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Chapter 14: Problem 36

(a) For a second-order reaction, what quantity, when graphed versus time, will yield a straight line? (b) How do the half-lives of first-order and second- order reactions differ?

Short Answer

Expert verified
(a) For a second-order reaction, graphing \(\frac{1}{[A]_t}\) versus time (\(t\)) will yield a straight line. (b) The half-life of a first-order reaction is constant and independent of the initial concentration, while the half-life of a second-order reaction depends on the initial concentration, increasing with decreasing reactant concentration.

Step by step solution

01

Understanding what a second-order reaction is

A second-order reaction is a type of chemical reaction where the reaction rate is proportional to the square of the concentrations of one or more reactants.
02

Write down the rate law for a second-order reaction

The rate law for a second-order reaction can be written as: \[rate = k[A]^2\] where \(rate\) is the rate of the reaction, \(k\) is the rate constant, and \([A]\) is the concentration of the reactant A.
03

Write down the integrated rate law for a second-order reaction

Integrating the rate law, we have the integrated rate law for a second-order reaction: \[\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}\] where \([A]_t\) is the concentration of A at time \(t\), and \([A]_0\) is the initial concentration of A.
04

Identify the quantity that yields a straight line

From the integrated rate law, we can see that when we graph \(\frac{1}{[A]_t}\) versus \(t\), we will get a straight line. The slope of this line is equal to the rate constant, \(k\). Question (b):
05

Write down the half-life expressions for first-order and second-order reactions

The half-life of a first-order reaction can be represented by: \[t_{1/2}^{(1)} = \frac{\ln 2}{k}\] The half-life of a second-order reaction can be represented by: \[t_{1/2}^{(2)} = \frac{1}{k[A]_0}\] where \(t_{1/2}^{(1)}\) and \(t_{1/2}^{(2)}\) are the half-lives for first-order and second-order reactions, respectively.
06

Compare the expressions for half-lives

In a first-order reaction, half-life is constant and independent of the initial concentration (\([A]_0\)). This means that as the reaction proceeds, the half-life remains the same. On the other hand, in a second-order reaction, half-life depends on the initial concentration (\([A]_0\)). As the reaction proceeds, the half-life will change. Specifically, as the concentration decreases, the half-life increases.
07

Summarize the differences between half-lives

To summarise, the main difference between the half-lives of a first-order and second-order reaction is that the half-life of a first-order reaction is constant and independent of the initial reactant concentrations, while the half-life of a second-order reaction changes as the reactant concentrations change- it increases with decreasing reactant concentration.

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Most popular questions from this chapter

(a) Explain the importance of enzymes in biological systems. (b) What chemical transformations are catalyzed (i) by the enzyme catalase, (ii) by nitrogenase?

The enzyme invertase catalyzes the conversion of sucrose, a disaccharide, to invert sugar, a mixture of glucose and fructose When the concentration of invertase is \(4.2 \times 10^{-7} M\) and the concentration of sucrose is \(0.0077 M\), invert sugar is formed at the rate of \(1.5 \times 10^{-4} \mathrm{M} / \mathrm{s} .\) When the sucrose concentration is doubled, the rate of formation of invert sugar is doubled also. (a) Assuming that the enzyme- substrate model is operative, is the fraction of enzyme tied up as a complex large or small? Explain. (b) Addition of inositol, another sugar, decreases the rate of formation of invert sugar. Suggest a mechanism by which this occurs.

NO catalyzes the decomposition of \(\mathrm{N}_{2} \mathrm{O}\), possibly by the following mechanism: $$ \begin{array}{r} \mathrm{NO}(\mathrm{g})+\mathrm{N}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{array} $$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Why is NO considered a catalyst and not an intermediate? (c) If experiments show that during the decomposition of \(\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism? If you think not, suggest what might be going on.

You study the effect of temperature on the rate of two reactions and graph the natural logarithm of the rate constant for each reaction as a function of \(1 / T\). How do the two graphs compare (a) if the activation energy of the second reaction is higher than the activation energy of the first reaction but the two reactions have the same frequency factor, and (b) if the frequency factor of the second reaction is higher than the frequency factor of the first reaction but the two reactions have the same activation energy? [Section \(14.5\)

A certain first-order reaction has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if (a) \(E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol} ;\) (b) \(E_{a}=125 \mathrm{~kJ} / \mathrm{mol}\) ?
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