Textbook Question
This reaction was monitored as a function of time: A → B + C A plot of ln[A] versus time yields a straight line with slope -0.0045/s. b. Write the rate law for the reaction.
Ch.14 - Chemical Kinetics
Chapter 14, Problem 54a
This reaction was monitored as a function of time: AB → A + B A plot of 1/[AB] versus time yields a straight line with a slope of +0.55/Ms.
a. What is the value of the rate constant (k) for this reaction at this temperature?
Verified step by step guidance1
Recognize that the plot of \( \frac{1}{[\text{AB}]} \) versus time indicates a second-order reaction.
For a second-order reaction, the integrated rate law is \( \frac{1}{[\text{AB}]} = kt + \frac{1}{[\text{AB}]_0} \), where \( k \) is the rate constant.
In the equation of a straight line \( y = mx + c \), the slope \( m \) corresponds to the rate constant \( k \) for a second-order reaction.
Given that the slope of the line is \(+0.55 \text{ M}^{-1}\text{s}^{-1}\), this value is the rate constant \( k \).
Therefore, the rate constant \( k \) for this reaction is \( 0.55 \text{ M}^{-1}\text{s}^{-1} \).
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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 of the form AB → A + B, the rate can be determined by the rate constant (k) and the concentration of AB raised to a power that reflects its order. Understanding the rate law is essential for calculating the rate constant from experimental data.
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Rate Law Fundamentals
Integrated Rate Laws
Integrated rate laws relate the concentration of reactants to time. For a second-order reaction, the integrated rate law is expressed as 1/[AB] = kt + 1/[AB]₀, where k is the rate constant, t is time, and [AB]₀ is the initial concentration. The linear relationship observed in the plot of 1/[AB] versus time indicates that the reaction follows second-order kinetics.
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Slope of a Line
In the context of a linear plot, the slope represents the rate constant (k) for the reaction. For the plot of 1/[AB] versus time, the slope is equal to k, which is given as +0.55/Ms in this case. This value directly provides the rate constant needed to describe the kinetics of the reaction.
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Related Practice
Textbook Question
This reaction was monitored as a function of time: A → B + C A plot of ln[A] versus time yields a straight line with slope -0.0045/s. c. What is the half-life?
Textbook Question
This reaction was monitored as a function of time: A → B + C A plot of ln[A] versus time yields a straight line with slope -0.0045/s. d. If the initial concentration of A is 0.250 M, what is the concentration after 225 s?
Textbook Question
This reaction was monitored as a function of time: AB → A + B A plot of 1/[AB] versus time yields a straight line with a slope of +0.55/Ms. b. Write the rate law for the reaction.
Textbook Question
This reaction was monitored as a function of time: AB → A + B A plot of 1/[AB] versus time yields a straight line with a slope of +0.55/Ms. c. What is the half-life when the initial concentration is 0.55 M?
Textbook Question
This reaction was monitored as a function of time: AB → A + B A plot of 1/[AB] versus time yields a straight line with a slope of +0.55/Ms.
d. If the initial concentration of AB is 0.250 M, and the reaction mixture initially contains no products, what are the concentrations of A and B after 75 s?