Question

Analyze the data and infer the rate order of each reaction with respect to each reactant. Clearly express the rate law for Reaction 1 and Reaction 2. Calculate the rate constant (with the correct units) for Reaction 1 and 2. Please show all work in a highly organized, very legible, easy to follow format.

Answer 1

Reaction 1 CH3CH2 BO + H2O CH3 CH2OH + HBO So kao rate = k [CH₃ CH₂ Br] [#₂0] karate constant for 8 - Trial 1, -2.11x10 .... (i) m = kak & [1.48x10272 (1.50 for Trrial 2, = kx [2.96x1721x (1.57°= 4.22* 168 ... (1) froom, eg (i) and (ii) a (ii) 9 10 & [1.48x1027x [UST KX [2.9681027* (1.5179 = 4.22 x 108 1.48 X a 109 => [] → () - (%2) : x=1 So, noe again froom, Total 3, z = kx [1.48x102] x [3.02] = 4.22x108 ....() From, equation (i) & (ili) xx [1.48x10]* [3.027° 4.22x108 xx [1.4841027² [1.51] y = 2.11 x 108 => [2] = [2] i.yal So, nate onders wn to both CH₃ CH₂ Bn & H₂O are 1. Now, from, eq. (1) 2.11x108 = kx (1.98x10²) (1.51) e ka 9.44x101 M's

Reaction (2) (CH3) 3 C Br + H₂O - (CH3) 3 COH + HBD Trrial 1 mi = kx (1,51x172] [1.56x1523° = 3.89x1o5...) Troial 2 = 6.78*105 ....ci r2 = k [3.02 x 162] [1.56 x 102 Froom eq i) and i) [1-6] *172] 9.39 [3.02x102] - 6.78 => [V2]X = [42] : x=1 Trial 3 mg = [1.51x1ő? [8:12 x 16 71° 3.39x105 .... ) from eq ci) & Ciii) - ]° 1 y=0 So, mate of the reaction is 1 aud, rate onders respectively 1&o w.nit. [CH3)3CBN and H₂O Now, from eq ci), 3.39x105 = kx (1.51x102) x 1 > 2. 24 x 10 3 = k : mate constant = 2.24x108 54 (Aus.)