Problem 32 - Chemistry: The Central Science

Question

The reaction between ethyl bromide \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Br}\right)\) and hydroxide ion in ethyl alcohol at \(330 \mathrm{~K}_{2} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Br}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{Br}^{-}(\)alc \()\), 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 \(0.100 M\), 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

Answer

(a) The value of the rate constant, \(k\), is approximately \(3.56 \times 10^{-6}\,(\mathrm{M}/\mathrm{s})/(\mathrm{M}\times\mathrm{M})\). (b) The units of the rate constant are \(\frac{\mathrm{M}}{\mathrm{s} \cdot \mathrm{M}^2}\). (c) The new rate of disappearance of ethyl bromide, after dilution, is approximately \(4.25 \times 10^{-8}\,\mathrm{M}/\mathrm{s}\), which is lower than the original rate.

Step 1: Identify Rate Law and Known Information

The reaction is first-order in both ethyl bromide and hydroxide ion, so the rate law for this reaction will take the form: \[Rate = k[\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{Br}][\mathrm{OH}^{-}]\] We are given the following information: \[[\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{Br}] = 0.0477\,\mathrm{M}\] \[[\mathrm{OH}^{-}] = 0.100\,\mathrm{M}\] \[Rate = 1.7 \times 10^{-7}\,\mathrm{M}/\mathrm{s}\]

Step 2: Calculate the Rate Constant

Now, we can plug in the known values and solve for the rate constant, \(k\): \[1.7 \times 10^{-7}\,\mathrm{M}/\mathrm{s} = k(0.0477\,\mathrm{M})(0.100\,\mathrm{M})\] \[k = \frac{1.7 \times 10^{-7}\,\mathrm{M}/\mathrm{s}}{(0.0477\,\mathrm{M})(0.100\,\mathrm{M})}\] \[k = 3.56 \times 10^{-6}\,(\mathrm{M}/\mathrm{s})/(\mathrm{M}\times\mathrm{M})\] So the rate constant, \(k\), is approximately \(3.56 \times 10^{-6}\,(\mathrm{M}/\mathrm{s})/(\mathrm{M}\times\mathrm{M})\). (b) Finding the units of the rate constant: We have already found the rate constant to be \(3.56 \times 10^{-6}\,(\mathrm{M}/\mathrm{s})/(\mathrm{M}\times\mathrm{M})\), so the units are: \[\frac{\mathrm{M}}{\mathrm{s} \cdot \mathrm{M}^2}\] (c) Effect of dilution on the rate of disappearance of ethyl bromide:

Step 3: Calculate the New Concentrations After Dilution

When an equal amount of pure ethyl alcohol is added, the volume of the solution doubles. Therefore, the new concentrations of ethyl bromide and hydroxide ion will be half of their original concentrations as the number of moles of the components does not change: \[ [\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{Br}]_\text{new} = \frac{1}{2} \times 0.0477\,\mathrm{M} = 0.02385\,\mathrm{M}\] \[[\mathrm{OH}^{-}]_\text{new} = \frac{1}{2} \times 0.100\,\mathrm{M} = 0.050\,\mathrm{M}\]

Step 4: Calculate the New Rate of Disappearance

Now, we can use the rate law again with the new concentrations: \[Rate_\text{new} = k[\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{Br}]_\text{new}[\mathrm{OH}^{-}]_\text{new}\] \[Rate_\text{new} = (3.56 \times 10^{-6}\,(\mathrm{M}/\mathrm{s})/(\mathrm{M}\times\mathrm{M}))(0.02385\,\mathrm{M})(0.050\,\mathrm{M})\] \[Rate_\text{new} = 4.25 \times 10^{-8}\,\mathrm{M}/\mathrm{s}\] The new rate of disappearance of ethyl bromide, after dilution, is approximately \(4.25 \times 10^{-8}\,\mathrm{M}/\mathrm{s}\), which is lower than the original rate.

Key Concepts

Rate Law

In chemical kinetics, the rate law is a crucial concept that describes how the concentration of reactants affects the rate of a reaction. For the reaction between ethyl bromide and hydroxide ion in ethyl alcohol, the rate law is determined based on its order concerning each reactant. Here, our reaction is first order with respect to both ethyl bromide \((\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{Br})\) and the hydroxide ion \((\mathrm{OH}^{-})\). Thus, the rate can be expressed as: \[Rate = k[\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{Br}][\mathrm{OH}^{-}]\]
This mathematical expression helps us understand that the rate is directly proportional to the product of the concentrations of the reactants, each raised to the power of their respective order, which in this case, is 1.
Hence, knowing which reactants influence the rate and to what extent, as specified in the rate law, allows for precise predictions about how changes in concentrations will affect the reaction speed.

Rate Constant

The rate constant, denoted as \(k\), is a fundamental parameter in the rate law equation that serves as a measure of how fast a reaction occurs under given conditions. It is crucial for calculating the reaction rate once the concentrations of the reactants are known.
For our specific reaction, we calculate \(k\) by reorganizing the rate law: \[k = \frac{Rate}{[\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{Br}][\mathrm{OH}^{-}]}\]
Substituting the values provided, we find: \[k = \frac{1.7 \times 10^{-7}\,\mathrm{M/s}}{0.0477\,\mathrm{M} \times 0.100\,\mathrm{M}} = 3.56 \times 10^{-6}\,\mathrm{M^{-1}\,s^{-1}}\]
The units of the rate constant vary depending on the overall order of the reaction. For a second-order reaction such as this one, the units are generally \(\mathrm{M^{-1}\,s^{-1}}\). This reflects the inverse relationship between the molarity of reactants and the time in which they react.
The rate constant is specific to a given reaction and is influenced by factors such as temperature, making it a vital component in the kinetic analysis of chemical reactions.

Effect of Concentration on Reaction Rate

The concentration of reactants is one of the primary factors affecting the rate of a chemical reaction. For our reaction, as dictated by the rate law, the rate is dependent on the concentration of both ethyl bromide and hydroxide ions.
When the initial concentrations are halved by adding an equal volume of pure ethyl alcohol, the new concentrations become \(0.02385\,\mathrm{M}\) for ethyl bromide and \(0.050\,\mathrm{M}\) for hydroxide ions.
Using the new concentrations in the rate law, we calculate the new rate as: \[Rate_\text{new} = k[\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{Br}]_\text{new}[\mathrm{OH}^{-}]_\text{new}\]
Substituting the known quantities: \[Rate_\text{new} = (3.56 \times 10^{-6}\,\mathrm{M^{-1}\,s^{-1}})(0.02385\,\mathrm{M})(0.050\,\mathrm{M}) = 4.25 \times 10^{-8}\,\mathrm{M/s}\]
Clearly, the rate decreases as the concentrations of reactants decrease. This illustrates the principle that lower concentrations generally lead to fewer molecular collisions per unit time, leading to a slower reaction.
Understanding this relationship enables chemists to control and optimize reaction conditions by manipulating the concentrations of reactants.