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 equation for the reaction

The given reaction is first order each in ethyl bromide and hydroxide ion. Thus, the rate equation for this reaction can be written as: \(rate = k[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}][\mathrm{OH}^-]\) Now we have to calculate the rate constant, k.

02

Calculate the rate constant, k

We have the rate of disappearance of ethyl bromide, which is equal to the rate of the reaction. We also have the initial concentrations of the reactants. Using these values, we can calculate the rate constant as: \(k = \frac{rate}{[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}][\mathrm{OH}^-]}\) \(k = \frac{1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}}{(0.0477 \mathrm{M})(0.100 \mathrm{M})}\) \(k ≈ 3.57 \times 10^{-6} \mathrm{M}^{-1} \mathrm{s}^{-1}\) The value of the rate constant, k, is approximately \(3.57 \times 10^{-6} \mathrm{M}^{-1} \mathrm{s}^{-1}\). Now that we have the rate constant, let's determine its units.

03

Determine the units of the rate constant

Looking at the rate equation and the units of the concentrations, we can determine the units of the rate constant by analyzing the dimensions: \(\frac{\mathrm{M} \cdot \mathrm{s}^{-1}}{\mathrm{M}^2} = \mathrm{M}^{-1} \cdot \mathrm{s}^{-1}\) Thus, the units of the rate constant, k, are \(\mathrm{M}^{-1} \cdot \mathrm{s}^{-1}\). Finally, we have to find how the rate of disappearance of ethyl bromide would change upon dilution.

04

Determine the change in the rate of disappearance of ethyl bromide upon dilution

When the solution is diluted by adding an equal volume of pure ethyl alcohol, the concentrations of both reactants are halved. So, the new concentrations are: \([C2H5Br]_{new} = 0.5 [C2H5Br]_{initial} = 0.5 (0.0477 M) = 0.02385 M\) \([OH^-]_{new} = 0.5 [OH^-]_{initial} = 0.5 (0.100 M) = 0.0500 M\) Now, we can calculate the new rate of disappearance of ethyl bromide using the rate equation and the rate constant: \(rate_{new} = k [C2H5Br]_{new} [OH^-]_{new}\) \(rate_{new} = (3.57 \times 10^{-6} M^{-1}s^{-1})(0.02385 M)(0.0500 M)\) \(rate_{new} ≈ 4.25 \times 10^{-9} M \cdot s^{-1}\) Upon dilution, the rate of disappearance of ethyl bromide would change to approximately \(4.25 \times 10^{-9} M \cdot s^{-1}\).

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