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 down the rate law expression for the reaction.

Since the given reaction is first-order in ethyl bromide and first-order in hydroxide ion, we can write down the rate law expression as: Rate \(= k[\mathrm{C}_{2} \mathrm{H}_{5}\mathrm{Br}][\mathrm{OH}^-]\) Where Rate is the reaction rate, k is the rate constant, and the concentrations of ethyl bromide and hydroxide ion are denoted by their respective chemical formulas.

02

Find the rate constant (k) for the reaction.

We are given the rate of disappearance of ethyl bromide as \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}\). We will plug in the values given for the concentrations of ethyl bromide and hydroxide ion, and then solve for k: \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s} = k(0.0477 \mathrm{M})(0.100 \mathrm{M})\) To find k, we simply divide the rate by the product of the concentrations: k \(= \frac{1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}}{(0.0477 \mathrm{M})(0.100 \mathrm{M})}\) k \(= 3.57 \times 10^{-6} \mathrm{M}^{-1} \mathrm{s}^{-1}\) So, the rate constant (k) for this reaction is \(3.57 \times 10^{-6} \mathrm{M}^{-1} \mathrm{s}^{-1}\).

03

Determine the units of the rate constant.

To determine the units of the rate constant (k), we can look at the units used in the rate law expression: Rate \(= k[\mathrm{C}_{2} \mathrm{H}_{5}\mathrm{Br}][\mathrm{OH}^-]\) The units of the rate are M/s, and the units of concentration are M. Therefore, the units of the rate constant k are: Units of k \(= \frac{\mathrm{M}/\mathrm{s}}{\mathrm{M} \times \mathrm{M}}\) Units of k \(= \mathrm{M}^{-1}\mathrm{s}^{-1}\) Thus, the units of the rate constant are M⁻¹s⁻¹.

04

Determine how the rate of disappearance of ethyl bromide would change if the solution is diluted.

If the solution is diluted by adding an equal volume of pure ethyl alcohol, the concentrations of both reactants will be halved: New concentration of ethyl bromide \(= \frac{1}{2} \times 0.0477 \mathrm{M} = 0.02385 \mathrm{M}\) New concentration of hydroxide ion \(= \frac{1}{2} \times 0.100 \mathrm{M} = 0.050 \mathrm{M}\) Now, we will use the rate law expression to find the new rate of disappearance of ethyl bromide with these new concentrations: New Rate \(= k[\mathrm{C}_{2} \mathrm{H}_{5}\mathrm{Br}][\mathrm{OH}^-]\) New Rate \(= (3.57 \times 10^{-6} \mathrm{M}^{-1} \mathrm{s}^{-1})(0.02385 \mathrm{M})(0.050 \mathrm{M})\) New Rate \(= 4.25 \times 10^{-8} \mathrm{M}/\mathrm{s}\) The new rate of disappearance of ethyl bromide, after diluting the solution, is \(4.25 \times 10^{-8} \mathrm{M}/\mathrm{s}\).

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