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Chapter 14: Problem 8

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

Short Answer

Expert verified
For each reaction order, the expected linear plots are: (a) Zero-order: Linear plot of \([A]_t\) vs. time (t). (b) First-order: Linear plot of \(\ln[A]_t\) vs. time (t). (c) Second-order: Linear plot of \(1/[A]_t\) vs. time (t).

Step by step solution

01

Recall the integrated rate laws for zero, first, and second-order reactions

For each type of reaction, we will use the appropriate integrated rate law to determine the relationship between the concentration of reactant A and time. (a) Zero-order reactions: For a zero-order reaction, the integrated rate law is \[ [A]_t = [A]_0 - kt \] (b) First-order reactions: For a first-order reaction, the integrated rate law is \[ \ln[A]_t = \ln[A]_0 - kt \] (c) Second-order reactions: For a second-order reaction, the integrated rate law is \[ \frac{1}{[A]_t} = \frac{1}{[A]_0} + kt \]
02

Determine the expected linear plots for each reaction order

Based on the integrated rate laws, let's determine the expected linear plot for each reaction order. (a) Zero-order reactions: The integrated rate law for a zero-order reaction is \[ [A]_t = [A]_0 - kt \] This equation represents a linear relationship between the concentration of A ([A]_t) and time (t). The slope of the line is -k, and the intercept is [A]_0. Hence, a graph of [A]_t versus time (t) will be linear for a zero-order reaction. (b) First-order reactions: The integrated rate law for a first-order reaction is \[ \ln[A]_t = \ln[A]_0 - kt \] This equation represents a linear relationship between the natural logarithm of the concentration of A (ln[A]_t) and time (t). The slope of the line is -k, and the intercept is ln[A]_0. Hence, a graph of ln[A]_t versus time (t) will be linear for a first-order reaction. (c) Second-order reactions: The integrated rate law for a second-order reaction is \[ \frac{1}{[A]_t} = \frac{1}{[A]_0} + kt \] This equation represents a linear relationship between the reciprocal of the concentration of A (1/[A]_t) and time (t). The slope of the line is k, and the intercept is 1/[A]_0. Hence, a graph of 1/[A]_t versus time (t) will be linear for a second-order reaction. To summarize the findings: (a) Zero-order: Linear plot of [A]_t vs. time (t). (b) First-order: Linear plot of ln[A]_t vs. time (t). (c) Second-order: Linear plot of 1/[A]_t vs. time (t).

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Key Concepts

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

Zero-Order Reaction
In the world of reaction kinetics, a zero-order reaction is one where the rate of reaction is constant and independent of the concentration of the reactant. This might seem counterintuitive since we often think that more reactant should mean a faster reaction. However, in zero-order reactions, the rate only depends on the rate constant, denoted as "k".

The mathematical expression that describes a zero-order reaction is the integrated rate law:
  • \( [A]_t = [A]_0 - kt \)
Here,
  • \([A]_t\) is the concentration of the reactant at time \(t\),
  • \([A]_0\) is the initial concentration, and
  • \(k\) is the rate constant.

When plotted on a graph with concentration \([A]_t\) against time \(t\), this relationship results in a straight line with a negative slope of \(-k\). The y-intercept of this line is \([A]_0\). This means that as time progresses, the reactant concentration decreases linearly, highlighting the constant rate of the reaction. Such reactions are often seen in enzymatic processes where the enzyme becomes saturated or in surface-catalyzed reactions where the surface is fully occupied.
First-Order Reaction
First-order reactions are quite common and are characterized by a rate that depends on the concentration of a single reactant. However, the dependence is logarithmic rather than linear, making the math a bit more interesting!

The integrated rate law for a first-order reaction is:
  • \( \ln[A]_t = \ln[A]_0 - kt \)
Here,
  • \(\ln[A]_t\) refers to the natural logarithm of the reactant concentration at time \(t\),
  • \(\ln[A]_0\) is the natural logarithm of the initial concentration, and
  • \(k\) is the rate constant.

If you graph \(\ln[A]_t\) against time \(t\), you'll see a straight line emerge. The slope of this line is \(-k\), and the y-intercept is \(\ln[A]_0\). This linear plot illustrates how the reaction rate decreases over time as the reactant is used up. First-order kinetics are typical in radioactive decay and many chemical reactions in solution where the concentration significantly affects the speed of the reaction.
Second-Order Reaction
Second-order reactions have a rate that typically depends on the concentration of one reactant squared or on two different reactants. Understanding these reactions involves looking at how the concentration changes with time.

The integrated rate law for a second-order reaction is more complex than those of zero and first order:
  • \( \frac{1}{[A]_t} = \frac{1}{[A]_0} + kt \)
In this equation,
  • \(\frac{1}{[A]_t}\) denotes the reciprocal of the reactant's concentration at time \(t\),
  • \(\frac{1}{[A]_0}\) is the reciprocal of the initial concentration, and
  • \(k\) is the rate constant.

When you plot \(\frac{1}{[A]_t}\) versus time \(t\), you'll find a linear relationship where the slope is \(k\) and the intercept is \(\frac{1}{[A]_0}\). This graph helps us understand how the reaction rate decreases more drastically as time progresses, requiring careful monitoring of conditions. Second-order reactions are often seen in processes involving gaseous reactions or in chemical systems where two different reactants are involved.

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Most popular questions from this chapter

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q) \quad \text { (slow) } \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q) \text { (fast) } \end{aligned} $$ (a) Write the chemical equation for the overall process. (b) Identify the intermediate, if any, in the mechanism. (c) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process.

The reaction \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)\) was performed and the following data obtained under conditions of constant \(\left[\mathrm{Cl}_{2}\right]\) : (a) Is the following mechanism consistent with the data? $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) & \longrightarrow \mathrm{NOCl}_{2}(g) \text { (fast) } \\ \mathrm{NOCl}_{2}(g)+\mathrm{NO}(g) & \longrightarrow 2 \mathrm{NOCl}(g) \text { (slow) } \end{aligned} $$ (b) Does the linear plot guarantee that the overall rate law is second order?

Cyclopentadiene \(\left(\mathrm{C}_{5} \mathrm{H}_{6}\right)\) reacts with itself to form dicyclopentadiene $\left(\mathrm{C}_{10} \mathrm{H}_{12}\right)\(. A \)0.0400 \mathrm{M}\( solution of \)\mathrm{C}_{5} \mathrm{H}_{6}$ was monitored as a function of time as the reaction $2 \mathrm{C}_{5} \mathrm{H}_{6} \longrightarrow \mathrm{C}_{10} \mathrm{H}_{12}$ proceeded. The following data were collected: $$ \begin{array}{cc} \hline \text { Time (s) } & {\left[\mathrm{C}_{5} \mathrm{H}_{6}\right](M)} \\\ \hline 0.0 & 0.0400 \\ 50.0 & 0.0300 \\ 100.0 & 0.0240 \\ 150.0 & 0.0200 \\ 200.0 & 0.0174 \\ \hline \end{array} $$ Plot \(\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time, $\ln \left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\( versus time, and \)1 /\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]$ versus time. (a) What is the order of the reaction? (b) What is the value of the rate constant?

The \(\mathrm{NO}_{x}\) waste stream from automobile exhaust includes species such as \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\). Catalysts that convert these species to \(\mathrm{N}_{2}\). are desirable to reduce air pollution. (a) Draw the Lewis dot and VSEPR structures of \(\mathrm{NO}, \mathrm{NO}_{2}\), and \(\mathrm{N}_{2}\) - (b) Using a resource such as Table \(8.4\), look up the energies of the bonds in these molecules. In what region of the electromagnetic spectrum are these energies? (c) Design a spectroscopic experiment to monitor the conversion of \(\mathrm{NO}_{x}\) into \(\mathrm{N}_{2}\), describing what wavelengths of light need to be monitored as a function of time.

Based on their activation energies and energy changes and assuming that all collision factors are the same, which of the following reactions would be fastest and which would be slowest? (a) \(E_{a}=45 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-25 \mathrm{~kJ} / \mathrm{mol}\) (b) \(E_{a}=35 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-10 \mathrm{~kJ} / \mathrm{mol}\) (c) \(E_{a}=55 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=10 \mathrm{~kJ} / \mathrm{mol}\)
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