Ch.14 - Chemical Kinetics

Problem 32c

Chapter 14, Problem 32c

The reaction between ethyl bromide (C₂H₅Br) and hydroxide ion in ethyl alcohol at 330 K, C₂H₅Br(alc) + OH^-(alc) → C₂H₅OH(l) + Br^-(alc), is first order each in ethyl bromide and hydroxide ion. When [C₂H₅Br] is 0.0477 M and [OH^-] is 0.100 M, the rate of disappearance of ethyl bromide is 1.7×10^-7 M/s. (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?

Verified step by step guidance

Identify the rate law for the reaction. Since the reaction is first order in both ethyl bromide \((\text{C}_2\text{H}_5\text{Br})\) and hydroxide ion \((\text{OH}^-)\), the rate law can be expressed as: \( \text{Rate} = k[\text{C}_2\text{H}_5\text{Br}][\text{OH}^-] \).

Understand the effect of dilution on concentration. When the solution is diluted by adding an equal volume of pure ethyl alcohol, the concentrations of both reactants \([\text{C}_2\text{H}_5\text{Br}]\) and \([\text{OH}^-]\) will be halved.

Determine the new concentrations after dilution. The initial concentration of \([\text{C}_2\text{H}_5\text{Br}]\) is 0.0477 M and \([\text{OH}^-]\) is 0.100 M. After dilution, these concentrations become 0.02385 M and 0.050 M, respectively.

Substitute the new concentrations into the rate law to find the new rate of disappearance of ethyl bromide. The new rate will be: \( \text{Rate}_{\text{new}} = k[0.02385][0.050] \).

Compare the new rate to the original rate. Since both concentrations are halved, the new rate will be \(\frac{1}{4}\) of the original rate, because the rate is proportional to the product of the concentrations of the reactants.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Rate Law

The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. For a reaction that is first order in each reactant, the rate can be described by the equation Rate = k[C2H5Br][OH-], where k is the rate constant. Understanding the rate law is essential for predicting how changes in concentration affect the reaction rate.

Dilution Effect

Dilution involves adding a solvent to a solution, which decreases the concentration of solutes. In this case, adding an equal volume of pure ethyl alcohol will halve the concentrations of both ethyl bromide and hydroxide ions. This change in concentration directly impacts the reaction rate, as the rate is dependent on the concentrations of the reactants.

First-Order Reactions

First-order reactions are those where the rate of reaction is directly proportional to the concentration of one reactant. In this scenario, since both reactants are first order, the overall rate will decrease when their concentrations are reduced due to dilution. This concept is crucial for understanding how the reaction rate will change in response to concentration adjustments.

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Related Practice

Textbook Question

The reaction between ethyl bromide (C₂H₅Br) and hydroxide ion in ethyl alcohol at 330 K, C₂H₅Br(alc) + OH^-(alc) → C₂H₅OH(l) + Br^-(alc), is first order each in ethyl bromide and hydroxide ion. When [C₂H₅Br] is 0.0477 M and [OH^-] is 0.100 M, the rate of disappearance of ethyl bromide is 1.7×10^-7 M/s. (a) What is the value of the rate constant?

Textbook Question

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl^- + I^- → OI^- + Cl^- . This rapid reaction gives the following rate data:

[OCl₄^-] (M) [I^-] (M) Initial Rate (M,s)

1.5 * 10-3 1.5 * 10-3

1.36 * 10-4 3.0 * 10-3 1.5 * 10-3 2.72 * 10-4

1.5 * 10-3 3.0 * 10-3 2.72 * 10-4

(a) Write the rate law for this reaction.

Textbook Question

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl - + I - ¡OI - + Cl - . This rapid reaction gives the following rate data:

[OCl₄^-] (M) [I^-] (M) Initial Rate (M,s)

1.5 * 10-3 1.5 * 10-3

1.36 * 10-4 3.0 * 10-3 1.5 * 10-3 2.72 * 10-4

1.5 * 10-3 3.0 * 10-3 2.72 * 10-4

(b) Calculate the rate constant with proper units.

Textbook Question

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl^- + I^- → OI^- + Cl^- . This rapid reaction gives the following rate data:

[OCl₄^-] (M) [I^-] (M) Initial Rate (M,s)

1.5 * 10-3 1.5 * 10-3

1.36 * 10-4 3.0 * 10-3 1.5 * 10-3 2.72 * 10-4

1.5 * 10-3 3.0 * 10-3 2.72 * 10-4 (c) Calculate the rate when [OCl^-] = 2.0 * 10-3 M and [I^-] = 5.0 * 10 - 4 M.