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 given information

Given: Reaction: \(C_{2}H_{5}Br(alc) + OH^-(alc) \longrightarrow C_{2}H_{5}OH(l) + Br^-(alc)\) Order of reaction: 1st order in ethyl bromide and 1st order in hydroxide ion. Initial concentration of ethyl bromide: \([C_{2}H_{5}Br] = 0.0477M\) Initial concentration of hydroxide ion: \([OH^-] = 0.100M\) Rate of disappearance of ethyl bromide: \(1.7 \times 10^{-7} \frac{M}{s}\)

02

Determine the value of the rate constant

Since the reaction is first order in ethyl bromide and hydroxide ion: Rate = k[C2H5Br][OH-] Plug in the given values into the equation: \(1.7 \times 10^{-7} \frac{M}{s} = k(0.0477M)(0.100M)\) Now solve for k: \(k = \frac{1.7 \times 10^{-7} \frac{M}{s}}{(0.0477M)(0.100M)}\) \(k \approx 3.57 \times 10^{-6} \frac{1}{Ms}\) (a) So, the value of the rate constant is approximately \(3.57 \times 10^{-6} \frac{1}{Ms}\).

03

Find the units of the rate constant

In the formula Rate = k[C2H5Br][OH-], the units of Rate = M/s, the units of [C2H5Br] = [OH-] = M. Let the units of k = x Now, we have: \(\frac{M}{s} = x\cdot M\cdot M \Rightarrow x = \frac{1}{Ms}\) So, the units of the rate constant are \(\frac{1}{Ms}\). (b)

04

Determine the change in the rate of disappearance of ethyl bromide if the solution is diluted

If the solution is diluted by adding an equal volume of pure ethyl alcohol, the concentrations of both ethyl bromide and hydroxide ions will be halved, since they are dissolved in a solution that now has twice the volume. New concentration of ethyl bromide: \(\frac{0.0477M}{2} = 0.02385M\) New concentration of hydroxide ion: \(\frac{0.100M}{2} = 0.050M\) Now use the rate equation with the new concentrations and the rate constant we found earlier: New Rate = \((3.57 \times 10^{-6} \frac{1}{Ms})(0.02385M)(0.050M)\) New Rate = \(4.25 \times 10^{-8} \frac{M}{s}\) (c) The new rate of disappearance of ethyl bromide is approximately \(4.25 \times 10^{-8} \frac{M}{s}\) when the solution is diluted.

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