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.
a. What is the value of the rate constant (k) for this reaction at this temperature?
Ch.14 - Chemical Kinetics
Chapter 14, Problem 54d
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?
Verified step by step guidance1
Identify the order of the reaction. Since a plot of \( \frac{1}{[\text{AB}]} \) versus time is a straight line, the reaction is second-order with respect to AB.
Use the second-order integrated rate law: \( \frac{1}{[\text{AB}]} = kt + \frac{1}{[\text{AB}]_0} \), where \( k \) is the rate constant, \( t \) is time, and \( [\text{AB}]_0 \) is the initial concentration.
Substitute the given values into the integrated rate law: \( \frac{1}{[\text{AB}]} = 0.55 \times 75 + \frac{1}{0.250} \).
Solve for \( [\text{AB}] \) to find the concentration of AB at 75 seconds.
Since the reaction is \( \text{AB} \rightarrow \text{A} + \text{B} \), the change in concentration of AB is equal to the increase in concentration of A and B. Calculate \( [\text{A}] = [\text{B}] = [\text{AB}]_0 - [\text{AB}] \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Rate Law and Reaction Order
The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. For the reaction AB → A + B, the straight line plot of 1/[AB] versus time indicates that the reaction is second-order with respect to AB. This means that the rate of reaction is proportional to the square of the concentration of AB, which is crucial for determining how concentrations change over time.
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Rate Law Fundamentals
Integrated Rate Law for Second-Order Reactions
For a second-order reaction, the integrated rate law is given by 1/[AB] = kt + 1/[AB]₀, where k is the rate constant, t is time, and [AB]₀ is the initial concentration. This equation allows us to calculate the concentration of AB at any time, which can then be used to find the concentrations of the products A and B, since they are produced in a 1:1 ratio.
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Stoichiometry of the Reaction
Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. In this case, for every mole of AB that reacts, one mole of A and one mole of B are produced. Therefore, knowing the change in concentration of AB allows us to directly calculate the concentrations of A and B after a given time, based on the initial concentration and the amount reacted.
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Related Practice
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
The decomposition of XY is second order in XY and has a rate constant of 7.02⨉10-3 M-1• s-1 at a certain temperature. a. What is the half-life for this reaction at an initial concentration of 0.100 M?