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Chapter 13: Problem 56

OBJECTIVE. Relate half-life and rate constant, and calculate concentration- time behavior from the half-life of a first-order reaction. A Calculate the half-life of a first-order reaction if the concentration of the reactant is \(0.0451 \mathrm{M}\) at 30.5 seconds after the reaction starts and is \(0.0321 \mathrm{M}\) at 45.0 seconds after the reaction starts. How many seconds after the start of the reaction does it take for the reactant concentration to decrease to \(0.0100 \mathrm{M} ?\)

Short Answer

Expert verified
The half-life is approximately 29.6 seconds, and it takes about 94.9 seconds for the concentration to decrease to 0.0100 M.

Step by step solution

01

Determine the Rate Constant

For a first-order reaction, the rate law can be expressed as \( ln \left(\frac{[A]_0}{[A]} \right) = kt \), where \([A]_0\) is the initial concentration and \([A]\) is the concentration at time \(t\). For two time points, we have: \[ ln \left(\frac{0.0451}{0.0321}\right) = k(45.0 - 30.5) \]\[ k = \frac{ln\left(\frac{0.0451}{0.0321}\right)}{14.5} \]Calculating, \( ln(1.4059) = 0.3395 \) and dividing by 14.5 gives \( k \approx 0.0234 \, \text{s}^{-1} \).
02

Calculate the Half-Life

The half-life \( t_{1/2} \) for a first-order reaction is given by the equation \( t_{1/2} = \frac{0.693}{k} \). Using the calculated rate constant:\[ t_{1/2} = \frac{0.693}{0.0234} \]Calculate this to find \( t_{1/2} \approx 29.6 \, \text{s} \).
03

Find Time for Concentration to Reach 0.0100 M

Using the rate law expression again:\[ ln \left(\frac{0.0451}{0.0100}\right) = k t \]Substitute \( k = 0.0234 \, \text{s}^{-1} \) and solve for \( t \):\[ ln(4.51) = 1.507 \]\[ t = \frac{1.507}{0.0234} \]Calculating this gives \( t \approx 64.4 \, \text{s} \).
04

Determine Time After Reaction Start

The calculated time \( t \approx 64.4 \, \text{s} \) is from the start of the observation at \( 30.5 \, \text{s} \). Add this to the initial time to find the total time after the reaction starts:\[ t_{total} = 64.4 + 30.5 \approx 94.9 \, \text{s} \].

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Key Concepts

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

Rate Constant
The rate constant, often denoted as \( k \), is a crucial parameter in the study of chemical kinetics, particularly for first-order reactions. It provides insight into the speed of a reaction and is specific to each chemical reaction and set of conditions, such as temperature. For first-order reactions, the rate constant can be determined using the equation:
\[ ln \left( \frac{[A]_0}{[A]} \right) = kt \]
where \([A]_0\) is the initial concentration, \([A]\) is the concentration at any time \( t \), and \( k \) is the rate constant.Key points include:
  • For first-order reactions, the rate is directly proportional to the concentration of one reactant.
  • The rate constant \( k \) carries the units of \( \, ext{s}^{-1} \) for a first-order reaction.
  • A larger \( k \) suggests a faster reaction rate, while a smaller \( k \) implies a slower reaction.
Understanding the rate constant allows chemists to predict how quickly a reaction will reach a certain extent under given conditions.
Half-Life
The half-life, denoted as \( t_{1/2} \), is the time required for the concentration of a reactant to decrease to half its initial value. For first-order reactions, the half-life is independent of the initial concentration of the reactants and is derived using the relationship:
\[ t_{1/2} = \frac{0.693}{k} \]Where \( k \) is the rate constant.Important points about half-life:
  • The half-life remains constant throughout the reaction for first-order kinetics.
  • It provides a practical measure to compare how quickly different reactions consume reactants.
  • The formula highlights the inverse relationship between the rate constant and the half-life: as \( k \) increases, \( t_{1/2} \) decreases, indicating a faster reaction.
Calculating the half-life helps chemists estimate how long a reaction will continue to proceed and is particularly useful in processes like radioactive decay.
Concentration-Time Behavior
In first-order reactions, concentration-time behavior describes how the concentration of a reactant decreases over time. This change is expressed by:
\[ [A] = [A]_0 \cdot e^{-kt} \]where \([A]_0\) is the initial concentration, \( [A] \) is the concentration at time \( t \), and \( k \) is the rate constant.Here are some interpretive points:
  • The concentration of reactants decreases exponentially over time in a first-order reaction.
  • This behavior allows for the calculation of the time required to reach a specific concentration (such as a particular threshold for product formation).
  • Knowing how concentration changes with time is essential for predicting reaction outcomes and optimizing reaction conditions in chemical processes.
Understanding concentration-time behavior enables chemists to control and manipulate reaction conditions to achieve desired results efficiently.

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Most popular questions from this chapter

OBJECTIVE. Calculate the concentration-time behavior for a first-order reaction from the rate law and the rate constant. When formic acid is heated, it decomposes to hydrogen and carbon dioxide in a first-order decay. $$ \mathrm{HCOOH}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) $$ At \(550^{\circ} \mathrm{C}\), the half-life of formic acid is 24.5 minutes. (a) What is the rate constant, and what are its units? (b) How many seconds are needed for formic acid, initially \(0.15 M\), to decrease to 0.015 M?

Sketch the energy-level diagram for an endothermic reaction with low activation energy and for an exothermic reaction with high activation energy.

OBJECTIVE. Predict the experimental rate law from the mechanism and differentiate among possible reaction mechanisms by examining experimental rate data. Nitrogen dioxide can react with ozone to form dinitrogen pentoxide and oxygen. $$ \begin{array}{l} 2 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{3}(\mathrm{~g}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \\ \text { rate }=k\left[\mathrm{NO}_{2}\right]\left[\mathrm{O}_{3}\right] \end{array} $$ A two-step mechanism has been proposed. Identify the rate-limiting step. $$ \begin{array}{l} \mathrm{NO}_{2}+\mathrm{O}_{3} \rightarrow \mathrm{NO}_{3}+\mathrm{O}_{2} \\ \mathrm{NO}_{3}+\mathrm{NO}_{2} \rightarrow \mathrm{N}_{2} \mathrm{O}_{5} \end{array} $$

OBJECTIVE. Calculate the concentration-time behavior for a first-order reaction from the rate law and the rate constant. The initial concentration of the reactant in a first-order reaction \(\mathrm{A} \rightarrow\) products is \(0.64 M\) and the half-life is 30 seconds. (a) Calculate the concentration of the reactant 60 seconds after initiation of the reaction. (b) How long would it take for the concentration of the reactant to decrease to one-eighth its initial value? (c) How long would it take for the concentration of the reactant to decrease to \(0.040 \mathrm{~mol} \mathrm{~L}^{-1}\) ?

OBJECTIVE. Write the rate law from an elementary step and determine its molecularity. Write the rate law and the molecularity for each of the following elementary reactions. (a) \(\mathrm{HCl} \rightarrow \mathrm{H}+\mathrm{Cl}\) (b) \(\mathrm{H}_{2}+\mathrm{Cl} \rightarrow \mathrm{HCl}+\mathrm{H}\) (c) \(2 \mathrm{NO}_{2} \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}\)
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