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Chapter 17: Problem 80

This problem is appropriate only if you have taken calculus. Show that for a zero-order reaction in a single reactant, A, that the integrated rate expression is given by the expression \([\mathrm{A}]=[\mathrm{A}]_{0}-k t\), and the halflife is given by \(t_{1 / j}=[\mathrm{A}]_{0} / 2 k\). What is the test plot for a zero-order reaction?

Short Answer

Expert verified
The integrated rate law is \([A] = [A]_0 - kt \), and the halflife is \(t_{1/2} = [A]_0/2k\). Plot \([A]\) versus time for a test plot.

Step by step solution

01

Understanding Zero-Order Reactions

A zero-order reaction is one whose rate does not depend on the concentration of the reactant(s). The rate can be expressed as \( \text{Rate} = k \), where \( k \) is the rate constant. The differential rate law is \( \frac{d[A]}{dt} = -k \).
02

Integrate the Rate Law

Integrate the differential rate law, \( \frac{d[A]}{dt} = -k \), with respect to time \( t \). We have:\[ \int d[A] = - \int k \, dt \]This gives \( [A] = -kt + C \), where \( C \) is the integration constant.
03

Solve for the Integration Constant

To find \( C \), use the initial condition at \( t = 0 \): \([A] = [A]_0\). Substitute to find \( C \):\[ [A]_0 = -k(0) + C \]So, \( C = [A]_0 \).
04

Write the Integrated Rate Law

Substitute \( C = [A]_0 \) back into the integrated expression:\[ [A] = -kt + [A]_0 \]Rearrange to get the integrated rate law:\[ [A] = [A]_0 - kt \]
05

Determine Halflife for Zero-Order Reaction

Halflife \( t_{1/2} \) is the time taken for \( [A] \) to decrease to half of its initial value. Set \( [A] = [A]_0/2 \) and solve:\[ [A]_0/2 = [A]_0 - kt_{1/2} \]\[ kt_{1/2} = [A]_0 - [A]_0/2 \]\[ kt_{1/2} = [A]_0/2 \]\[ t_{1/2} = \frac{[A]_0}{2k} \]
06

Determine the Test Plot for Zero-Order Reaction

In a zero-order reaction, plotting concentration \( [A] \) versus time \( t \) yields a straight line with a slope of \(-k\), confirming the zero-order kinetics.

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Key Concepts

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

Integrated Rate Law
In the realm of chemical kinetics, understanding how concentrations change over time is crucial. The *integrated rate law* comes into play when calculating these changes in concentration. Specifically for a zero-order reaction, the rate of reaction is constant, indicated by the expression \( [A] = [A]_0 - kt \). Here,
  • \([A]_0\) is the initial concentration of the reactant.
  • \(k\) is the rate constant.
  • \(t\) represents time.
For zero-order reactions, the concentration of the reactant decreases linearly with time, resulting in a straight-line graph when plotted against time. This linear relationship, often expressed as \( y = mx + b \) in mathematical terms, showcases \([A]\) as \(y\), * \(t\) as \(x\),* \(-k\) as the slope \(m\), * and \([A]_0\) as the intercept \(b\).

This makes it relatively easy to determine changes in concentration without needing to measure the concentration repeatedly.
Halflife
When studying zero-order reactions, the concept of *halflife* becomes essential. The halflife
  • ((\(t_{1/2}\))
is the time it takes for a reactant to reduce to half of its initial concentration. In zero-order reactions, unlike first or second-order reactions, the halflife depends directly on the initial concentration and is given by \( t_{1/2} = \frac{[A]_0}{2k} \).
  • This equation highlights that as the initial concentration \([A]_0\) increases, the halflife also increases.
  • Conversely, a higher rate constant \(k\) results in a shorter halflife, signifying faster reactions.
Understanding halflife offers valuable insights into how long a reaction may take under certain conditions, allowing for better prediction and control over chemical processes in industries and research.
Calculus
The use of *calculus* is central in deriving the integrated rate law for zero-order reactions. Calculus allows us to move from knowing a rate at an instantaneous moment to understanding how quantities change over a period. Start by considering the differential form
  • (\( \frac{d[A]}{dt} = -k \))
which tells us that the rate of change of concentration with respect to time is constant and negative, indicating a depletion in \([A]\). By integrating this equation with respect to time, we move from rates to absolute concentrations. This step is crucial because it allows us to form a direct relationship between concentration and time. Without calculus, understanding these dynamic processes over time would be impossible, making calculus an indispensable tool in chemical kinetics.
Differential Equations
At the heart of zero-order reactions is the concept of *differential equations*. These are mathematical equations that relate a function with its derivatives, providing a framework for describing how systems change. In a zero-order reaction, the rate law is given by the differential equation
  • (\( \frac{d[A]}{dt} = -k \)).
This form implies a constant rate of change regardless of the concentration.
Solving this equation via integration helps us determine how concentrations change from their initial values over time.
The solution, \( [A] = [A]_0 - kt \), is a straightforward linear relationship,
  • where manipulation of differential equations allows us to predict concentrations at any given time.
These equations are not only vital in chemistry but in many fields of science and engineering where modeling changes in a system is required.

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Most popular questions from this chapter

Asample of uranite is found to have a \({ }_{82}^{206} \mathrm{~Pb} /{ }_{92}^{-98} \mathrm{U}\) mass ratio of \(0.395 .\) Estimate the age of the uranite. The half-life of the conversion of uranium-238 to lead-206 is \(4.51 \times 10^{9}\) years.

The U.S. Public Health Service requires that milk fresh from a pasteurizer may contain no more than 20000 bacteria per milliliter. It has been reported that the number of bacteria in milk stored in a refrigerator at \(4^{\circ} \mathrm{C}\left(40^{\circ} \mathrm{F}\right)\) may double in 39 hours. If a milk sample had 20000 bacteria per milliliter after pasteurization, what is the bacteria count per milliliter after 10 days at this temperature?

The reaction described by the equation \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{HCl}(g)\) was studied at \(300 \mathrm{~K}\), and the following initial-rate data were collected: \begin{tabular}{ccc} \hline & & Initial rate of formation \\ Run & {\(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right]_{0} / \mathrm{M}\)} & of \(\mathbf{C}_{2} \mathrm{H}_{4}(g) / \mathrm{M} \cdot \mathrm{s}^{-1}\) \\ \hline 1 & \(0.93\) & \(2.40 \times 10^{-30}\) \\ 2 & \(0.55\) & \(4.00 \times 10^{-30}\) \\ 9 & \(0.90\) & \(6.54 \times 10^{-30}\) \\ \hline \end{tabular} Determine the rate law and the value of the rate constant for the reaction.

Peroxydisulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}(a q)\) decomposes in aqueous solution according to the equation $$ \begin{aligned} \mathrm{S}_{2} \mathrm{O}_{8}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \\ 2 \mathrm{SO}_{4}^{2-}(a q)+\frac{1}{2} \mathrm{O}_{2}(g)+2 \mathrm{H}^{+}(a q) \end{aligned} $$ Given the following data from an experiment with \(\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]_{0}=0.100 \mathrm{M}\) in a solution with \(\left[\mathrm{H}^{+}\right]\) fixed at \(0.100 \mathrm{M}\), determine the reaction rate law and calculate the value of the rate constant: \begin{tabular}{cc} \hline\(t / \mathrm{min}\) & {\(\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-1} / \mathrm{M}\right.\)} \\ \hline 0 & \(0.100\) \\ 17 & \(0.050\) \\ 34 & \(0.025\) \\ 51 & \(0.012\) \\ \hline \end{tabular}

The reaction described by the chemical equation $$ 3 \mathrm{BrO}^{-}(a q) \rightarrow \operatorname{BrO}_{3}^{-}(a q)+2 \mathrm{Br}^{-}(a q) $$ is second order in \(\mathrm{BrO}^{-}(a q)\) in basic solution with a rate constant equal to \(0.056 \mathrm{M}^{-1} \cdot \mathrm{s}^{-1}\) at \(80^{\circ} \mathrm{C}\). If \(\left[\mathrm{BrO}^{-}\right]_{0}\) \(=0.212 \mathrm{M}\), what will \(\left[\mathrm{BrO}^{-}\right]\) be \(1.00\) minute later?
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