Question

Determine the rate constant for each of the following firstorder reactions: (a) \(2 \mathrm{~A} \rightarrow \mathrm{B}+\mathrm{C}\), given that the concentration of A decreases to one-fourth its initial value in \(61 \mathrm{~min}\); (b) \(2 \mathrm{~A} \longrightarrow\) \(\mathrm{B}+\mathrm{C}\), given that \([\mathrm{A}]_{0}=0.021 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) and that after \(45 \mathrm{~s}\) the concentration of \(\mathrm{B}\) increases to \(0.00125 \mathrm{~mol} \cdot \mathrm{L}^{-1}\); (c) \(2 \mathrm{~A} \rightarrow 3 \mathrm{~B}+\) \(\mathrm{C}\), given that \([\mathrm{A}]_{0}=0.060 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) and that after \(10.8\) min the concentration of \(\mathrm{B}\) rises to \(0.040 \mathrm{~mol} \cdot \mathrm{L}^{-1}\). In each case, write the rate law for the rate of loss of \(\mathrm{A}\).

Short answer

For each reaction, use the first-order kinetics formula. For (a), the rate constant \( k \) is calculated using the time of 3660 seconds and the concentration decrease to one-fourth. For (b), compute \( k \) using the concentration increase of B and the given initial concentration of A over 45 seconds. And for (c), calculate \( k \) with the concentration increase of B and given \( [A]_0 \) over 648 seconds. Express the rate of loss of A with the rate law \( \text{rate} = k[A] \) for a first-order reaction.

Step-by-step solution

Identify the Order of the Reaction

First-order reactions have a rate that is directly proportional to the concentration of one of the reactants. The given reactions are first-order reactions, and the rate constant can be determined using the formula for first-order reaction kinetics: \( k = \frac{\ln(\frac{[A]_0}{[A]})}{t} \) where \( k \) is the rate constant, \( [A]_0 \) is the initial concentration, \( [A] \) is the concentration at time \( t \) (in seconds), and \( \ln \) is the natural logarithm.

Determine the Rate Constant for Reaction (a)

For part (a), the concentration of A decreases to one-fourth its initial value in 61 minutes. Converting 61 minutes into seconds gives us 3660 seconds. Substituting the given values into the first-order reaction formula: \( k = \frac{\ln(\frac{1}{1/4})}{3660 \text{ s}} = \frac{\ln(4)}{3660 \text{ s}} \). Calculate the value of the rate constant \( k \) to get the result for reaction (a).

Calculate the Rate Constant for Reaction (b)

For part (b), use the increase in concentration of B, which is \( 0.00125 \text{ mol} \cdot \text{L}^{-1} \) in 45 seconds to determine the change in concentration of A because the ratio of A to B forming is 2:1. This gives us the change in concentration of A as \( 2 \times 0.00125 \text{ mol} \cdot \text{L}^{-1} \). Since \( [A]_0 = 0.021 \text{ mol} \cdot \text{L}^{-1} \), the remaining concentration of A after 45 seconds is \( [A] = [A]_0 - 2 \times 0.00125 \text{ mol} \cdot \text{L}^{-1} \) . Now calculate \( k \) using this concentration and \( t = 45 \text{ s} \) in the first-order reaction formula.

Compute the Rate Constant for Reaction (c)

For part (c), since B increases to \( 0.040 \text{ mol} \cdot \text{L}^{-1} \) from 3 times the decrease in A, the change in concentration of A is \( \frac{1}{3} \times 0.040 \text{ mol} \cdot \text{L}^{-1} \). Use \( [A] = [A]_0 - \frac{1}{3} \times 0.040 \text{ mol} \cdot \text{L}^{-1} \) and \( t = 10.8 \text{ min} = 648 \text{ s} \) to calculate \( k \) with the first-order reaction formula.

Write the Rate Law for the Rate of Loss of A

The rate law for the rate of loss of A in a first-order reaction is given by \( \text{rate} = k[A] \) where \( [A] \) is the concentration of reactant A at any time \( t \) and \( k \) is the rate constant.

Key concepts

Rate Constant Calculation

The rate constant in a chemical reaction is a critical factor that determines the speed at which the reaction proceeds. It is essential for understanding how quickly reactants are converted into products under certain conditions. For first-order reactions, the rate constant, denoted by the symbol \( k \), can be calculated using the formula:

\[ k = \frac{\ln(\frac{[A]_0}{[A]})}{t} \]

where \( [A]_0 \) is the initial concentration of the reactant, \( [A] \) is the concentration at a later time \( t \), and \( \ln \) indicates the natural logarithm. This mathematical relationship asserts that the rate at which the concentration of a reactant diminishes is directly proportional to its current concentration.

When solving problems related to the rate constant, it's important always to pay attention to units. Time, for example, should be consistent throughout the calculation and typically is in seconds. Additionally, the concentration of reactants should be adjusted based on the stoichiometry of the reaction if intermediates or products are given, as seen in part (b) and part (c) of the original exercise. Ensuring correct use of the formula will lead to accurate values of the rate constant, which is pivotal in understanding the kinetics of the reaction.

Rate Law

The rate law tells us how the rate of a chemical reaction is affected by the concentration of the reactants. For first-order reactions, the rate law is astonishingly simple. It states that the rate of reaction is directly proportional to the concentration of one reactant. Here is what the rate law generally looks like:

\[ \text{rate} = k[A] \]

In this equation, \( k \) is the rate constant, while \( [A] \) represents the reactant concentration at any given moment. This direct proportionality implies that if we were to double the concentration of the reactant \( A \), the rate of the reaction would also double. Moreover, the rate law provides a foundation for understanding how different factors such as temperature, catalysts, or changes in reactant concentration might influence the speed of a reaction. It's important to note that the rate law cannot be deduced from the balanced chemical equation alone; it must be determined experimentally. However, once the reaction order has been identified, as in the given exercise where it’s first-order, we can then express the rate law succinctly based on the known dependencies.

Reaction Order

Reaction order is a term used in kinetics to describe the power to which the concentration of a reactant is raised in the rate law. The overall reaction order is the sum of these exponents for all reactants involved in the rate-determining step. In a first-order reaction, this order is one, meaning the rate of reaction depends linearly on the concentration of a single reactant. Essentially, this tells us that the reaction rate will change in direct proportion to changes in the concentration of the reactant.

Understanding the order of a reaction is important because it provides insights into the mechanism of the reaction and how different concentrations or conditions might affect the rate at which products form. The reaction order can also affect how the half-life of a reaction -- the time it takes for half of the reactant to be consumed -- behaves. In first-order reactions, the half-life is constant, not dependent on the initial concentration, which is a unique and helpful characteristic in calculating reaction kinetics. The original exercise requires the comprehension of these concepts to calculate the rate constants and to understand the rate at which the reactant \( A \) is used up.