Question

26. Questions 26-27 refer to the following information. The reaction of acetone and bromine can be acid catalyzed as shown in the above reaction. The following initial rates of the bromination of acetone using an acid catalyst were measured at various initial concentrations as recorded in the table below. Trial [CH, COCH] Br] [H'] Initial Rate (Ms) catalyst 0.30 0.30 0.30 0.40 0.40 0.0500.050 0.10 0.050 0.050 0.050 0.050 0.050 0.10 0.20 5.7 x10- 5.7 x10 1.2 x104 3.I x10 7.6 × 10-5 4 What is the order with respect to the catalyst for this reaction? (A) Catalysts cannot be assigned an order of reaction. (B) zero order (C) first order (D) second order A new trial is run and the concentrations of the bromine are measured. The results are shown in the table below 27. 0 360. 0.300.300.30 0.15 | -- Based on the concentration of the Br2 at 360. s, what are the concentrations of the CH3COCH3 and H' at 360. s? [CH COCH 0.45 0.30 0.15 0.15 0.45 0.45 0.45 0.15 (D)

Answer 1

Rate law of the above reaction R = K[CH_3 COCH_3]^x [Br_2]^y [H^+]^z where x, y, z are order of reaction with respect to Ch_3 COCH_3, BR_2 and H^+ respectively Trial 4: 3.1 times 10^-4 = K(0.40)^x (0.050)^y (0.20)^z trial 5: 7.6 times 10^-5 = K(0.40)^x (0.050)^y (0.050)^z Trial 5 division Trial 4 we get 7.6 times 10^-5/3.1 times 10^-4 = K(0.40)^x (0.050)^y (0.050)^z /K(0.40)^x (0.050)^y (0.20)^z 1/4 = (1/4)^z or, z = 1 So, order with respect to catalyst: first order Trial 2 division Trial 1 5.7 times 10^-4 /5.7 times 10^-4 = K(0.30)^x (0.10)^y (0.050)^z /K(0.30)^x (0.050)^y (0.050)^z = 2^y = 1 = 2^0 y = 0 Trial 4: Trial 3 3.1 times 10^-4 /1.2 times 10^-4 = K(0.40)^x (0.050)^y (0.20)^z /K(0.30)^x (0.050)^y (0.10)^z = (4/3)^x times 2 (4/3)^x = 3.1/2.4 = (4/3)^1 => x = 1 Rate law R = K[CH_3 COCH_3] [H^+] Time(s): [CH_3 CO CH_3] [H^+] Time(s): [CH_3 COCH_3] [H^+] [-Br_2] Since