additional Example

20. Homework  Practice pg. 504: #1, 2  Practice pg. 506: #1, 2  Concept Review: “Systems at Equilibrium”: #1-14  Use your time wisely….the concept review will be due before you know it! We will finish section 14.2 next time…