Question

This reaction was monitored as a function of time: $$\mathrm{AB} \longrightarrow \mathrm{A}+\mathrm{B}$$ A plot of 1\(/[\mathrm{AB}]\) versus time yields a straight line with slope \(-0.055 / \mathrm{M} \cdot \mathrm{s} .\) \begin{equation} \begin{array}{l}{\text { a. What is the value of the rate constant }(k) \text { for this reaction at this }} \\ {\text { temperature? }}\\\\{\text { b. Write the rate law for the reaction. }} \\ {\text { c. What is the half-life when the initial concentration is } 0.55 \mathrm{M} \text { ? }}\\\\{\text { d. If the initial concentration of } \mathrm{AB} \text { is } 0.250 \mathrm{M}, \text { and the reaction mixture }} \\ {\text { initially contains no products, what are the concentrations of } \mathrm{A} \text { and } \mathrm{B}} \\ {\text { after } 75 \mathrm{s} \text { ? }}\end{array} \end{equation}

Short answer

a. The rate constant (k) is 0.055 M/s. b. The rate law is rate = k[AB]. c. The half-life with an initial concentration of 0.55 M is t_(1/2) = ln(2)/k. d. After 75 s, the concentrations of A and B will be equal and can be determined using the first order integrated rate law.

Step-by-step solution

Determine the rate constant (k)

The reaction is first order, as indicated by the linearity of a plot of 1/[AB] versus time. The slope of the line is the negative of the rate constant. Thus, the rate constant is k = 0.055 M/s.

Write the rate law for the reaction

For a first-order reaction, the rate law is given by the formula: rate = k[AB], where [AB] is the concentration of reactant AB.

Calculate the half-life for the initial concentration of 0.55 M

The half-life of a first-order reaction is given by t_(1/2) = ln(2)/k. Using the rate constant from Step 1, calculate the half-life.

Calculate the concentrations of A and B after 75 s

Use the integrated rate law for a first-order reaction, [AB] = [AB]0 * e^(-kt), to determine the remaining concentration of AB, and the amount of A and B formed at time t = 75 s, assuming stoichiometric 1:1 production of A and B from AB.

Key concepts

Rate Constant

In chemical kinetics, the rate constant (represented by the symbol k) is a crucial factor that quantifies the speed of a chemical reaction. For a first-order reaction, it has units of reciprocal seconds (s^-1) and is determined from the slope of a plot of the inverse of the reactant concentration \( 1/[AB] \) over time. When the plot yields a straight line, it reveals that the reaction follows first-order kinetics. From the exercise, we determine that the rate constant is \( k = 0.055 \text{ M/s} \) by taking the absolute value of the slope of the line.

Understanding the k value is vital since it allows us to predict how fast a reaction will occur under specific conditions. A larger k means a faster reaction, whereas a smaller k indicates a slower reaction. It's also worth noting that the rate constant is affected by temperature; as the temperature increases, typically, so does the rate constant.

Rate Law

The rate law for a chemical reaction expresses the relationship between the reaction rate and the concentrations of reactants. It takes the form of an equation: \( \text{rate} = k[\text{Reactants}]^n \) where k is the rate constant, the reactants are raised to their reaction orders (denoted by \( n \) that can be zero, a whole number, or a fraction), and the rate is the change in concentration over time. In the case of a first-order reaction, the rate law is simplified to \( \text{rate} = k[AB] \) where \( [AB] \) is the concentration of the single reactant AB. This linear relationship implies that if the concentration of AB doubles, the rate of the reaction also doubles.

The exercise asks to write the rate law for our specified reaction, and by following the definition of first-order reaction kinetics, we establish that the rate of the reaction is directly proportional to the concentration of AB.

Reaction Half-life

The reaction half-life, \( t_{1/2} \), is the period of time it takes for the concentration of a reactant to decrease to half of its original value. Intriguingly, for a first-order reaction, the half-life is constant and independent of the initial concentration. It can be calculated using the formula: \( t_{1/2} = \frac{\ln(2)}{k} \).

In the provided exercise, with an initial concentration \( [AB]_0 = 0.55 \text{ M} \) and the rate constant \( k = 0.055 \text{ M/s} \) discovered earlier, the half-life can be computed using the given formula. Using the natural logarithm of 2 and the calculated rate constant, students can find the specific half-life for this reaction—this value signifies how much time is required for the concentration of AB to reduce to 0.275 M under these conditions.

Integrated Rate Law

While the rate law gives an instantaneous picture of how reactant concentration affects the reaction rate, the integrated rate law tells us how reactant concentration changes over time. For a first-order reaction, the integrated rate law is: \( [AB] = [AB]_0 \cdot e^{-kt} \), where \( [AB]_0 \) is the initial concentration, \( [AB] \) is the concentration at time \( t \), and \( e \) is the base of the natural logarithm. This equation shows that the concentration of the reactant decreases exponentially over time.

In our exercise, applying the integrated rate law allows us to calculate the concentrations of A and B after 75 s. As AB dissociates into A and B in a 1:1 ratio, knowing the remaining concentration of AB at time \( t \) also indicates the amount of A and B formed—when \( [AB] \) decreases, the concentrations of A and B increase equivalently. Thus, students can apply the provided initial concentration of AB (\( 0.250 \text{ M} \) in our problem) and the time (75 s) to the formula to find the concentrations of products A and B at that moment. As the concentration of AB decreases, it's crucial to appreciate the exponential nature of this decline—which is not linear, unlike the plot for the rate constant determination.