showed that doubling the concentration of acetone increased the rate of the

# Showed that doubling the concentration of acetone

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showed that doubling the concentration of acetone increased the rate of the reaction. Varying the rate of [Br 2 ] appeared to have no effect on the rate of reaction as shown when comparing experiment one to experiment two. D. Calculating x: -.004616 = k [.004] x [.8] y [.2] z ←DATA FROM EXP. 1 -.0045 k [.008] x [.8] y [.2] z DATA FROM EXP. 2 1.02578 = .5 x log 1.02578 = x log .5 x = log 1.02578 / log .5 x = -3 Calculating y: -.0045 = k [.008] x [.8] y [.2] z ←DATA FROM EXP. 2 -.001016 k [.008] x [1.6] y [.2] z ←DATA FROM EXP. 4 4.4 = .5 y log 4.4 = y log .5 y = log 4.4 / log .5 y = -2 Calculating z: -.0045 = k [.008] x [.8] y [.2] z ←DATA FROM EXP. 2 -.00198 k [.008] x [.8] y [.4] z ←DATA FROM EXP. 3 .44 = 2 z log .44 = z log 2 z = log .44 / log 2 z = -1 Rate Law from Calculations: Rate = k[Br 2 ] -3 [Acetone] -2 [H + ] Negative exponents in this case do not seem logical given previous knowledge. Assuming that this was a mathematical or calculation error, the exponents are to be made positive. E. Experiment # k value 1 -5.6 x 10 5 M -5 s -1 2 -6.9 x 10 4 M -5 s -1 3 -1.5 x 10 4 M -5 s -1 4 -3.9 x 10 3 M -5 s -1 Experiment One: Given the rate law is: Rate = k[Br 2 ] 3 [Acetone] 2 [H + ] -0.004616 = k [.004] 3 [.8] 2 [.2] k = -563476.56 M -5 s -1 = -5.6 x 10 5 M -5 s -1 Experiment Two: Given the rate law is: Rate = k[Br 2 ] 3 [Acetone] 2 [H + ] -.0045 = k[.008] 3 [.8] 2 [.2] k = -68664.55 M -5 s -1 = -6.9 x 10 4 M -5 s -1  #### You've reached the end of your free preview.

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