CHEM 0120 1st Edition Lecture 5Outline of Current Lecture I. Expressing Rates of ReactionsII. Integrated Rate LawsIII. Example 13.5IV. Half-LifeV. Elementary Reactions Current LectureI. Expressing rates of reactions aA+bB  cC + dD Rate = −1a×Δ[A]Δt = −1b×Δ[B]Δt=1c×Δ[C]Δt=1d×Δ[D]Δt*Rates are expressed as positive quantities Example (Chapter 13 #44): What is the average rate of reaction in this time interval?2FeCl3 + SnCl2  2FeCl2 + SnCl4 *redox reaction* Fe3+ Fe2+ [Fe3+]I = 0.03586M  4 minutes  [Fe3+]f =0.02715M= -Δ[FeCl 3]Δt= −(0.02715 – 0.03586 M)4 minutes=−(−0.00871)4 minutes=0.00218 molL × secondII. Integrated Rate Laws – relate concentration (or change in concentration) to time and also provide an alternative method for determining rate of reactionA. First Order2N2O5 (g)  2NO2 (g) + 02(g)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.Rate = k [N2O5]Rate = −Δ [N 2 O 5 ]Δt k [N2O5] = −Δ [N 2 O 5 ]ΔtGeneral: ln ⁡([A]t[A]o)=−kt Applied: ln ⁡([N 2 O 5]t[N 2O 5]o)=−ktLinear: ln[A]t =ln[A]o−kt y = b - mx Slope= −rate cons tant−k*straight line = first order B. Second Order1[A]t=1[A]o+kty = b + mx Slope = rate constantk C. Zero Order[A]t =[A]o−kty = b - mx Slope = −rate constant−kIII. Example 13.5: Using Integrated Rate Law2N2O5 (g)  4NO2 (g) + O2 (g) : first order reaction [N2O5 ]o = 1.65 x 10−2 M[N2O5 ]t = unknownk = 4.80 x 10−4t = 6.00 x 102sln[N2O5 ]t = ln[N2O5 ]o –ktln[N2O5 ]t = ln [1.65 x 10−2 M] – [(4.80 x 10−4) (6.00 x 102s)]ln[N2O5 ]t = -4.393[N2O5 ]t = e−4.393[N2O5 ]t = 0.0123 MIV. Half – Lives (t ½) – time it takes for a reactant to decrease to ½ of its initial valueA. First order reaction [A] t ½ = ½ [A]oln([A]t[A]o)=−ktA1/2[¿]o[¿¿[A]o)=−kt¿ln ¿ln(12)=−kt−0.693=−kt 1/20.693=kt 1/20.693k=t 1/2B. Second Order Reactiont12=1k[A]t *Half-life increasesC. Zero Order Reaction [A]2 k=t 1 /2 *Rate is slowerV. Elementary Reactions – a single molecular event when determining m & n; we use initial rate method, integrated rate method, but not stoichiometric… unless it is an elementary reaction because then the one step is the rate determining step Unimolecular reaction : Rate = k[A]Bimolecular Reaction : Rate = k[A]2Termolecular Reaction : Rate =