Problem 58 - Chemistry The Science in Context

Question

Two reactions in which there is a single reactant have nearly the same magnitude rate constant. One is first order; the other is second order. a. If the initial concentrations of the reactants are both \(1.0 \mathrm{mM},\) which reaction will proceed at the higher rate? b. If the initial concentrations of the reactants are both 2.0 \(M,\) which reaction will proceed at the higher rate?

Step-by-Step Solution

Answer

Answer: At the initial concentration of 1.0 mM, the first-order reaction will proceed at a higher rate compared to the second-order reaction.

Step 1: Calculate the rate of the first-order reaction

For the first-order reaction, the rate is given by: Rate\(_1\) = \(k_1 [A]\) Since the initial concentration is 1.0 mM, the rate becomes: Rate\(_1\) = \(k_1\times 1.0 \ \mathrm{mM}\)

Step 2: Calculate the rate of the second-order reaction

For the second-order reaction, the rate is given by: Rate\(_2\) = \(k_2 [A]^2\) With the initial concentration of 1.0 mM, the rate becomes: Rate\(_2\) = \(k_2 (1.0 \ \mathrm{mM})^2\)

Step 3: Compare the two rates

Given that \(k_1≈k_2\) and both initial concentrations are 1.0 mM, the rates become: Rate\(_1\) = \(k_1\times 1.0 \ \mathrm{mM}\) Rate\(_2\) = \(k_2\times 1.0 \ \mathrm{mM^2}\) Since \(k_1≈k_2\) and the concentration of A is the same for both reactions, the first-order reaction will have a higher rate for an initial concentration of 1.0 mM. Therefore, at 1.0 mM initial concentration, the first-order reaction will proceed at a higher rate. #Phase 2: Comparing the Rates for 2.0 M Initial Concentrations #

Step 1: Calculate the rate of the first-order reaction

For the first-order reaction, the rate is given by: Rate\(_1\) = \(k_1 [A]\) Since the initial concentration is 2.0 M, the rate becomes: Rate\(_1\) = \(k_1\times 2.0 \ \mathrm{M}\)

Step 2: Calculate the rate of the second-order reaction

For the second-order reaction, the rate is given by: Rate\(_2\) = \(k_2 [A]^2\) With the initial concentration of 2.0 M, the rate becomes: Rate\(_2\) = \(k_2 (2.0 \ \mathrm{M})^2\)

Step 3: Compare the two rates

Given that \(k_1≈k_2\) and both initial concentrations are 2.0 M, the rates become: Rate\(_1\) = \(k_1\times 2.0 \ \mathrm{M}\) Rate\(_2\) = \(k_2\times 4.0 \ \mathrm{M^2}\) Since \(k_1≈k_2\), the second-order reaction rate will be twice as fast as the first-order reaction rate for an initial concentration of 2.0 M, when compared to their rates at 1.0 mM initial concentration. Therefore, at 2.0 M initial concentration, the second-order reaction will proceed at a higher rate.

Key Concepts

First-Order Reaction

In reaction kinetics, a first-order reaction is one where the rate of reaction is directly proportional to the concentration of one reactant. In mathematical terms, this can be expressed as: Rate = \(k[A]\). Here, \(k\) is the rate constant, and \([A]\) is the concentration of the reactant.

• These reactions have a linear relationship between the rate and reactant concentration.
• In practical terms, if you double the concentration of the reactant, the rate of reaction doubles.
• Graphically, plotting the natural logarithm of the reactant concentration versus time yields a straight line.

This type of reaction is often seen in processes like radioactive decay and certain simple chemical reactions. The simple dependency on a single reactant makes predicting the rate of reaction quite straightforward in first-order kinetics.

Second-Order Reaction

A second-order reaction is one where the rate is proportional to the square of the concentration of a single reactant, or to the product of the concentrations of two different reactants. The rate expression is written as: Rate = \(k[A]^2\), for a single reactant.

• This relationship indicates that small changes in the reactant concentration result in much larger changes in the rate.
• For example, doubling the concentration of the reactant quadruples the rate of reaction.
• This type of reaction is common in bimolecular reactions, which involve two molecules coming together.

When plotted, the inverse of the reactant concentration versus time gives a straight line, which helps in determining the reaction order and the rate constant. This non-linear relationship makes predicting changes in reaction rate a bit more complex compared to first-order reactions.

Rate Constant

The rate constant, \(k\), is a critical component in the rate equations for both first and second-order reactions. It is a proportionality factor that bridges the gap between reaction rate and reactant concentration.

• Its units depend on the order of the reaction:
  • In first-order reactions, \(k\) has units of s⁻¹.
  • In second-order reactions, \(k\) has units of M⁻¹s⁻¹.
• The magnitude of \(k\) provides insight into how fast a reaction proceeds.
• While independent of concentration, \(k\) can be affected by factors like temperature and catalyst presence.

Understanding \(k\) helps chemists control and optimize reaction conditions, ensuring reactions proceed at desired rates even when reactant concentrations or conditions change. Its role in the rate equation emphasizes how fundamental it is to the study of kinetics.