Presentation on theme: "CHEMICAL EQUILIBRIUM Unit 11, Part II"— Presentation transcript:
1 CHEMICAL EQUILIBRIUM Unit 11, Part II 4/15/2017CHEMICAL EQUILIBRIUM Unit 11, Part IIPb2+(aq) Cl–(aq) PbCl2(s)Dr. Mihelcic AP Chemistry 11
2 What is equilibrium?Definition (dictionary.com): a state of rest or balance due to the equal action of opposing forcesChemical Equilibrium: A process where a forward and reverse reaction occur at equal ratesNot all chemical reactions are reversible!!!
3 General Characteristics of Equilibrium DYNAMIC (in constant motion)REVERSIBLEcan be approached from either direction (reaction can run in the forward direction or the reverse direction)Cobalt Chloride Complex EquilibriumPink to blueCo(H2O)6Cl2 Co(H2O)4Cl H2OBlue to pinkCo(H2O)4Cl H2O Co(H2O)6Cl2
4 Characteristics of Dynamic Equilibrium Fe SCN FeSCN2++After a period of time, the concentrations of reactants and products are constant.The forward and reverse reactions continue after equilibrium is attained.
5 Examples of Chemical Equilibria Phase changes such as H2O(s) H2O(liq)
6 Examples of Chemical Equilibria Formation of stalactites and stalagmitesCaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq)
7 Graphing Dynamic Equilibrium Equilibrium achievedProduct conc. increases and then becomes constant at equilibriumEquilibrium achieved when product and reactant concentrations remain constant!!Reactant conc. declines and then becomes constant at equilibrium
8 The Equilibrium Expression, Keq Keq = equilibrium constant (for a given T)Brackets "[ ]" = concentration (molarity)"a, b, c, and d" = coefficients from balanced equationThe "c" in Kc = concentration(Kc = a special Keq based on concentration)
9 The Equilibrium Constant for the Synthesis of HI Note that Equilibrium Constants have NO units!!Equilibrium achievedIn the equilibrium region
10 SOLIDS – (s) after the formula There are two cases when a species is not shown in the equilibrium expression:#1:SOLIDS – (s) after the formula#2:pure LIQUIDS – (l) after the formula
11 Writing Equilibrium Expressions Solids and liquids NEVER appear in equilibrium expressions.NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
12 Writing Equilibrium Expressions Solids and liquids NEVER appear in equilibrium expressions.S(s) + O2(g) SO2(g)Burning Sulfur in Oxygen
13 Examples of Equilibrium Expressions: The oxidation-reduction reaction occurring between iron(III) chloride and tin(II) chloride:2 FeCl3 (aq) + SnCl2 (aq) 2 FeCl2 (aq) + SnCl4 (aq)The replacement of silver ions by copper:Cu (s) + 2 Ag+ (aq) Cu2+ (aq) + 2 Ag (s)
14 Equilibrium Constants K is independent of:Original amounts of reactants or productsSize of containerPressureK is dependent on:TemperatureStates of reactants and products
15 Can tell if a reaction is product-favored or reactant-favored The Meaning of KCan tell if a reaction is product-favored or reactant-favoredFor N2(g) H2(g) 2 NH3(g)If K is large, the conc. of products is much greater than that of the reactants at equilibrium.The reaction is strongly product-favored(right side of equation is favored)
16 The Meaning of KFor AgCl(s) Ag+(aq) + Cl-(aq)Kc = [Ag+] [Cl-] = 1.8 x 10-5If K is small, the conc. of the products is much less than that of the reactants at equilibrium.The reaction is strongly reactant-favored (left side of equation is favored).AgCl(s) Ag+(aq) + Cl-(aq)is reactant-favored.
17 Product- or Reactant Favored 4/15/2017Product- or Reactant FavoredProduct-favoredK >> 1Reactant-favoredK << 1Dr. Mihelcic AP Chemistry 1
18 Determining an Equilibrium Constant if all Concentrations are known: Kc = [ NH3]2[N2] [ H2]3Can substitute numbers into concentrations; if all concentrations = 2.0M,Then Kc = (2)2 \ (2) x (2)3 = ¼ or .25(no units on K)
19 Example: For NH4Cl(s) NH3(g) + HCl(g) at 500oC, at equilibrium there are 2.0 mol of ammonia, 2.0 mol of hydrochloric acid and 1.0 mol of ammonium chloride present in a 5.0 L container. Calculate the equilibrium constant of the system at this temperature.
20 Example:For the system 2SO3(g) 2SO2(g) + O2(g) at a given temperature the equilibrium concentrations are [SO2] = [O2] = 0.10M and [SO3] = 0.20M. Calculate the Kc at this temperature.
21 Example: For N2(g) + O2(g) 2NO(g) Kc = 1.0 x 10-30 at 25°C Calculate the equilibrium concentration of NO(g) if at equilibrium the concentration of N2(g) is 0.04 M and that of O2 is 0.01 M.
22 LE CHATELIER’S PRINCIPLE “If a system at equilibrium is stressed, the system tends to shift its equilibrium position to counter the effect of the stress.”
23 How does a “stress” influence equilibrium? The impact of addition of reactantson reaction rate
25 The “stresses” (factors) that can cause changes at equilibrium Changes in amount of speciesA. Add reactant; system shifts to right (produces more products)B. Add product; system shifts to left (produces more reactants)C. Remove reactant; system shifts to leftD. Remove product; system shifts to right
26 CH4(g) + 2S2(g) CS2(g) + 2H2S(g) Example: Predict the direction of shift of the following concentration changes on the reaction:CH4(g) + 2S2(g) CS2(g) + 2H2S(g)(A) Some S2(g) is added.(B) Some CS2(g) is added.(C) Some H2S(g) is removed.(D) Some argon gas (an inert gas) is added.
27 Changes in Pressure/Volume If P goes down (same as V goes up), system shifts to increased # of moles of gasIf P goes up (same as V goes down), system shifts to decreased # of moles of gas
28 Changes in Temperature Write heat as a product (exothermic) or reactant (endothermic).System shifts to get rid of added heat: will shiftLEFT for exo reactions as T goes upRIGHT for endo reactions as T goes up
29 Changes in Amount of Pure Solid/Liquid or Inert Substance If a substance is NOT in Keq, changes will have NO EFFECT on equilibrium!
30 (a) CH4(g) + 2S2(g) CS2(g) + 2H2S(g) Example Predict the effect of increasing pressure (decreasing volume) on each of the following reactions.(a) CH4(g) + 2S2(g) CS2(g) + 2H2S(g)(b) H2(g) + Br2(g) 2HBr(g)(c) CO2(g) + C(s) 2CO(g)(d) PCl5(g) PCl3(g) + Cl2(g)
31 How can the reaction be shifted to the right? Le Chatelier ExampleEXAMPLE:CO(g) + 2 H2 (g) <---> CH3OH (g)How can the reaction be shifted to the right?
32 Example 16.17What conditions would be most ideal for shifting the following equilibrium as far to the right as possible?N2(g) + 3H2(g) 2NH3(g) ∆H = negative
33 Haber-Bosch Process for NH3 N2(g) + 3 H2(g) 2 NH3(g) + heatK = 3.5 x 108 at 298 K