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Define the half-life of a reaction. Explain on the molecular level why the half-life of a first-order reaction is constant.

Short Answer

Expert verified
The half-life is the time taken for a reactant's concentration to halve. In first-order reactions, it is constant because the rate depends only on the concentration.

Step by step solution

01

Define Half-Life

The half-life of a reaction is the time required for the concentration of a reactant to decrease to half of its initial concentration.
02

Understand First-Order Reactions

For a first-order reaction, the rate of the reaction is directly proportional to the concentration of the reactant. This can be expressed with the rate law: \[ \text{Rate} = k[A] \] where \( k \) is the rate constant and \([A]\) is the concentration of the reactant.
03

Derive the Half-Life Formula for First-Order Reactions

The integrated rate law for a first-order reaction is: \[ [A_t] = [A_0] e^{-kt} \] where \([A_t]\) is the concentration at time \(t\), \([A_0]\) is the initial concentration, and \(k\) is the rate constant. To find the half-life \(t_{1/2}\), set \([A_{t_{1/2}}] = \frac{[A_0]}{2} \) and solve for \( t \):\[ \frac{[A_0]}{2} = [A_0] e^{-kt_{1/2}} \]Divide both sides by \([A_0]\):\[ \frac{1}{2} = e^{-kt_{1/2}} \]Take the natural logarithm of both sides:\[ \text{ln} \frac{1}{2} = -kt_{1/2} \]Since \(\text{ln} \frac{1}{2} = - \text{ln} 2\), we have:\[ - \text{ln} 2 = -kt_{1/2} \]Therefore, solving for \( t_{1/2}\):\[ t_{1/2} = \frac{\text{ln} 2}{k} \]
04

Explain the Constant Half-Life on the Molecular Level

In a first-order reaction, the fraction of reactants that react in a given time period is constant because the rate depends only on the concentration. As molecules react and the concentration decreases, the rate decreases proportionally so that it always takes the same amount of time for the concentration to halve.

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Key Concepts

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

first-order reaction kinetics
First-order reaction kinetics describe reactions where the reaction rate is directly proportional to the concentration of one reactant. This means as the reactant's concentration decreases, so does the rate of the reaction.

To visualize this, consider the reaction rate law for a first-order reaction: \[\text{Rate} = k[A]\],
where \[k\] is the rate constant and \[A\] is the concentration of the reactant.
Imagine you start with a certain concentration of reactant. As the reaction progresses and the quantity of reactant diminishes, the reaction rate falls proportionately. This linear relationship keeps the study of these reactions simple but powerful in understanding how time and concentration are interconnected.
rate constant
The rate constant, symbolized as \[k\], is a crucial factor in the kinetics of a reaction. It determines the speed of the reaction at any given concentration of reactant.

Think of \[k\] as a multiplier in the rate equation: \[\text{Rate} = k[A]\].
Its value depends on factors like temperature and the presence of a catalyst. Higher temperatures or the addition of a catalyst typically increase the rate constant, making reactions proceed faster.

For a first-order reaction, \[k\] is calculated using the formula for the half-life: \[t_{1/2} = \frac{\text{ln} 2}{k}\] going further, knowing \[k\] allows you to predict how long it will take for a reaction to reach a certain level of completion, which is invaluable in both laboratory and industrial settings.
integrated rate law
The integrated rate law for a first-order reaction helps to connect the concentration of reactants at a given time directly to the initial concentration and the time elapsed. This is particularly useful for calculating how much reactant remains after a certain period.

For a first-order reaction, the integrated rate law is given by: \[ [A_t] = [A_0] e^{-kt}\]. Here, \[ [A_t]\] is the concentration of the reactant at time \[t\], and \[ [A_0]\] is the initial concentration.

You can use this equation to see how the concentration changes over time. By inputting various time values into the equation, you can plot a decay curve, which showcases the exponential decrease in reactant concentration. This is essential for understanding how quickly a reaction proceeds and helps in predicting the duration required to reach specific concentrations.
reaction rate
The reaction rate is a measure of how quickly a reactant is consumed or a product is formed in a chemical reaction. For first-order reactions, the rate is directly tied to the concentration of one reactant.

This relationship is expressed as: \[ \text{Rate} = k[A] \], meaning that if the concentration of \[A\] changes, the rate changes proportionally.

The concept is intuitive: imagine you have a pile of chemical \[A\]; the larger the pile, the more 'events' (reactions) you can observe per unit of time. Hence, as \[A\] decreases, so does the observable rate of these reactions. Understanding this concept sheds light on why the half-life of first-order reactions remains constant, as a fixed fraction of the reactant is consumed uniformly over equal time intervals.

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Most popular questions from this chapter

Archaeologists can determine the age of an artifact made of wood or bone by measuring the amount of the radioactive isotope \({ }^{14} \mathrm{C}\) present in the object. The amount of this isotope decreases in a first-order process. If \(15.5 \%\) of the original amount of \({ }^{14} \mathrm{C}\) is present in a wooden tool at the time of analysis, what is the age of the tool? The half- life of \({ }^{14} \mathrm{C}\) is \(5730 \mathrm{yr}\).

How does an increase in temperature affect the rate of a reaction? Explain the two factors involved.

The citric acid cycle is the central reaction sequence in the cellular metabolism of humans and many other organisms. One of the key steps is catalyzed by the enzyme isocitrate dehydrogenase and the oxidizing agent \(\mathrm{NAD}^{+}\). In yeast, the reaction is eleventh order. Rate \(=k[\) enzyme \(]\) [isocitrate] \(^{4}[\mathrm{AMP}]^{2}\left[\mathrm{NAD}^{+}\right]^{m}\left[\mathrm{Mg}^{2+}\right]^{2}\) What is the order with respect to \(\mathrm{NAD}^{+} ?\)

16.103 Even when a mechanism is consistent with the rate law, later work may show it to be incorrect. For example, the reaction between hydrogen and iodine has this rate law: rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]\). The long-accepted mechanism had a single bimolecular step; that is, the overall reaction was thought to be elementary: \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)\) In the 1960 s, however, spectroscopic evidence showed the presence of free I atoms during the reaction. Kineticists have since proposed a three-step mechanism: (1) \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) [fast] (2) \(\mathrm{H}_{2}(g)+\mathrm{I}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{I}(g)\) [fast] (3) \(\mathrm{H}_{2} \mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g) \quad[\) slow \(]\) Show that this mechanism is consistent with the rate law.

The decomposition of NOBr is studied manometrically because the number of moles of gas changes; it cannot be studied colorimetrically because both \(\mathrm{NOBr}\) and \(\mathrm{Br}_{2}\) are reddish brown: $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ Use the data below to answer the following: (a) Determine the average rate over the entire experiment. (b) Determine the average rate between 2.00 and \(4.00 \mathrm{~s}\). (c) Use graphical methods to estimate the initial reaction rate. (d) Use graphical methods to estimate the rate at \(7.00 \mathrm{~s}\). (e) At what time does the instantaneous rate equal the average rate over the entire experiment? $$ \begin{array}{cc} \text { Time (s) } & \text { [NOBr] (mol/L) } \\ \hline 0.00 & 0.0100 \\ 2.00 & 0.0071 \\ 4.00 & 0.0055 \\ 6.00 & 0.0045 \\ 8.00 & 0.0038 \\ 10.00 & 0.0033 \end{array} $$

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