Rates of reaction, redox reactions? 10 points?

Question

An structural isomer of bromobutane (C4H9Br) can be hydrolysed using aqueous sodium hydroxide to produce butanol. This can be represented by the following equation:

C4H9Br(l) + OH-(aq) > C4H9OH(l) + Br-(aq)

This reaction was investigated experimentally and the following results
were obtained:

Experiment A -
Initial [bromobutane], mol dm-3 = 0.01
Initial [OH-], mol dm-3 = 0.01
Initial rate, mol dm-3s-1 = 4.3 x 10 -4^

Experiment B
Initial [bromobutane], mol dm-3 = 0.01
Initial [OH-], mol dm-3 = 0.02
Initial rate, mol dm-3s-1 = 8.6 x 10-4^

Experiment C -
Initial [bromobutane], mol dm-3 = 0.02
Initial [OH-], mol dm-3 = 0.02
Initial rate, mol dm-3s-1 = 1.7 x 10-3^

(b) (i) Deduce the overall order of the reaction and write the rate equation for this reaction

(ii) What is meant by the rate-limiting step of a reaction mechanism?

(iii) What can you say about the rate-limiting step for the above reaction?

(iv) Suggest which isomer of bromobutane is likely to react via this rate-limiting step. Explain your answer fully.

Answer 1

Look at the three experiments. Note that A and B have the same concn bromobutane but OH is twice as much in B as in A. Note the rate change is by a factor of two also. 2^^x = 2. Of course x = 1 so the reaction is first order with respect to OH^-. Do the same for bromobenzene. Then the overall order is the sum of the two orders.
The rate equation is
rate = k[bromomenzene][OH^-]

Answer 2

why do the same for bromomenzene? it sayes to do so for bromobutane?

Answer 3

My typo. My brain got bromobutane and bromobenzen mixed up and bromobutane got lost.
You have the order for OH. Find the order for bromobutane the same way. Then
rate = k[bromobutane]^x[OH^-]^y
where x and y are the orders. I've already shown y is 1. You need to do x. The overall order is x+y.

Answer 4

Thank you soo much! Really appreciate it. Could you also help with iv please? Thank you.