Integrated rate law

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	0.702	0.796	8.63×10-3
2	0.702	1.59	8.63×10-3
3	1.40	0.796	1.72×10-2
4	1.40	1.59	1.72×10-2

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

Compare expts 1 and 3: [(CH₃)₃CBr] doubles while [OH-] remains constant. The rate doubles telling us that the reaction is FIRST ORDER in (CH₃)₃CBr.

Compare expts. 1 and 2: [OH-] doubles while [CH₃)₃CBr] remains constant. The rate does not change telling us that the reaction is ZERO ORDER in OH-

Rate law: Rate = k[CH₃)₃CBr]

To find k (rate constant), use the rate law and any experiment in the table above. I'll use expt. 2

8.63x10-3 M/s = k (0.702 M)

k = 8.63x10^-3 M/s / 0.702 M

k = 0.0123 s^-1