Question

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?

Short answer

(a) The value of the rate constant is approximately \(3.57 \times 10^{-6} \frac{1}{Ms}\). (b) The units of the rate constant are \(\frac{1}{Ms}\). (c) The new rate of disappearance of ethyl bromide after dilution is approximately \(4.25 \times 10^{-8} \frac{M}{s}\).

Step-by-step solution

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: \(1.7 \times 10^{-7} \frac{M}{s}\)

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 \(3.57 \times 10^{-6} \frac{1}{Ms}\).

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)

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.

Key concepts

Reaction Order

In chemical kinetics, understanding the reaction order is crucial as it specifies how the concentration of reactants affects the rate of the reaction. The reaction order is determined through experimentation and signifies the power to which the concentration of a reactant is raised in the rate equation. For the reaction between ethyl bromide and hydroxide ion, the reaction is first order each in ethyl bromide and hydroxide ion. This means that the rate of reaction is directly proportional to the concentration of ethyl bromide and hydroxide ion. So, if you double the concentration of either reactant, the rate of reaction will also double. It's vital to note that the reaction order is not necessarily related to the stoichiometric coefficients in the balanced chemical equation.

Rate Constant

The rate constant, denoted as \(k\), is a crucial part of the rate equation in chemical kinetics and acts as a proportionality factor. It provides information about the reaction's speed under specific conditions, and it remains constant for a given reaction at a fixed temperature. The rate constant can be calculated using the rate equation:

• For the reaction: \[ \text{Rate} = k [C_{2}H_{5}Br][OH^-] \]

Given the rate and the concentrations of the reactants, you can solve for \(k\):

• \[ k = \frac{1.7 \times 10^{-7} \: M/s}{0.0477 \: M \times 0.100 \: M} \]

The calculated value of \(k\) is approximately \(3.57 \times 10^{-6} \: M^{-1}s^{-1}\). Remember, the value of the rate constant is dependent upon the reaction order and thus, its units differ depending on the overall order of the reaction.

Rate Equation

The rate equation, also known as the rate law, is a mathematical representation that expresses the relationship between the concentration of reactants and the rate of a chemical reaction. For any given reaction, the rate equation is formulated based on its reaction order.

• For the reaction between ethyl bromide and hydroxide ion, the rate equation is: \[ \text{Rate} = k [C_{2}H_{5}Br][OH^-] \]

This indicates the rate of disappearance of ethyl bromide is proportional to the product of the concentrations of ethyl bromide and hydroxide ion. The coefficients in the rate equation, which reflect the reaction order with respect to each reactant, need not correspond to the stoichiometric coefficients found in the balanced equation. Moreover, the rate equation is empirical, derived from experimental observations rather than theoretical predictions.

Dilution Effect

Dilution affects the concentration of reactants and thereby impacts the reaction rate. When a solution is diluted, its volume increases, leading to decreased concentrations of the reactants present. In the specific reaction between ethyl bromide and hydroxide ion, if the solution is diluted by doubling the volume, the concentration of each reactant is halved. Hence, the new concentrations become \([C_{2}H_{5}Br] = 0.02385 \: M\) and \([OH^-] = 0.050 \: M\).Substituting these new concentrations into the rate equation yields a significantly reduced rate of reaction:

• New Rate = \((3.57 \times 10^{-6} \: M^{-1}s^{-1})(0.02385 \: M)(0.050 \: M)\)

The calculated new rate is approximately \(4.25 \times 10^{-8} \: M/s\), demonstrating how dilution directly reduces the frequency of effective collisions between reactant molecules, thus slowing down the rate of the chemical reaction.