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A reaction of the form \(a \mathrm{~A} \longrightarrow\) Products is second order with a rate constant of \(0.413 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s}) .\) What is the half-life, in seconds, of the reaction if the initial concentration of \(\mathrm{A}\) is \(5.25 \times\) \(10^{-3} \mathrm{~mol} / \mathrm{L} ?\)

Short Answer

Expert verified
The half-life of the reaction is approximately 461.5 seconds.

Step by step solution

01

Understanding Reaction Order and Equation

For a second-order reaction, the rate law can be written as \( r = k[A]^2 \), where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant. The formula for the half-life of a second-order reaction is \( t_{1/2} = \frac{1}{k[A]_0} \), where \([A]_0\) is the initial concentration of the reactant.
02

Substitute Known Values into the Half-Life Formula

We know \( k = 0.413 \, \mathrm{L}/(\mathrm{mol} \cdot \mathrm{s}) \) and \( [A]_0 = 5.25 \times 10^{-3} \; \text{mol/L} \). Substitute these values into the half-life formula: \[ t_{1/2} = \frac{1}{0.413 \cdot 5.25 \times 10^{-3}}. \]
03

Calculate the Half-Life

Perform the calculation: \[ t_{1/2} = \frac{1}{0.413 \times 5.25 \times 10^{-3}} = \frac{1}{0.00216675} \approx 461.5 \text{ seconds.} \] Thus, the half-life of the reaction is approximately 461.5 seconds.

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Key Concepts

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

Half-Life Calculation
The concept of half-life is crucial in understanding how chemical reactions progress. In the context of a second-order reaction, the half-life refers to the time it takes for the concentration of a reactant to reduce to half of its initial value. This is different from first-order reactions where the half-life is constant and independent of concentration. For second-order reactions, the half-life equation is given by:
  • \( t_{1/2} = \frac{1}{k[A]_0} \)
where \( t_{1/2} \) represents the half-life, \( k \) is the rate constant, and \( [A]_0 \) is the initial concentration of the reactant. Notice that the half-life is inversely proportional to the initial concentration. This means the half-life will change as the reaction proceeds, making it shorter as the concentration decreases. Understanding this relationship helps to appreciate the dynamics of second-order reactions.
Rate Law
The rate law for a reaction describes how the rate is related to the concentration of reactants. For a second-order reaction, the rate law can be expressed simply as:\( r = k[A]^2 \).
Here:
  • \( r \) is the reaction rate, which indicates how fast the reactants are converted to products.
  • \( k \) represents the rate constant, which is a proportionality constant related to the speed of the reaction under specific conditions.
  • \( [A] \) is the concentration of the reactant.
This squared dependency on concentration highlights that the rate of a second-order reaction significantly increases as the reactant concentration rises. It underscores the sensitivity of second-order reactions to changes in concentration, which in practice means doubling the concentration quadruples the rate.
Initial Concentration
Initial concentration, denoted as \([A]_0\), refers to the concentration of the reactant at the start of the reaction. It plays a pivotal role in predicting how the reaction will behave over time, especially for second-order reactions.
In the equation for the half-life of a second-order reaction, \( t_{1/2} = \frac{1}{k[A]_0} \), the initial concentration \([A]_0\) directly affects the half-life. This means if you start with a higher concentration, the half-life becomes shorter. This influence of initial concentration is a distinctive feature of second-order reactions compared to first-order, where the half-life remains constant regardless of the initial concentration.Understanding the concept of initial concentration allows us to manipulate and control the rate and duration of reactions in practical applications such as pharmaceuticals and chemical manufacturing.
Reaction Rate Constant
The reaction rate constant, \( k \), is a fundamental factor in the kinetics of chemical reactions. It encapsulates information about the intrinsic speed of a reaction at a given temperature and is specific to the type of reaction being studied. In the rate law expression for a second-order reaction \( r = k[A]^2 \), \( k \) comprises the units \( L/(mol \cdot s) \), distinguishing it from first-order reactions where the units are typically \( s^{-1} \).
The value of \( k \) not only illuminates how swift a reaction progresses but also plays an integral role in calculating the half-life. For the given problem, using \( k = 0.413 \mathrm{~L}/(\mathrm{mol} \cdot \mathrm{s}) \), provided a direct path to finding the half-life by influencing the denominator in the half-life formula. Tuning \( k \) enables scientists to modify reaction conditions, paving the way to diverse industrial and laboratory applications.

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Most popular questions from this chapter

Draw a potential-energy diagram for an uncatalyzed exothermic reaction. On the same diagram, indicate the change that results on the addition of a catalyst. Discuss the role of a catalyst in changing the rate of reaction.

The hypothetical reaction \(\mathrm{A}+2 \mathrm{~B} \longrightarrow\) Products has the rate law Rate \(=k[\mathrm{~A}]^{2}[\mathrm{~B}]^{3}\). If the reaction is run two separate times, holding the concentration of A constant while doubling the concentration of \(\mathrm{B}\) from one run to the next, how would the rate of the second run compare to the rate of the first run?

Identify the molecularity of each of the following elementary reactions. a. \(\mathrm{NO}+\mathrm{O}_{3} \longrightarrow \mathrm{NO}_{2}+\mathrm{O}_{2}\) b. \(\mathrm{NOCl}_{2}+\mathrm{NO} \longrightarrow 2 \mathrm{NOCl}\) C. \(\mathrm{O}_{3} \longrightarrow \mathrm{O}_{2}+\mathrm{O}\) d. \(\mathrm{H}+\mathrm{H}+\mathrm{N}_{2} \longrightarrow \mathrm{H}_{2}+\mathrm{N}_{2}^{*}\)

At high temperature, the reaction $$\mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g)$$ is thought to occur in a single step. What should be the rate law in that case?

Methyl isocyanide, \(\mathrm{CH}_{3} \mathrm{NC}\), isomerizes, when heated, to give acetonitrile (methyl cyanide), \(\mathrm{CH}_{3} \mathrm{CN}\). $$\mathrm{CH}_{3} \mathrm{NC}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CN}(g)$$ The reaction is first order. At \(230^{\circ} \mathrm{C}\), the rate constant for the isomerization is \(6.3 \times 10^{-4} / \mathrm{s}\). What is the half- life? How long would it take for the concentration of \(\mathrm{CH}_{3} \mathrm{NC}\) to decrease to \(50.0 \%\) of its initial value? to \(25.0 \%\) of its initial value?

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