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 rate constant (k) for this reaction is \(3.57 \times 10^{-6} \mathrm{M}^{-1} \mathrm{s}^{-1}\). \(b)\) The units of the rate constant are \(\mathrm{M}^{-1}\mathrm{s}^{-1}\). \(c)\) The new rate of disappearance of ethyl bromide, after diluting the solution, is \(4.25 \times 10^{-8} \mathrm{M}/\mathrm{s}\).

Step-by-step solution

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.

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 \(3.57 \times 10^{-6} \mathrm{M}^{-1} \mathrm{s}^{-1}\).

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.

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 \(4.25 \times 10^{-8} \mathrm{M}/\mathrm{s}\).

Key concepts

First-Order Reaction

Understanding a first-order reaction is simple when you think about how the reaction rate depends on the concentration of one of the reactants. In chemical kinetics, a first-order reaction means that the rate is directly proportional to the concentration of one reactant.

In this case, the reaction between ethyl bromide (\(\mathrm{C}_{2} \mathrm{H}_{5}\mathrm{Br}\)) and hydroxide ion (\(\mathrm{OH}^-\)) in ethyl alcohol is actually first-order with respect to both reactants. This implies:

• If you double the concentration of ethyl bromide, the rate of the reaction also doubles.
• The same doubling effect occurs if you double the concentration of hydroxide ion.

Overall, the reaction depends linearly on the concentrations of both substances. This dual dependence creates a second-order reaction overall, which is the sum of the orders with respect to each reactant.

The rate law expression for this reaction can be written as:
\[\text{Rate} = k[\mathrm{C}_{2} \mathrm{H}_{5}\mathrm{Br}][\mathrm{OH}^-]\]
Here, \(k\) is the rate constant, which we'll explore further in the next section.

Rate Constant

The rate constant, often denoted as \(k\), is a crucial factor in the rate law equation. It provides the proportionality between the reaction rate and the concentrations of the reactants.

The value of \(k\) for a reaction doesn't change with different concentrations but may vary with temperature or solvent conditions. In this reaction, we calculated \(k\) with the given rate and concentrations:
\[k = \frac{1.7 \times 10^{-7} \text{ M/s}}{(0.0477 \text{ M})(0.100 \text{ M})} = 3.57 \times 10^{-6} \text{ M}^{-1} \text{s}^{-1}\]
Here’s why this matters:

• The units of \(k\) tell us the overall order of the reaction. In this case, the units are \(\text{M}^{-1}\text{s}^{-1}\), confirming it is second-order overall.
• Knowing \(k\) allows you to predict how altering concentrations will affect the reaction rate.

This consistency is important when planning reactions or adjusting processes in a lab setting.

Dilution Effect

Dilution plays a significant role in reaction rates, especially for reactions that involve reactant concentration in their rate laws. When you dilute a solution by adding solvent, the concentrations of all solutes in the mixture decrease proportionally.

Here's how dilution impacts our reaction:

• If you add an equal volume of ethyl alcohol to the original solution, both \([\mathrm{C}_{2} \mathrm{H}_{5}\mathrm{Br}]\) and \([\mathrm{OH}^-]\) are halved.
• This reduction leads to a decrease in the rate of the reaction, calculated using the new concentrations:

\[\text{New Rate} = (3.57 \times 10^{-6} \text{ M}^{-1} \text{s}^{-1})(0.02385 \text{ M})(0.050 \text{ M}) = 4.25 \times 10^{-8} \text{ M/s}\]
So, the rate of disappearance of ethyl bromide after dilution decreases significantly. This illustrates how changes in concentration directly influence reaction speed, crucial for controlling processes in chemical engineering and laboratory settings.