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Wednesday, March 19 th : “A” Day Thursday, March 20 th : “B” Day Agenda

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

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Wednesday, March 19 th : “A” Day Thursday, March 20 th : “B” Day Agenda

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  1. Wednesday, March 19th: “A” DayThursday, March 20th: “B” DayAgenda • 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…

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

  3. Quiz14.1: “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)

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

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

  6. ** Rules for determining Keq ** • Write a balanced chemical equation. • Make sure that the reaction is at equilibrium before you write a chemical equation. • 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.

  7. **Rules for determining Keq ** • Complete the equilibrium expression • Raise each substance’s concentration to the power equal to the substance’s coefficient in the balanced chemical equation.

  8. The Equilibrium Constant, Keq • Limestone caverns form as rainwater, slightly acidified by H3O+, dissolves calcium carbonate. • The reverse reaction also takes place, depositing calcium carbonate and forming stalactites and stalagmites. CaCO3(s) + 2 H3O+(aq) Ca2+(aq) + CO2(g) + 3 H2O(l) • When the rates of the forward and reverse reactions become equal, the reaction reaches chemical equilibrium.

  9. The Equilibrium Constant, Keq • Write the equilibrium constant expression for limestone reacting with acidified water at 25°C according to the balanced equation: • CaCO3(s) + 2 H3O+(aq) Ca2+(aq) + CO2(g) + 3 H2O(l) • (left out) (left out) • Keq = [Ca2+] [CO2] • [H3O+]2 • Keq 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.

  10. Calculating Keq from Concentrations of Reactants and Products ( Sample Problem) • Calculate Keq for the following reaction: Br2 (g) 2 Br (g) At equilibrium [Br2] = 0.99 M, [Br] = 0.020 M • Write Equilibrium expression: [Br] 2 [Br2] • Plug in values = (0.020) 2and Solve… 0.99 Keq = 4.0 X 10-4

  11. Calculating Keqfrom 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 Keq

  12. Sample Problem A, pg. 504, cont. 3.3 X 10-2 M 1.19 X 10-4M 1.19 X 10-4M K eq = [HCO3-] [H3O+] [H2CO3] • 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 Keq = 4.3 X 10-7

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

  14. Keq Shows if the Reaction is Favorable • The synthesis of ammonia is very favorable at 25°C and has a large Keq value.

  15. Keq Shows if the Reaction is Favorable • However, the reaction of oxygen and nitrogen to give nitrogen monoxide is not favorable at 25°C. K eq = [NO]2= 4.5 X 10-31at 25˚C [N2] [O2] • It’s a good thing that this reaction is not favorable or we would have trouble breathing!

  16. Calculating Concentrations ofProducts and Reactants from Keq; Sample Problem B, pg. 506 • Keqfor the equilibrium below is 1.8 × 10−5at a temperature of 25°C. Calculate [NH4+] when [NH3] = 6.82 × 10−3 • Write equilibrium expression: K eq = [NH4+] [OH-] [NH3] • NH4+ and OH- ions are produced in equal numbers, so [NH4+] = [OH-] = X 1.8 X 10-5= ___X2 ___ 6.82 X 10-3 [NH4+] = 3.5 X 10-4

  17. Calculating Concentrations ofProducts and Reactants from Keq; Additional Example • Find [H2] equilibrium at 700K when [CH3OH] = 0.25, [CO] = 0.0098, Keq = 290 CO (g) + 2 H2 (g) CH3OH (g) • Write equilibrium expression: Keq = [CH3OH] [CO] [H2]2 • Plug in values: 290 = 0.25__ (.0098)[H2]2 • Cross multiply: 2.8 [H2]2 = 0.25 Divide each side by 2.8: [H2]2 = .089 [H2] = 0.30

  18. 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…

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