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(a) Define the following symbols that are encountered in rate equations for the generic reaction \(\mathrm{A} \longrightarrow \mathrm{B}:[\mathrm{A}]_{0}, t_{1 / 2}[\mathrm{~A}]_{t}, k\) (b) What quantity, when graphed versus time, will yield a straight line for a first-order reaction? (c) How can you calculate the rate constant for a first- order reaction from the graph you made in part (b)?

Short Answer

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(a) For the reaction A → B: \([\mathrm{A}]_{0}\) is the initial concentration of A, \(t_{1 / 2}\) is the half-life of the reaction, \([\mathrm{A}]_{t}\) is the concentration of A at time \(t\), and \(k\) is the rate constant. (b) Plotting \(ln([\mathrm{A}]_{t})\) against time (\(t\)) for a first-order reaction will yield a straight line. (c) Calculate the rate constant (\(k\)) by finding the absolute value of the negative slope from the graph obtained in part (b).

Step by step solution

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(a) Define the symbols

For the generic reaction A → B, the rate equation symbols are as follows: 1. \([\mathrm{A}]_{0}\): Initial concentration of reactant A at the beginning of the reaction (\(t = 0\)). 2. \(t_{1 / 2}\): Half-life of the reaction, which is the time required for the concentration of reactant A to reduce to half its initial value. 3. \([\mathrm{A}]_{t}\): Concentration of reactant A at a specific time \(t\). 4. \(k\): Rate constant, which is a proportionality constant for the reaction rate and the concentrations of the reactants.
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(b) Straight line graph for a first-order reaction

For a first-order reaction, the integrated rate law is given by: \[ln([\mathrm{A}]_{t}) = ln([\mathrm{A}]_{0}) - kt\] When plotting the natural logarithm of the concentration of reactant A, \(ln([\mathrm{A}]_{t})\), against time, \(t\), for a first-order reaction, we will obtain a straight line.
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(c) Calculating the rate constant from the graph

To calculate the rate constant, \(k\), for a first-order reaction from the graph in part (b), do the following: 1. Find the slope of the straight line obtained by plotting \(ln([\mathrm{A}]_{t})\) against \(t\). The slope will be equal to the negative value of the rate constant. 2. The slope can be calculated using the formula: \[\text{Slope} = \frac{\Delta y}{\Delta x}\] where \(y\) represents the natural logarithm of the concentration of reactant A and \(x\) is time. 3. To find the rate constant, \(k\), take the absolute value of the slope: \[k = |- \text{Slope}|\] By following these steps, you can calculate the rate constant for a first-order reaction from the straight line graph obtained in part (b).

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Key Concepts

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

Understanding Rate Equations in Chemical Kinetics
In the context of chemical kinetics, rate equations play a crucial role as they express the relationship between the rate of a chemical reaction and the concentrations of reactants. For a basic reaction where substance A transforms into substance B, the rate equation encapsulates how quickly this conversion occurs. With the symbols introduced such as \( [A]_{0} \) for the initial concentration and \( [A]_{t} \) for the concentration at time \( t \), we can predict and analyze the reaction's progress over time. \
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Particularly in first-order reactions, where the rate is directly proportional to the concentration of one reactant, these equations help in determining crucial parameters like the reaction rate constant \( k \). This constant reveals the speed at which reactant A is consumed and thus dictates how fast the reaction moves forward. Understanding these fundamental relationships allows chemists to predict outcomes, control reaction speeds, and even optimize industrial processes.
Digging Deeper into Chemical Kinetics
Chemical kinetics encompasses the study of reaction rates and the factors affecting them. It helps us understand the speed of reactions, the steps involved, and how different conditions like temperature, pressure, and concentration impact the rate. \
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In kinetic analysis, the rate constant \( k \) surfaces as a critical component that ties the chemical behavior to time. Whether a reaction zooms to completion in a flash or leisurely proceeds over years, the rate constant offers quantitative insight. It's influenced by the particular nature of the reactants, the activation energy required, and environmental factors. From catalytic processes in the automotive industry to enzyme reactions in biological systems, kinetics principles mold our understanding of how chemical reactions proceed.
The Significance of Reaction Half-Life
The reaction half-life, denoted as \( t_{1/2} \) in kinetics, is a concept that beautifully illustrates the intuitive nature of first-order reactions. This value represents the time required for the concentration of a reactant to reach half of its initial concentration \( [A]_{0} \). For first-order reactions, the half-life remains constant regardless of the starting concentration, which is a unique property distinguishing it from other reaction orders. \
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Half-lives are especially valuable in fields like pharmacology where they guide dosage intervals for medications. Understanding half-life is also essential in environmental science for predicting how long pollutants last in nature. By mastering this concept, one gains an indispensable tool for maneuvering through chemical processes and their temporal aspects.
Integrated Rate Law and Its Application
The integrated rate law for first-order reactions bridges the gap between theoretical kinetics and practical data analysis. It presents the concentration of a reactant at any time \( t \) as a function of its original concentration and the elapsed time. The equation \[\ln([A]_{t}) = \ln([A]_{0}) - kt\] showcases this relationship. What's powerful about this law is its ability to transform nonlinear concentration data into a linear plot, allowing for straightforward interpretation and calculation of the rate constant. \
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The slope of the line in a plot of \( \ln([A]_{t}) \) versus time gives the rate constant \( k \) directly, minus the sign. This way of graphical analysis is both intuitive and practical, transforming kinetic study from abstract to accessible. Whether in academic research or industrial practice, the integrated rate law is an analytical cornerstone for chemists working with first-order reactions.

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Most popular questions from this chapter

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The following data were collected for the rate of disappearance of \(\mathrm{NO}\) in the reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) : $$ \begin{array}{llll} \hline \text { Experiment } & {[\mathrm{NO}](M)} & {\left[\mathrm{O}_{2}\right](M)} & \text { Initial Rate }(M / s) \\ \hline 1 & 0.0126 & 0.0125 & 1.41 \times 10^{-2} \\ 2 & 0.0252 & 0.0125 & 5.64 \times 10^{-2} \\ 3 & 0.0252 & 0.0250 & 1.13 \times 10^{-1} \end{array} $$ (a) What is the rate law for the reaction? (b) What are the units of the rate constant? (c) What is the average value of the rate constant calculated from the three data sets? (d) What is the rate of disappearance of \(\mathrm{NO}\) when \([\mathrm{NO}]=0.0750 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.0100 \mathrm{M} ?(\mathrm{e})\) What is the rate of disappearance of \(\mathrm{O}_{2}\) at the concentrations given in part ( \(\mathrm{d}\) )?

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