The reaction of C4H9Br with hydroxide ion proceeds with the formation of C4H9OH

Question

The reaction of C4H9Br with hydroxide ion proceeds with the formation of C4H9OH
C4H9Br(aq) + OH%u2013(aq) -> C4H9OH(aq) + Br%u2013(aq)

The following data were obtained at 55oC (concentrations are in mol/L)

What will be the initial rate in mol/L s be in Experiement 4?

The correct answer is 3.0 * 10^-3 , but can you please show all work leading to this answer? Thank you.

Exp [(CH3)3CBr]0 mol/l [OH-]0 mol/l Initial Rate mol/l s 1 .1    .1 1.0x10^-3 2 .2 .1 2.0x1-^-3 3 .1 .2 1.0x10^-3 4 .3 .2 ???

Explanation / Answer

if we compare exp-1 and 2

then rate becomes double when concen of [(CH3)3Br]0 is doubled so

rate is directly propotional to conc. of [(CH3)3Br]0

and on comapring 1 and 3 we get that the rate does not depend on the conc of [OH-]

r=K [(CH3)3Br]

so in the final experiment rate will be 3 times of the exp -1

so the answer is 3*10^-3

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