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Chapter 14: Problem 40

(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 line? (b) What is the slope of the straight line from part (a)? (c) Does the half-life of a second-order reaction increase, decrease, or remain the same as the reaction proceeds?

Short Answer

Expert verified
(a) For a generic second-order reaction A → B, the quantity that, when graphed versus time, will yield a straight line is \(\frac{1}{[A]}\). (b) The slope of the straight line from part (a) is the rate constant \(k\). (c) The half-life of a second-order reaction increases as the reaction proceeds.

Step by step solution

01

Derive the integrated rate law for a second-order reaction

For a second-order reaction A → B, the rate law can be written as: \[r = k[A]^2\] To find the integrated rate law, first, rewrite the rate law in terms of concentration and differential: \[\frac{d[A]}{dt} = -k[A]^2\] Now, separate variables and integrate both sides: \[\int \frac{d[A]}{[A]^2} = -k \int dt\] \[\frac{-1}{[A]} = -kt + C\]
02

Find the quantity that gives a straight line

To find the quantity that, when plotted against time, will give a straight line, we want a linear relationship between that quantity and time. In the integrated rate law, we have: \[\frac{1}{[A]} = kt + \frac{1}{[A]_0}\] The relationship between \(\frac{1}{[A]}\) and \(t\) is linear. Therefore: (a) The graph of \(\frac{1}{[A]}\) versus time (\(t\)) will yield a straight line for a second-order reaction.
03

Determine the slope of the straight line

From the above linear equation of the form \(y = mx + b\), where \(y\) is \(\frac{1}{[A]}\), and \(x\) is \(t\), and the slope (m) is: (b) The slope of the straight line from part (a) is \(k\), the rate constant of the reaction.
04

Analyze the half-life for second-order reactions

The equation for the half-life (\(t_{1/2}\)) for a second-order reaction is: \[ t_{1/2} = \frac{1}{k[A]_0} \] As the reaction progresses, the concentration of A, [A]₀, decreases. Since the half-life is inversely proportional to the initial concentration, as [A]₀ decreases, the half-life will increase. Therefore: (c) The half-life of a second-order reaction increases as the reaction proceeds.

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Most popular questions from this chapter

Consider the hypothetical reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow 2 \mathrm{C}+\mathrm{D}\) . The following two-step mechanism is proposed for the reaction: $$ \begin{array}{l}{\text { Step } 1 : \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{X}} \\ {\text { Step } 2 : \mathrm{A}+\mathrm{X} \longrightarrow \mathrm{C}+\mathrm{D}}\end{array}$$ \(X\) is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (ii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).

Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) or silica \(\left(\mathrm{Si} \mathrm{O}_{2}\right)\) (a) Why is this an effective way of utilizing the catalyst material compared to having powdered metals? ( b) How does the surface area affect the rate of reaction?

Enzymes are often described as following the two-step mechanism: $$ \begin{array}{c}{\mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{ES} \text { (fast) }} \\ {\mathrm{ES} \longrightarrow \mathrm{E}+\mathrm{P} \quad(\text { slow })}\end{array}$$ where \(\mathrm{E}=\) enzyme, \(\mathrm{S}=\) substrate,\(\mathrm{ES}=\) enzyme-substrate complex, and \(\mathrm{P}=\) product.(a) If an enzyme follows this mechanism, what rate law is expected for the reaction? (b) Molecules that can bind to the active site of an enzyme but are not converted into product are called enzyme inhibitors. Write an additional elementary step to add into the preceding mechanism to account for the reaction of E with I, an inhibitor.

For the elementary process \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)\) the activation energy \(\left(E_{a}\right)\) and overall \(\Delta E\) are 154 \(\mathrm{kJ} / \mathrm{mol}\) and 136 \(\mathrm{kJ} / \mathrm{mol}\) , respectively. (a) Sketch the energy profile for this reaction, and label \(E_{a}\) and \(\Delta E\) . (b) What is the activation energy for the reverse reaction?

(a) The gas-phase decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}, \mathrm{SO}_{2} \mathrm{Cl}_{2}(g)\) \(\longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g),\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2} .\) At 600 \(\mathrm{K}\) the half-life for this process is \(2.3 \times 10^{5} \mathrm{s}\) . What is the rate constant at this temperature? (b) At 320 "C the rate constant is \(2.2 \times 10^{-5} \mathrm{s}^{-1} .\) What is the half-life at this temperature?
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