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Chapter 15: Problem 53

This reaction was monitored as a function of time: \(A \longrightarrow B+C\) A plot of \(\ln [\mathrm{A}]\) versus time yields a straight line with slope \(-0.0045 / \mathrm{s}\) a. What is the value of the rate constant \((k)\) for this reaction at this temperature? b. Write the rate law for the reaction. c. What is the half-life? d. If the initial concentration of \(\mathrm{A}\) is \(0.250 \mathrm{M},\) what is the concentration after 225 s?

Short Answer

Expert verified
a. The rate constant \(k\) is 0.0045 / \mathrm{s}. b. The rate law is \(\text{rate} = k [\mathrm{A}]\). c. The half-life is \(t_{1/2} = \frac{\ln(2)}{k} = \frac{0.693}{0.0045 / \mathrm{s}} \). d. After 225 seconds, the concentration of \(A\) is \(0.250 \mathrm{M} \cdot e^{-0.0045 / \mathrm{s} \cdot 225 \mathrm{s}}\).

Step by step solution

01

Determining the Rate Constant

The slope of the plot of \(\ln [\mathrm{A}]\) versus time is equal to the negative of the rate constant \(k\) for a first-order reaction. Since the slope is given as \(-0.0045 / \mathrm{s}\), the value of the rate constant \(k\) is \(0.0045 / \mathrm{s}\).
02

Writing the Rate Law

For a first-order reaction, the rate law can be expressed as \(\text{rate} = k [\mathrm{A}]\). The rate at which \(A\) is consumed is directly proportional to its concentration at any point in time.
03

Calculating the Half-Life

The half-life of a first-order reaction is given by the equation \(t_{1/2} = \frac{\ln(2)}{k}\). Substituting the determined rate constant \(k = 0.0045 / \mathrm{s}\) into this formula will give the half-life.
04

Computing the Concentration after Time

For a first-order reaction, the concentration of \(A\) over time is given by \([\mathrm{A}] = [\mathrm{A}_0]e^{-kt}\), where \([\mathrm{A}_0]\) is the initial concentration. Using the initial concentration \(0.250 \mathrm{M}\) and time \(225 \mathrm{s}\), along with the rate constant \(k = 0.0045 / \mathrm{s}\), we can calculate \([\mathrm{A}]\) after 225 seconds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Constant
Understanding the rate constant is essential in the study of chemical reaction kinetics. It's a measure of the speed at which a reaction proceeds. Specifically, in the context of a first-order reaction such as
\(A \longrightarrow B + C\),
the rate constant \(k\) is determined from the slope of a plot of \(\ln [A]\) versus time. Since the given slope is \(-0.0045 / \mathrm{s}\), the rate constant is \(0.0045 / \mathrm{s}\). This value indicates how rapidly reactant \(A\) is converted to products \(B\) and \(C\) per second under the given conditions.
Rate Law
The rate law expresses the relationship between the reaction rate and the concentrations of reactants. For a first-order reaction like the one we're discussing, the rate law is given by:
\(\text{rate} = k [A]\).
This mathematical model states that the rate at which reactant \(A\) is consumed is directly proportional to its instantaneous concentration. As \(A\) decreases, the rate at which it reacts diminishes accordingly. It's a simple yet powerful way of quantifying how reactants change over time.
Chemical Reaction Half-Life
Half-life, symbolized by \(t_{1/2}\), is the time required for half of the reactant to be consumed in a chemical reaction. For first-order reactions, the half-life is independent of the initial concentration and solely depends on the rate constant \(k\). The equation
\(t_{1/2} = \frac{\ln(2)}{k}\)
allows us to calculate the half-life. Using the rate constant \(0.0045 / \mathrm{s}\), we find a specific numerical value representing the time in which the concentration of reactant \(A\) decreases by half. This concept is vital for predicting how long a reaction will take to reach a certain point.
Reaction Concentration Over Time
In first-order kinetics, the concentration of a reactant decreases exponentially over time. The key equation representing this behavior is:
\([A] = [A_0]e^{-kt}\).
Here, \([A_0]\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time elapsed. This formula allows us to predict the remaining concentration of reactant \(A\) at any given time. By plugging in the values for \([A_0]\), \(k\), and \(t\), we can determine the concentration of \(A\) after 225 seconds, illustrating how it diminishes as the reaction progresses towards completion.

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Most popular questions from this chapter

Iodine atoms combine to form \(\mathrm{I}_{2}\) in liquid hexane solvent with a rate constant of \(1.5 \times 10^{10} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\). The reaction is second order in I. since the reaction occurs so quickly, the only way to study the reaction is to create iodine atoms almost instanta- neously, usually by photochemical decomposition of \(\mathrm{I}_{2} .\) Suppose a flash of light creates an initial [I] concentration of \(0.0100 \mathrm{M} .\) How long will it take for \(95 \%\) of the newly created iodine atoms to recombine to form \(\mathrm{I}_{2} ?\)

The rate constant \((k)\) for a reaction was measured as a function of temperature. A plot of \(\ln k\) versus \(1 / T(\) in \(\mathrm{K})\) is linear and has a slope of \(-1.01 \times 10^{4} \mathrm{~K}\). Calculate the activation energy for the reaction.

Consider a simple reaction in which reactant A forms products: \(\mathrm{A} \longrightarrow\) products What is the rate law if the reaction is zero order with respect to A? First order? Second order? For each case, explain how a doubling of the concentration of A would affect the rate of reaction.

In this chapter, we have seen a number of reactions in which a single reactant forms products. For example, consider the following first-order reaction: \(\mathrm{CH}_{3} \mathrm{NC}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CN}(g)\) However, we also learned that gas-phase reactions occur through collisions. a. One possible explanation for how this reaction occurs is that two molecules of \(\mathrm{CH}_{3} \mathrm{NC}\) collide with each other and form two molecules of the product in a single elementary step. If that were the case, what reaction order would you expect? b. Another possibility is that the reaction occurs through more than one step. For example, a possible mechanism involves one step in which the two \(\mathrm{CH}_{3} \mathrm{NC}\) molecules collide, resulting in the "activation" of one of them. In a second step, the activated molecule goes on to form the product. Write down this mechanism and determine which step must be rate determining in order for the kinetics of the reaction to be first order. Show explicitly how the mechanism predicts first-order kinetics.

Suppose that the reaction \(\mathrm{A} \longrightarrow\) products is exothermic and has an activation barrier of \(75 \mathrm{~kJ} / \mathrm{mol} .\) Sketch an energy diagram showing the energy of the reaction as a function of the progress of the reaction. Draw a second energy curve showing the effect of a catalyst.
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