(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?

Short Answer

Expert verified
(a) For a generic second-order reaction, plotting \(\frac{1}{[A]_t}\) against time (\(t\)) will yield a straight line. (b) The slope of the straight line is the rate constant, \(k\). (c) The half-life of a first-order reaction remains constant, independent of the reactant concentration, while the half-life of a second-order reaction depends on the initial concentration of the reactant.

Step by step solution

01

Part (a): Determine the y-axis quantity for the graph

We start by writing the rate equation for a second-order reaction: \(rate = k \times [A]^2\) Where \(k\) is the rate constant, and \([A]\) represents the concentration of reactant A. We are asked to find a quantity that will yield a straight line when graphed against time. To do this, we can convert the rate equation into an integrated rate law. For a second-order reaction, the integrated rate law is: \(\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt\) Where \([A]_t\) is the concentration of A at a specific time, \([A]_0\) is the initial concentration of A, and \(t\) is time. Here, if we plot \(\frac{1}{[A]_t}\) against time (\(t\)), we will get a straight line.
02

Part (b): Find the slope of the straight line

To find the slope of the straight line from part (a), we can rewrite the integrated rate law in the form of a linear equation: \(\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}\) This equation is in the form of \(y = mx + c\), where \(y = \frac{1}{[A]_t}\), \(x = t\), \(m = k\), and \(c = \frac{1}{[A]_0}\). The slope of the straight line, or the coefficient of \(t\), is the rate constant, \(k\).
03

Part (c): Comparison of half-lives between first-order and second-order reactions

The half-life, \(t_{1/2}\), is the amount of time it takes for the concentration of the reactant to reduce to half its initial value. The half-life of a reaction depends on the order of the reaction. For a first-order reaction, the half-life is given by: \(t_{1/2}^{(1)} = \frac{0.693}{k}\) Here, we can see that the half-life is constant and does not depend on the initial concentration. For a second-order reaction, the half-life is given by: \(t_{1/2}^{(2)} = \frac{1}{k[A]_0}\) In this case, the half-life depends on the initial concentration of the reactant. As the initial concentration increases, the half-life becomes smaller. In summary, the half-life of a first-order reaction remains constant, independent of the reactant concentration, while the half-life of a second-order reaction depends on the initial concentration of the reactant.

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Most popular questions from this chapter

The following mechanism has been proposed for the reaction of \(\mathrm{NO}\) with \(\mathrm{H}_{2}\) to form \(\mathrm{N}_{2} \mathrm{O}\) and \(\mathrm{H}_{2} \mathrm{O}\) : $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{NO}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g) \\ \mathrm{N}_{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{aligned} $$ (a) Show that the elementary reactions of the proposed mechanism add to provide a balanced equation for the reaction. (b) Write a rate law for each elementary reaction in the mechanism. (c) Identify any intermediates in the mechanism. (d) The observed rate law is rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]\). If the proposed mechanism is correct, what can we conclude about the relative speeds of the first and second reactions?

(a) Two reactions have identical values for \(E_{a} .\) Does this ensure that they will have the same rate constant if run at the same temperature? Explain. (b) Two similar reactions have the same rate constant at \(25^{\circ} \mathrm{C}\), but at \(35^{\circ} \mathrm{C}\) one of the reactions has a larger rate constant than the other. Account for these observations.

Enzymes are often described as following the two-step mechanism: $$ \begin{array}{l} \mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{ES} \quad(\text { fast }) \\ \mathrm{ES} \longrightarrow \mathrm{E}+\mathrm{P} \quad(\text { slow }) \end{array} $$ 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 \(\mathrm{E}\) with \(\mathrm{I}\), an inhibitor.

The reaction $2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\( \)\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ was studied with the following results:

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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