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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

Platinum nanoparticles of diameter \(\sim 2 \mathrm{~nm}\) are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of \(3.924 \AA\). (a) Estimate how many platinum atoms would fit into a \(2.0-\mathrm{nm}\) sphere; the volume of a sphere is \((4 / 3) \pi r^{3}\). Recall that \(1 \dot{A}=1 \times 10^{-10} \mathrm{~m}\) and \(1 \mathrm{~nm}=1 \times 10^{-9} \mathrm{~m}\). (b) Estimate how many platinum atoms are on the surface of a \(2.0\)-nm Pt sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one Pt atom can be estimated from its atomic diameter of \(2.8 \mathrm{~A}\). (c) Using your results from (a) and (b), calculate the percentage of \(\mathrm{Pt}\) atoms that are on the surface of a \(2.0-\mathrm{nm}\) nanoparticle. (d) Repeat these calculations for a \(5.0\)-nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

(a) For a generic second-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\), what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part (a)? (c) How do the half-lives of first-order and second-order reactions differ?

Heterogeneous catalysts that perform hydrogenation reactions, as illustrated in Figure 14.24, are subject to "poisoning," which shuts down their catalytic ability. Compounds of sulfur are often poisons. Suggest a mechanism by which such compounds might act as poisons.

The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\), at \(70^{\circ} \mathrm{C}\) is \(6.82 \times\) \(10^{-3} \mathrm{~s}^{-1}\). Suppose we start with \(0.0250 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g})\) in a volume of \(2.0 \mathrm{~L}\). (a) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0 \mathrm{~min}\) ? (b) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010\) mol? (c) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70{ }^{\circ} \mathrm{C}\) ?

The enzyme carbonic anhydrase catalyzes the reaction \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HCO}_{3}{ }^{-}(a q)+\mathrm{H}^{+}(a q)\). In water, without the enzyme, the reaction proceeds with a rate constant of \(0.039 \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(1.0 \times 10^{6} \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.
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