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What effect does doubling the initial concentration of a reactant have on the half-life in a reaction that is second order in the reactant?

Short Answer

Expert verified
Answer: Doubling the initial concentration of a reactant in a second-order reaction will cause the half-life to decrease by half.

Step by step solution

01

Write the rate law for a second-order reaction

For a second-order reaction, the rate law is given by: Rate = k[A]虏 where Rate is the reaction rate, k is the rate constant, and [A] is the concentration of the reactant A.
02

Write the integrated rate law for a second-order reaction

To find the integrated rate law, we need to integrate the rate law with respect to time. For second-order reactions, the integrated rate law is: 1/[A] - 1/[A]鈧 = kt where [A]鈧 is the initial concentration of the reactant A, [A] is the concentration at time t and k is the rate constant.
03

Derive the half-life formula for a second-order reaction

Half-life (t鈧/鈧) is the time it takes for the concentration of a reactant to fall to half its initial value. Thus, at t = t鈧/鈧, [A] = [A]鈧 / 2. We can substitute this into the integrated rate law equation to derive the half-life formula: 1/([A]鈧/2) - 1/[A]鈧 = kt鈧/鈧
04

Simplify the half-life formula and solve for t鈧/鈧

Now we will simplify the half-life formula: 2/[A]鈧 - 1/[A]鈧 = kt鈧/鈧 1/[A]鈧 = kt鈧/鈧 t鈧/鈧 = 1/(k[A]鈧) The half-life formula is: t鈧/鈧 = 1/(k[A]鈧)
05

Analyze the effect of doubling the initial concentration on half-life

Now, let's see how the half-life changes when we double the initial concentration: Original half-life: t鈧/鈧 = 1/(k[A]鈧) New half-life with doubled initial concentration: t鈧/鈧' = 1/(k(2[A]鈧)) t鈧/鈧' = 1/(2k[A]鈧) Comparing the original half-life with the new half-life after doubling the initial concentration, we see that: t鈧/鈧' = (1/2)t鈧/鈧 Thus, doubling the initial concentration of a reactant in a second-order reaction will cause the half-life to decrease by half.

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Key Concepts

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

Rate Law
In chemical kinetics, the rate law is a mathematical equation that describes the rate of a chemical reaction. For second-order reactions, the rate law takes the form:
  • Rate = k[A]虏
Here, 'Rate' is the speed at which the reaction proceeds, 'k' is the rate constant unique to each reaction, and '[A]' represents the concentration of reactant A. The squared concentration signifies a second-order reaction, where the reaction rate is proportional to the square of the reactant's concentration. Understanding the rate law helps predict how changes in concentration affect reaction speed, laying the foundation for deeper analysis involving integrated rate laws and other kinetics concepts. This would be particularly useful for anyone looking to control reaction speeds in industrial or laboratory settings.
Integrated Rate Law
The integrated rate law combines the information from the rate law to describe how reactant concentrations change with time. For a second-order reaction, it is derived by integrating the rate law equation:
  • 1/[A] - 1/[A]鈧 = kt
In this formula, [A]鈧 is the initial concentration of reactant A, [A] is its concentration at time t, and 'k' is once again the rate constant. The integrated rate law simplifies tasks such as determining the time required for a reactant to reach a specific concentration. It also provides insights into the kinetics of the reaction, showing how concentration decreases over time. Being familiar with this concept is crucial for tasks like predicting reaction progress and estimating time needed for completing a reaction.
Half-Life
The half-life of a reaction helps determine how long it takes for the concentration of a reactant to reduce to half its initial amount. For second-order reactions, the half-life formula is:
  • t鈧/鈧 = 1/(k[A]鈧)
This formula reveals that the half-life is inversely proportional to the initial concentration [A]鈧. Thus, unlike first-order reactions, the half-life varies with initial concentration in a second-order reaction. When you double the initial concentration, the half-life becomes half of the original half-life, illustrating the dynamic relationship between concentration and reaction time. Understanding this is essential for practical applications such as determining the shelf life of a chemical or managing reactants in a lab setting.
Reaction Kinetics
Reaction kinetics involves the study of the speeds of chemical processes and how they change under different conditions. In second-order reactions, kinetics explores how variations in reactant concentration and other factors affect the rate and overall progress of a reaction.
  • Factors such as temperature, concentration, and catalysts play a significant role in reaction kinetics.
  • Kinetics provides a detailed map of how a reaction proceeds over time.
This field aids in understanding the mechanisms behind chemical reactions, allowing for the development of more efficient industrial processes and innovations in fields like pharmaceuticals. Learning about reaction kinetics equips students and professionals with skills to manipulate and optimize reactions for desired outcomes.

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Most popular questions from this chapter

Why doesn't a quadrupling of the rate correspond to a reaction order of \(4-\) for example, Rate \(\propto[\mathrm{NO}]^{4} ?\)

The rate constant for the reaction of ozone with oxygen atoms was determined at four temperatures. Calculate the activation energy and frequency factor \(A\) for the reaction $$\mathrm{O}(g)+\mathrm{O}_{3}(g) \rightarrow 2 \mathrm{O}_{2}(g)$$ given the following data: $$\begin{array}{cc}T(\mathrm{K}) & k\left[\mathrm{cm}^{3} /(\text { molecule } \cdot \mathrm{s})\right] \\\250 & 2.64 \times 10^{-4} \\\\\hline 275 & 5.58 \times 10^{-4} \\\\\hline 300 & 1.04 \times 10^{-3} \\\\\hline 325 & 1.77 \times 10^{-3} \\\\\hline\end{array}$$

Using data in Appendix \(4,\) calculate \(\Delta H^{\circ}\) for the reaction $$\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{NO}_{2}(g)$$

Kinetic data for the reaction \(\mathrm{Cl}_{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{ClO}(g)\) are summarized in the following table. $$\begin{array}{cc}\text { Time }(\mu \mathrm{s}) & {\left[\mathrm{Cl}_{2} \mathrm{O}_{2}\right](\mathrm{M})} \\\\\hline 0 & 6.60 \times 10^{-8} \\\\\hline 172 & 5.68 \times 10^{-8} \\\\\hline 345 & 4.89 \times 10^{-8} \\\\\hline 517 & 4.21 \times 10^{-8} \\\\\hline 690 & 3.62 \times 10^{-8} \\\\\hline 862 & 3.12 \times 10^{-8} \\\\\hline\end{array}$$ Determine the value of the rate constant. b. Determine \(t_{1 / 2}\) for the decomposition of \(\mathrm{Cl}_{2} \mathrm{O}_{2}\).

The chemistry of smog formation includes \(\mathrm{NO}_{3}\) as an intermediate in several reactions. a. If \(\Delta\left[\mathrm{NO}_{3}\right] / \Delta t=-2.2 \times 10^{5} \mathrm{m} M / \mathrm{min}\) in the following reaction, what is the rate of formation of \(\mathrm{NO}_{2} ?\) $$\mathrm{NO}_{3}(g)+\mathrm{NO}(g) \rightarrow 2 \mathrm{NO}_{2}(g)$$ b. What is the rate of change of \(\left[\mathrm{NO}_{2}\right]\) in the following$$ \begin{array}{l} \text { reaction if } \Delta\left[\mathrm{NO}_{3}\right] / \Delta t=-2.3 \mathrm{mM} / \mathrm{min} \text { ? } \\ \qquad 2 \mathrm{NO}_{3}(g) \rightarrow 2 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \end{array}$$

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