The reaction of tert-butyl-bromide (CH3)3CBr with water is represented by the eq

Question

The reaction of tert-butyl-bromide (CH3)3CBr with water is represented by the equation:

(CH3)3CBr + H2O (CH3)3COH + HBr

The following data were obtained from three experiments using the method of initial rates:

a. What is the order with respect to (CH3)3CBr?

b. What is the order with respect to H2O?

c. What is the overall order of the reaction?

d. Write the rate equation.

e. Calculate the rate constant k for the reaction

Initial [(CH3)3CBr] mol L-1 Initial [H2O] mol L-1 Initial rate mol L-1 min-1 Experiment 1   5.0 x 10^-2 2.0 x 10^-2 2.0 x 10-6 Experiment 2 5.0 x 10-2 4.0 x 10-2 2.0 x 10-6 Experiment 3 1.0 x 10-1 4.0 x 10-2 4.0 x 10-6

Explanation / Answer

To calculate the order of reaction with respec to individual reactant we take the values where the concentration of other reactant is same.

a) For (CH3)CBr , we will consider Experiment 2 and 3.

(Rate)2 / (Rate)3 = {initial [CH3)3CBr] 2 / initial [CH3)3CBr] 3}^x

2.0 x 10-6 / 4.0 x 10-6 = [ 5.0 x 10-2 / 1.0 x 10-2]^x

1/2 = (5)^x

log0.5 = xlog5

-0.3010 = x (0.698)

x = -0.432 (order with respect to [(CH3)3CBr] )

c) Let the order with respect to water = y

We will consider experiment 1 and experiment 2

so rate of reaction will depend upon water only

(rate)1 / (rate)2 = ([H2O]1 / pH2O]2)^y

As per given rates it is not affected by change in concentration of water

so order with respect to water = 0

c) overall order = 0 + (-0.432) = -0.432

d) rate of reaction = K [(CH3)3CBr]^-0.432

e) rate consatant = Rate / [(CH3)3CBr^-0.432

Rate constant = 2.0 x 10-6 / (5.0 x 10-2 )^-0.432

Rate constant = 5.47 X 10^-7

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