The following initial rate data are for the reaction of tertiary butyl bromide with hydroxide ion at 55 C :

Question

The following initial rate data are for the reaction of tertiary butyl bromide with hydroxide ion at 55 ^oC:

(CH₃)₃CBr + OH^-  (CH₃)₃COH + Br^-

Experiment	[(CH3)3CBr]o, M	[OH-]o, M	Initial Rate, Ms-1
1	0.806	0.362	9.35×10-3
2	0.806	0.724	9.35×10-3
3	1.61	0.362	1.87×10-2
4	1.61	0.724	1.87×10-2

Complete the rate law for this reaction in the box below.

Use the form k[A]^m[B]ⁿ , where '1' is understood for m or n and concentrations taken to the zero power do not appear. Don't enter 1 for m or n

Rate =

From these data, the rate constant is ____ s^-1.

Answer 1

Comparing experiments 1 and 2, we see that [(CH3)3CBr]o, M remains constant while [OH-] doubles. The rate remains the same telling us the reaction is ZERO order with respect to [OH-].

Comparing experiments 2 and 3, we see that [(CH3)3CBr]o, M doubles and since the reaction is independent of the [OH-], it doesn't matter what happens to it. The rate also doubles telling us the reaction is FIRST order with respect to [(CH3)3CBr]o, M. (You could also compare exp. 2 and 4 where [(CH3)3CBr]o, M doubles and [OH-] remains constant. Rate still doubles, etc.)

Rate = k[(CH3)3CBr]

To find the value of k, simply choose any experiment and solve the rate equation for k. I'll choose exp.1...

9.35×10^-3 Ms^-1 = k (0.806 M)

k = 9.35x10^-3 Ms^-1 / 0.806 M

k = 0.0116 s^-1