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The reaction between ethyl bromide \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right)\) and hydroxide ion in ethyl alcohol at \(330 \mathrm{~K}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}(a l c)+\) \(\mathrm{OH}^{-}(a l c) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{Br}^{-}(a l c)\), is first order each in ethyl bromide and hydroxide ion. When \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right]\) is \(0.0477 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.100 \mathrm{M}\), the rate of disappearance of ethyl bromide is \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}\). (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) 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?

Step by step solution

01

Write the rate law

For a reaction that is first order with respect to both ethyl bromide and hydroxide ion, we can write the rate law as: Rate = k * [C2H5Br] * [OH-] We are given the rate of disappearance of ethyl bromide, which is 1.7 脳 10鈦烩伔 M/s, and the initial concentrations of C2H5Br and OH- as 0.0477 M and 0.100 M, respectively.

02

Solve for the rate constant (k)

We have all the information required to solve for k. Using the rate law: k = Rate / ([C2H5Br] * [OH-]) k = (1.7 脳 10鈦烩伔 M/s) / (0.0477 M * 0.100 M) k 鈮� 3.57 脳 10鈦烩伓 M鈦宦箂鈦宦� So, the value of the rate constant is approximately 3.57 脳 10鈦烩伓 M鈦宦箂鈦宦�. #b) Units of the rate constant# The units of the rate constant can be derived from the rate law: Rate = k * [C2H5Br] * [OH-] Since Rate has units of M/s and concentrations have units of M, we can find the units for k as follows: M/s = k * (M * M) => k = (M/s) / (M * M) => k = M鈦宦箂鈦宦� Therefore, the units of the rate constant are M鈦宦箂鈦宦�. #c) Change in the rate of disappearance of ethyl bromide if the solution is diluted#

03

Calculate the new concentrations after dilution

If we dilute the solution by adding an equal volume of pure ethyl alcohol, the concentrations of both C2H5Br and OH- will be halved: [C2H5Br'] = [C2H5Br] / 2 = 0.0477 M / 2 = 0.02385 M [OH-'] = [OH-] / 2 = 0.100 M / 2 = 0.050 M

04

Determine the new rate of disappearance of ethyl bromide

Using the rate law and the new concentrations, we can calculate the new rate of disappearance of ethyl bromide: Rate' = k * [C2H5Br'] * [OH-'] Rate' = (3.57 脳 10鈦烩伓 M鈦宦箂鈦宦�) * (0.02385 M) * (0.050 M) Rate' 鈮� 4.25 脳 10鈦烩伔 M/s After dilution, the rate of disappearance of ethyl bromide would be approximately 4.25 脳 10鈦烩伔 M/s.

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