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

1 0.626 0.800 7.14×10^-3
2 0.626 1.60 7.14×10^-3
3 1.25 0.800 1.43×10^-2
4 1.25 1.60 1.43×10^-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.

Rate =  k [ (CH₃)₃CBr ]

k = 0.011 s⁻¹

Explanation:

The rate law in general form for this reaction is:

rate = k[A]^m x [B]^n

and we want to determine m and n.

Notice the experiments conducted involved variying the concentration of one of the reactants while keeping the other constant. By comparing two experiments in this way and noticing the effect on the reaction rate, one can deduce the order with respect to that reactant.

Take experiments 1 and 2 :

Doubling the concentration of OH⁻ while keeping (CH₃)₃CBr constant had  no effect on the initial rate.

This implies the order with respect to [OH⁻] is zero.

From experiments 1 and 3 we  see that doubling the [ (CH₃)₃CBr ] while keeping [OH-] the same doubles the initial rate.

We now have the orders of the two reactants, and the rate law will be:

rate = k [ (CH₃)₃CBr ] [ OH⁻]⁰  = k [ (CH₃)₃CBr ]

We could use another comparison to check our answer. For example comparing 3 and 4,  maintaining  [ (CH₃)₃CBr ]  constant and doubling  [ OH⁻] has no effect on the rate which confirms the order respect to  [ OH⁻]  is zero.

The rate constant we can determine it from any of the experiment by solving for k, for example from experiment 1:

7.14 x 10⁻³ M/s  = k x 0.626 M   ⇒ k = 0.011 s⁻¹

answer; economic order quantity;