1Week 9 CHEM 1310 - Sections L and M 1Exam #2 Results Class Average = 77 (Great Job!)Week 9 CHEM 1310 - Sections L and M 2Exam #2 Results 4 Perfect Scores!2Week 9 CHEM 1310 - Sections L and M 3Ch 6: Chemical Equilibrium What is Equilibrium? Equilibrium Constant, K Equilibrium Expressions Involving Pressures Activity Heterogeneous Equilibria Applications of Equilibrium Constant Solving Equilibrium Problems Le Chatelier’s Principle - very important Equilibria Involving Real GasesWeek 9 CHEM 1310 - Sections L and M 4What is Equilibrium?Equilibrium is the phenomenon that occurs whenthe rate of the forward reaction equalsthe rate of the reverse reaction.3Week 9 CHEM 1310 - Sections L and M 5What is Equilibrium?ExampleH2(g) + I2(g) 2 HI(g)Forward RxnReverse RxnForward Rxn: Product = HIReverse Rxn: Products = H2 and I2 At equilibrium, the concentrations of all reactants and products remain constant with time. [HI] = constant; [H2] = constant; [I2] = constantWeek 9 CHEM 1310 - Sections L and M 6What is Equilibrium?Notice how the concentrations of products for the forward and reverse reactions are not necessarily equalat equilibrium!4Week 9 CHEM 1310 - Sections L and M 7What does equilibrium look like in a chemical system?N2O4 2 NO2colorlessbrownEquilibriumClosed system reaches the same equilibrium concentrationswhether the reaction starts with the N2O4 or the NO2!Equilibrium CharacteristicsWeek 9 CHEM 1310 - Sections L and M 8The Equilibrium ConstantaA + bB cC + dDEquilibriumConstantK = [C]c x [D]d [A]a x [B]b CharacteristicsExponents are coefficients from balanced chemical equation.Units for K will vary depending upon coefficients.5Week 9 CHEM 1310 - Sections L and M 9The Equilibrium ConstantcC + dD aA + bBEquilibriumConstantK = [C]c x [D]d [A]a x [B]b CharacteristicsReversing the reactants and products inverts the equilibriumexpression. Thus,Kforward = 1KreverseWeek 9 CHEM 1310 - Sections L and M 10Law of Mass ActionK is constant despite different initial and equilibriumconcentrations of reactants and products!6Week 9 CHEM 1310 - Sections L and M 11Equilibrium equations can be reversed, scaled orcombined.aA + bB cC + dDForward:K1 = [C]c x [D]d[A]a x [B]bK2 = [C]c x [D]d[A]a x [B]bK1K2Reverse:cC + dD aA + bBBy defn: K1 x K2 = 1Manipulation of Equilibrium EqnsWeek 9 CHEM 1310 - Sections L and M 12Equilibrium equations can be reversed, scaled orcombined.PCl3 + Cl2 PCl5K1 = [PCl5][PCl3] x [Cl2]K2 = KScaled:ExampleK2 = (K1)22 PCl3 + 2 Cl2 2 PCl5K[PCl5]2[PCl3]2 x [Cl2]2Manipulation of Equilibrium EqnsWhen stoichiometry is scaled, the resulting K is raised to the power of the scale factor7Week 9 CHEM 1310 - Sections L and M 13Equilibrium equations can be reversed, scaled orcombined.K1K3 = Equation #1Equation #2-Equation #3K2K1K2K1K3 = Equation #1Equation #2+Equation #3K2K1x K2SubtractionAdditionSubtraction of Equilibrium EqnsWeek 9 CHEM 1310 - Sections L and M 14What is the relationship between Q and K?aA + bB cC + dD Reaction Quotient vs. Equilibrium ConstantQ = [C]c x [D]d[A]a x [B]bHolds whether at equilibrium or not!K = [C]c x [D]d[A]a x [B]bHolds at equilibrium only!K and the Reaction Quotient, Q8Week 9 CHEM 1310 - Sections L and M 15What is the relationship between Q and K?aA + bB cC + dDWhen Q = K = [C]c x [D]d[A]a x [B]bEquilibrium occurs Q vs KWeek 9 CHEM 1310 - Sections L and M 16What is the relationship between Q and K?aA + bB cC + dDWhen Q = K = [C]c x [D]d[A]a x [B]bEquilibrium occurs Q < KWHEN[A] and [B] >>> [C] and [D]Forward rxn proceedsQ vs K9Week 9 CHEM 1310 - Sections L and M 17What is the relationship between Q and K?aA + bB cC + dDWhen Q = K = [C]c x [D]d[A]a x [B]bEquilibrium occurs Q > KReverse rxn proceedsWHEN [C] and [D] >>> [A] and [B] Q vs KWeek 9 CHEM 1310 - Sections L and M 18Thus, knowing K and calculating Q for any given statehelps us predict which way a chemical reactionwill proceed!aA + bB cC + dDWhen Q < K reaction proceeds to the rightWhen Q = K equilibrium occursWhen Q > K reaction proceeds to the leftQ vs K10Week 9 CHEM 1310 - Sections L and M 19Equilibrium equations for gaseous reactions can bewritten in terms of concentrations orpartial pressures.Why? Recall… PV = nRTP = P = M (RT) nVRTPressure is proportional to molar concentration. Equilibrium Equations for GasesWeek 9 CHEM 1310 - Sections L and M 20Equilibrium and Partial PressureEquilibrium expressions can be written in terms of the partial pressures of the gases instead of theirmolar concentrationsN2(g) + 3H2(g) 2NH3(g)K = P2NH3PN2P3H2xIn text, Kp denotesequilibrium constantexpressed in termsof partial pressures11Week 9 CHEM 1310 - Sections L and M 21How can we express the equilibrium constant whenthe reactants and products are in different phases?Si3N4(s) + 4 O2(g) 3 SiO2(s) + 2 N2O(g)Rule #1. Express gases as partial pressuresRule #2. Express solute in solution as molar conc.Rule #3. Express pure solids/liquids as “1”.Rule #4. Products multiplied in the numerator reactants multiplied in the denominatorHeterogeneous EquilibriaWeek 9 CHEM 1310 - Sections L and M 22How can we express the equilibrium constant whenthe reactants and products are in different phases?Si3N4(s) + 4 O2(g) 3 SiO2(s) + 2 N2O(g)Rule #1. Express gases as partial pressuresRule #2. Express solute in solution as molar conc.Rule #3. Express pure solids/liquids as “1”.Rule #4. Products multiplied in the numerator reactants multiplied in the denominatorK =N2OP2x 13O2P4x 1K =N2OP2O2P4Heterogeneous Equilibria12Week 9 CHEM 1310 - Sections L and M 23• Know how to write equilibrium expressions• Know how to calculate K and mathematicallymanipulate K• Be able to calculate Q (via conc or partialpressures) and relate Q to K• Be able to calculate K for gases in equilibrium• Know how to express heterogeneous equilibriaWhat To Study and Know…Week 9 CHEM 1310 - Sections L and M 24[1] 3.5 x 1025[2] 7.0 x 10-252 SO2 + O22SO3K = 7.0 x 1025Calculate K for SO3SO2 + 0.5 O2[3] 1.2 x 10-13[4] 1.4 x 10-26PRS Question13Week 9 CHEM 1310 - Sections L and M 252 SO2 + O22SO3K = 7.0 x 1025Calculate K for SO3SO2 + 0.5 O2Kreverse = 1KforwardTo solve this problem:1st:2nd: Molar ratio is half, so take the square root of KreverseKreverse = 1.4 x 10-26K = 1.2 x 10-13Answer = #3PRS QuestionWeek 9 CHEM 1310 - Sections L and M 26Which reaction