Problem 32 - Chemistry The Central Science In Si Units

Question

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

Answer

(a) Rate constant, \( k = 3.56 \times 10^{-5} \text{ M}^{-1}\text{s}^{-1} \). (b) Units are \( \text{M}^{-1}\text{s}^{-1} \). (c) Rate decreases to \( 4.24 \times 10^{-8} \text{ M/s} \) after dilution.

Step 1: Write the rate law expression

Since the reaction is first order with respect to both ethyl bromide and hydroxide ion, the rate law can be expressed as: \[\text{Rate} = k \cdot \left[ \text{C}_2\text{H}_5\text{Br} \right] \cdot \left[ \text{OH}^{-} \right]\] where \( k \) is the rate constant.

Step 2: Solve for the rate constant, k

Given that \( \left[ \text{C}_2\text{H}_5\text{Br} \right] = 0.0477 \, \text{M} \), \( \left[ \text{OH}^{-} \right] = 0.100 \, \text{M} \), and the rate of disappearance of ethyl bromide is \( 1.7 \times 10^{-7} \, \text{M/s} \), use the rate law expression and solve for \( k \): \[1.7 \times 10^{-7} = k \times 0.0477 \times 0.100 \]\[k = \frac{1.7 \times 10^{-7}}{0.0477 \times 0.100} = \frac{1.7 \times 10^{-7}}{0.00477} \approx 3.56 \times 10^{-5} \, \text{M}^{-1} \text{s}^{-1}\]

Step 3: Determine the units of the rate constant

For a second-order reaction (first order in each reactant), the units of the rate constant \( k \) are \( \text{M}^{-1} \text{s}^{-1} \). This matches the units obtained in the previous step.

Step 4: Analyze the effect of dilution on the rate

When the solution is diluted by adding an equal volume of ethyl alcohol, the concentration of each reactant is halved. The new concentrations will be \( \left[ \text{C}_2\text{H}_5\text{Br} \right] = 0.02385 \, \text{M} \) and \( \left[ \text{OH}^{-} \right] = 0.0500 \, \text{M} \).

Step 5: Calculate the new rate after dilution

Substitute the new concentrations into the rate law to find the new rate: \[\text{New Rate} = k \cdot (0.02385) \cdot (0.0500)\]Using \( k \approx 3.56 \times 10^{-5} \, \text{M}^{-1} \text{s}^{-1} \): \[\text{New Rate} = 3.56 \times 10^{-5} \cdot 0.02385 \cdot 0.0500 \approx 4.24 \times 10^{-8} \, \text{M/s}\]

Step 6: Interpret the result

The rate of disappearance of ethyl bromide after dilution is reduced to \( 4.24 \times 10^{-8} \, \text{M/s} \). This shows that the rate decreases when the solution is diluted, as both reactant concentrations are reduced.

Key Concepts

Rate Law

The rate law is an equation that relates the rate of a chemical reaction to the concentration of its reactants. The general form of the rate law is:\[ \text{Rate} = k \cdot [A]^m \cdot [B]^n \]where:

• \(k\) is the rate constant, a unique value for each reaction at a given temperature.
• \([A]\) and \([B]\) are the concentrations of reactants.
• \(m\) and \(n\) are the orders of the reaction with respect to each reactant.

In the provided exercise, the reaction between ethyl bromide (\(\text{C}_2\text{H}_5\text{Br}\)) and hydroxide ion is first order in both reactants. This means both reactants affect the rate equally and proportionately to their concentrations.If you know the concentrations of reactants and the reaction rate, you can calculate the rate constant \(k\) using this formula, as it binds all the components of the rate law together.

Reaction Order

Reaction order tells you how the concentration of a reactant impacts the rate of the reaction. It is the sum of the exponents in the rate law: - First-order reaction: Rate depends on the concentration of one reactant. - Second-order reaction: Rate depends on the concentration of one reactant squared or the concentration of two reactants multiplied together. For the given exercise, the reaction order is two, since the reaction is first order with respect to both ethyl bromide and hydroxide ion. This means changes in concentration of either reactant proportionately affect the reaction rate. When a solution is diluted, the concentrations of reactants decrease, which reduces the rate of reaction. This aligns with the definition where total reaction order determines how sensitive a reaction is to changes in reactant concentrations.

Rate Constant

The rate constant \(k\) is a proportionality constant in the rate law equation. It provides essential information about the reaction kinetics and varies with temperature, meaning it's specific to the conditions under which a reaction occurs.To find \(k\), you rearrange the rate law equation to solve for \(k\):\[ k = \frac{\text{Rate}}{[A]^m \cdot [B]^n} \]In the provided reaction, this calculation was performed using given concentrations and the observed reaction rate. The units of the rate constant depend on the reaction order:

• For a first-order reaction, \(k\) has units of \(s^{-1}\).
• For a second-order reaction, like the one in the exercise, \(k\) has units of \(M^{-1} s^{-1}\).

The calculated rate constant allows predictions of reaction rates under similar conditions, offering insights into the speeding or slowing factors in a chemical process.