Question

You are working in a chemistry laboratory and want to determine the mechanism for the following reaction: CH3CH2CH2Br + -OH → CH3CH2CH2OH + Br- a. Your first step is to determine the rate law for the reaction. In the table below are some of the conditions you will use. Fill out the rest of the table with the missing conditions needed to determine the rate law for the reaction. [CH3CH2CH2Br] mol/L [-OH] mol/L 2.0 2.0 2.0 4.0 b. Copy your (-OH) answers from part a into the table below. You determine that the reaction is first order in CH3CH2CH2Br and first order in OH. Fill out the table with the initial rates that you expect. [CH3CH2CH2Br] mol/L [-OH] mol/L Initial rate mol/(L·min) 2.0 2.0 1.0 2.0 4.0 c. What is the rate law for the reaction? d. From the rate law that you determined, what can you tell about the mechanism of the reaction? e. Draw the mechanism using curved arrows.

Answer 1

The chemical reaction. I an oli oleh CH CH CH BS + OH CH CH CHOH + Br a) o Giren on est [MCH₂ CH₂ Br Imol/ [OH-] molle au To 2.0 . 4.0 4.0 2.0 AN b) Given 0321 10. [CHACH OHB) mol/l [onl] melde. Initial rate moll min) 2.0 2-05 4.0 2.0 1.0 2.0 4.0 2 .0 c) Hwhat is the rate Laut for the reaction? Rate you = K[CH₂ CH₂ CH Br* [OH-] 1 Rate & k ah erh CH Br] [ort 1 O t here. x - 1

Note - To find the value of x and to in 11 Rate lauta RL (Rate) - k52.07 (2.00/= 1.0 equation, where [CH₂ CH CH Br] = [2.0] [OH-] = [2.07 On Similiarly, Re( Rate) - k [2-01"[4.039 Geanation galing where [chet Char] = [2.07 [ou] = [4,0] R₂ (Rate) = t [4.07 2.078-2.0 Bequation where (CH₂ CH CH Br] = [4.0] [h] [2:07 Dividing by and 2 by equation. R K [4.07² [2.07 = 2.0 K 12.07(2077 1.0 HRU) (R2) - K[2.07 (4:07 252.019(2.017 1 2.0 1.0

from here, the value of Calculated a sol and y are 1 y = 1 The rate of reaction is proportional to the concentration of [CH CH CH ₂ Br] and [ot]. This also tells des about how the reaction rate depends on the concentration of each - component that is, the overall wrate law for the reaction. We then compare our experimental rate Lawet with the state Laws predicted by our possible mechanisms, our experimental rate Lawe may agree with the rate Law predicted by a proposed mechanism e) Mechanism slow slet CH-CH-CH₂ - Br + oH CH-CH₂-cthy -ots Intermediate fast step CH-cp-ot & Br

( The structure for the transition state CH halos Transition state Diagram Xo Exothermic Reaction Energy Reaction Coordinate

roll A 3c. initial l.om M om change - Air equilibrium 1-bc Bw Given 1 x = 0.7 x = 0.3 At equilibrium 0.74 o.am Graph 08 0.6+ 0.5t 0.4- 0.3 62 1 2 3 4 5 6 7 8 9 10 Time (minutes)

0.8 fort [e] 0-3 0.2 Oh 1 2 3 4 5 6 7 8 9 10 Time (minutes) A3c ( Reaction) equilibrium comptant - [67- (ko) [A To calculate the value ofk ko [0.97 - 1.0414 [c]=[0-9 [0.7 [A7= [0]

A 30 [A] = 1.2M No, the reaction is not at equilibrium. [6] = 0.5M Reaction Quotient (Q) = [] 3 [A7 = [0.57% = 0.1041 [1:2] & Here & <Kc The reaction will shift in forward direction . because have less products, than system at equilibrium. Note- & (Reaction Quotient) = The reaction quotient (Q) is a measure of the relative amounts of products and greactants present in a reaction at a given time.