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Chapter 14: Problem 8

Which of the following linear plots do you expect for a reaction \(\mathrm{A} \longrightarrow\) products if the kinetics are (a) zero order, (b) first order, or (c) second order? [Section 14.4\(]\)

Short Answer

Expert verified
Zero-order: [A] vs. t; First-order: ln[A] vs. t; Second-order: 1/[A] vs. t.

Step by step solution

01

Understanding Zero-Order Kinetics

For a zero-order reaction, the rate of reaction is constant and does not depend on the concentration of reactants. The integrated rate law for a zero-order reaction is given by: \[ [A] = [A]_0 - kt \]where \([A]\) is the concentration of \(A\) at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant. The plot of \([A]\) versus time \(t\) should be a straight line with a negative slope of \(-k\).
02

Understanding First-Order Kinetics

For a first-order reaction, the rate of reaction depends on the concentration of the reactant \(A\). The integrated rate law is:\[ \ln [A] = \ln [A]_0 - kt \]A plot of \(\ln [A]\) versus time \(t\) will yield a straight line. The slope of the line is \(-k\), indicating a linear relationship on a semi-logarithmic scale.
03

Understanding Second-Order Kinetics

For a second-order reaction, the rate depends on the square of the concentration of \(A\). The integrated rate law is:\[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \]In this case, a plot of \(\frac{1}{[A]}\) versus time \(t\) should be a straight line. The slope of this line is equal to \(k\).

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Key Concepts

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

Zero-Order Kinetics
Zero-order kinetics describe a scenario where the rate of reaction does not change with the concentration of the reactant. This means that the reaction proceeds at a constant rate until the reactant is depleted. The plot that represents a zero-order reaction is a straight line when the concentration of the reactant \([A]\) is plotted against time \(t\). The formula for this relationship is given by the integrated rate law:
  • \([A] = [A]_0 - kt\)
In this equation, \([A]_0\) is the initial concentration, and \(k\) is the rate constant with a negative slope of -k. This linear plot reflects a direct decrease in concentration over time. Thus, the higher the value of \(k\), the steeper the decline in concentration, signaling a faster reaction rate.
First-Order Kinetics
First-order kinetics represent reactions whose rate is directly proportional to the concentration of one reactant. These are quite common in chemical reactions. For first-order kinetics, a plot of the natural logarithm of the reactant concentration \(\ln[A]\) against time is used to gain a linear relationship. The integrated rate law for a first-order reaction is:
  • \(\ln [A] = \ln [A]_0 - kt\)
This equation shows that the natural logarithm of concentration decreases linearly over time. The negative slope \(-k\) indicates the rate at which the reaction occurs. This logarithmic relationship ensures that even with decreasing concentrations, the reaction continues to proceed at a reproducible rate.
Second-Order Kinetics
In second-order kinetics, the reaction rate is proportional to the square of the concentration of the reactant, or to the product of the concentrations of two reactants. These reactions tend to proceed more slowly compared to first-order reactions because the change in concentration must be more significant to affect the rate. The integrated rate law for second-order kinetics is expressed as:
  • \(\frac{1}{[A]} = \frac{1}{[A]_0} + kt\)
For a second-order reaction, plotting \(\frac{1}{[A]}\) against time will yield a straight line. Here, the slope is \(k\), which signifies that a greater change in concentration results in greater alterations in reaction rate. This type of kinetic behavior is typical in reactions where two molecules of the same or different substances must collide for the reaction to progress.

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Most popular questions from this chapter

(a) What factors determine whether a collision between two molecules will lead to a chemical reaction? (b) Does the rate constant for a reaction generally increase or decrease with an increase in reaction temperature? (c) Which factor is most sensitive to changes in temperature-the frequency of collisions, the orientation factor, or the fraction of molecules with energy greater than the activation energy?

A reaction \(A+B \longrightarrow C\) obeys the following rate law: Rate \(=k[A]^{3}\). (a) If [B] is doubled, how will the rate change? Will the rate constant change? (b) What are the reaction orders for \(\mathrm{A}\) and \(\mathrm{B}\) ? What is the overall reaction order? (c) What are the units of the rate constant?

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: \(\mathrm{OCl}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{OI}^{-}+\mathrm{Cl}^{-}\). This rapid reaction gives the following rate data: \begin{tabular}{lcc} \hline\(\left[0 \mathrm{Cl}^{-}\right](M)\) & {\(\left[I^{-}\right](M)\)} & Initial Rate \((M / s)\) \\ \hline \(1.5 \times 10^{-3}\) & \(1.5 \times 10^{-3}\) & \(1.36 \times 10^{-4}\) \\ \(3.0 \times 10^{-3}\) & \(1.5 \times 10^{-3}\) & \(2.72 \times 10^{-4}\) \\ \(1.5 \times 10^{-3}\) & \(3.0 \times 10^{-3}\) & \(2.72 \times 10^{-4}\) \\ \hline \end{tabular} (a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when \(\left[\mathrm{OCl}^{-}\right]=2.0 \times 10^{-3} \mathrm{M}\) and \(\left[1^{-}\right]=5.0 \times 10^{-4} \mathrm{M}\).

The isomerization of methyl isonitrile \(\left(\mathrm{CH}_{3} \mathrm{NC}\right)\) to acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) was studied in the gas phase at \(215^{\circ} \mathrm{C},\) and the following data were obtained: \begin{tabular}{rr} \hline Time (s) & {\(\left[\mathrm{CH}_{3} \mathrm{NC}\right](M)\)} \\ \hline 0 & 0.0165 \\ 2000 & 0.0110 \\ 5000 & 0.00591 \\ 8000 & 0.00314 \\ 12,000 & 0.00137 \\ 15,000 & 0.00074 \\ \hline \end{tabular} (a) Calculate the average rate of reaction, in \(M / \mathrm{s}\), for the time interval between each measurement. (b) Calculate the average rate of reaction over the entire time of the data from \(t=0\) to \(t=15,000 \mathrm{~s}\). \((\mathbf{c})\) Which is greater, the average rate between \(t=2000\) and \(t=12,000 \mathrm{~s}\), or between \(t=8000\) and \(t=15,000 \mathrm{~s} ?\) (d) Graph [CH \(\left._{3} \mathrm{NC}\right]\) versus time and determine the instantaneous rates in \(M / \mathrm{s}\) at \(t=5000 \mathrm{~s}\) and \(t=8000 \mathrm{~s}\).

The oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is accelerated by \(\mathrm{NO}_{2}\). The reaction proceeds according to: $$ \begin{array}{l} \mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \end{array} $$ (a) Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of \(\mathrm{SO}_{2}\) by \(\mathrm{O}_{2}\) to give \(\mathrm{SO}_{3}\). (b) Do we consider \(\mathrm{NO}_{2}\) a catalyst or an intermediate in this reaction? (c) Would you classify NO as a catalyst or as an intermediate? (d) Is this an example of homogeneous catalysis or heterogeneous catalysis?
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