The reaction between ethyl bromide (C2H5Br) and hydroxide ion in ethyl alcohol i

Question

The reaction between ethyl bromide (C2H5Br) and hydroxide ion in ethyl alcohol is shown below at 330 K. C2H5Br(alc) + OH (alc) C2H5OH(l) + Br(alc) The reaction is first order each in ethyl bromide and hydroxide ion. When [C2H5Br] is 0.0458 M and [OH ] is 0.187 M, the rate of disappearance of ethyl bromide is 3.1 10-7 M/s. (1)What is the value of the rate constant? (2)How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution? (a)1/4 the rate (b)1/2 the rate (c)2 times the rate (d)4 times the rate (e)not change at all

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Explanation / Answer

(a)
Rate = k [C2H5Br][OH -]

Plug in all the knowns and solve for k:
k= 3.1 * 10^-7 /(.0458*.187) = 3.62 * 10^-5 M^-1s^-1

(b)
The ethyl alcohol will simply dilute the concentrations of each reactant to 1/2 of the original concentrations.
If the reaction is 1st order with respect to each.
The original reaction rate would be 1/2*1/2 = 1/4 the original rate.

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