CHEM 1212 1nd Edition Lecture 12 Outline of Last Lecture I. Linear formatsII. Half-LifeIII. Collision TheoryOutline of Current Lecture I. Practice ProblemsII. Chemical EquilibriumCurrent Lecture***The ozone of the stratosphere can be decomposed by reaction with nitrogen dioxide (commonly called nitric oxide) from high-flying jet aircraft.O3(g) + NO(g)  NO2(g) + O2 (g)The rate expression is rate = k[O3][NO]. Which of the following mechanisms is consistent with the observed rate expression? A&BA. NO + O3  NO3 + O slow NO3 + O  NO2 + O2fastB. NO + O3  NO2 + O2slow, one stepC. O3  O2 + O slow O + NO  NO2fastD. NO  N + O slow O + O3  2O2fast O2 + N  NO2fastChapter 16 – Chemical Equilibrium- Equilibrium is based on understanding kinetics- The Equilibrium Stateo Consider the reaction A B Ratef = kf[A] Rater = kr[B]o At equilibrium, rate of the forward reaction = the rate of the reverse reaction Kf/kr = [B]/[A]o At equilibrium, the reaction does not stopo In the kinetic region, the system is approaching equilibrium Kf/kr = Kco Equilibrium constant values do not have units Kc = [B]/[A] (ratio of equilibrium concentration of the species)***The reaction below represents a single elementary step in equilibrium:These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.2SO2(g) + O2(g)  2SO3(g)- What is the rate law expression for the forward reaction? Ratef = kf[SO2]2[O2]- What is the rate law expression for the reverse reaction? Rater = kr[SO3]2- What is the value of Kc? Kc = .15 o Kf = 0.15s-1o Kr = 1.00s-1- Chemical equilibrium can be established wither from the forward or the reverse direction- Reconsider this gaseous reaction:o 2SO2(g) + O2(g)  2SO3 (P = MRT)o Kp = PSO32/PSO22 x PO2 atm ([SO3]RT)2/([SO2]RT)2 x ([O2]RT) = Kc x (RT)-1 Kp = Kc(RT)Δn- Δ = (# of moles of gaseous products) – (# of moles of gaseous reactants)o Solids do not show up in equilibrium constant expressions- Meaning of Kco The numerical value of Kc distinguishes whether the reaction (as written) is product favored or reactant favored. ABo What must be true for the value of Kc to be less than one?  Reactant favored Concentration of [A] must be greater than [B]o What must be true in order for Kc to be greater than one?  Product favored Concentration of B must be greater than A- The reaction Quotient, Qo 2SO2(g) + O2(g)  2SO3(g)o Q = (SO3)2/(SO2)2 x (O2)o Kc < Q  reaction favors production of reactants to reach equilibriumo Kc = Q  at equilibriumo Kc > Q  Reaction favors production of products to reach equilibrium- ICE Tables and Kc o PCl5(g)  PCl3(g) + O2(g)PCl5PCl3O2Initial Concentration 1.0 M 0 0Change in Concentration-0.60 M +0.60 M +0.60 MEquilibrium Concentration0.40 M 0.60 M 0.60