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Physical Chemistry 20130410 week 2 Wednesday April 10, 2013 page 1 Competing first order reactions A k1 k2 k -1 k -2 C and A D and C A and D A k ' s are rate constants → → → → D K2 K1 A C ⇌ ⇌ K 1 ,K 2 are equilibrium constants [C] K 1 [A] [C] = = K 2 [D] [D] [A] also k 1 [C] = k 2 [D] since dividing the formula for [C] by the formula for [D] for rate of reaction simplifies to k1/k2. If k1/k2 >>1 then [C]>>[D] so C is favored kinetically. If K1/K2<<1 then [D]>>[D] so D is favored thermodynamically. See handout for 1,3 butadiene. HBr is a very strong acid. The intermediate of a reaction is always at the bottom of the dip. 1,4 addition is favored thermodynamically because it has lower energy. 1,2 addition is favored kinetically because the hump isn’t as high. determination of rate laws -has to be done experimentally r=k[A]α[B]β[L]λ How do we determine values for α, β, λ, and k? 1. half-life method rate = k[A]n a. If n=1 then t1/2 is independent of [A]0 b. If n≠1 then t 1/2 = 2n-1 -1 1 t 1/2 = (n-1)k [A]n-1 0 logt 1/2 =log 2n-1 -1 (n-1)[A]n-1 0 k take log 2n-1 -1 -(n-1)log[A]0 (n-1)k ka=k A Graph logt 1/2 against log[A]0 and the slope is -(n-1)=1-n and log 2n-1 -1 is the y intercept. ka=k A (n-1)k That’s only for n≠1. Start with concentration [A]0 and wait until [A] is [A]0/2. The amount of time elapsed is the half life. This is where the data comes from for the graph to determine rate. The fractional life method (tα) Define: tα = time required for [A]0 to drop to α[A]0 so for t1/2 α=1/2. α=1/2 for half life 0<α<1 If r=k[A]n logt ∝ = α= [A] [A]0 tα= -lnα k log∝1-n -1 -(n-1)log[A]0 (n-1)k n≠1 n≠1 t α =fractional life α usually set to 0.75 2. Powell plot r=k[A]n α is dimensionless parameter α= For n=1 ϕ = kt ϕ=fee=k[A]n-1 0 t ϕ is dimensionless [A] ( ) [A]0 (α)1-n-1 = (n-1)ϕ lnα = -kt = -ϕ [A] [A]0 1-n =1+[A]n-1 0 (n-1)kt n≠1 n≠1 n=1 ϕ = k[A]0n-1t logϕ = logk[A]0n-1+logt logϕ-logt = logk[A]0n-1 right side is constant so left side is constant Slide plot of data left and right over master plot to get a match. 3. initial rate method r0 = k[A]0α[B]0β[C]0λ A+B+C-> product(s) # is experiment number Do two experiments, with [B]0 and [C]0 the same in both but different [A]0 #2 [B]#1 0 =[B]0 #2 [C]#1 0 = [C]0 #2 [A]#1 0 ≠[A]0 (r0 )#1 k[A]α0 [B]β0 [C]λ0 experiment 1 Divide the rate laws of the two experiments: = = (r0 )#2 k[A]α [B]β [C]λ experiment 2 0 0 0 That gives: (r0 )#1 =([A]0 )α (r0 )#2 where [A]0 is a ratio of the two different [A]0