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(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

The oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is catalyzed by \(\mathrm{NO}_{2}\). Thereaction proceeds as follows: $$ \begin{aligned} &\mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g) \\ &2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \end{aligned} $$ (a) Show that the two reactions can be summed to give the overall oxidation of \(\mathrm{SO}_{2}\) by \(\mathrm{O}_{2}\) to give \(\mathrm{SO}_{3}\). (Hint: The top reaction must be multiplied by a factor so the \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) cancel out.) (b) Why do we consider \(\mathrm{NO}_{2}\) a catalyst and not an intermediate in this reaction? (c) Is this an example of homogeneous catalysis or heterogeneous catalysis?

(a) Define the following symbols that are encountered in rate equations: \([\mathrm{A}]_{0}, t_{1 / 2}[\mathrm{~A}]_{t}, k .(\mathrm{b})\) What quantity, when graphed versus time, will yield a straight line for a firstorder reaction?

(a) What is meant by the term molecularity? (b) Why are termolecular elementary reactions so rare? (c) What is an intermediate in a mechanism?

You perform a series of experiments for the reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C}\) and find that the rate law has the form rate \(=k[\mathrm{~A}]^{x}\). Determine the value of \(x\) in each of the following cases: (a) There is no rate change when [A] is tripled. (b) The rate increases by a factor of 9 when [A] is tripled. (c) When [A] is doubled, the rate increases by a factor of 8 . [Section 14.3]

The following data were collected for the rate of disappearance of \(\mathrm{NO}\) in the reaction $2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow$ \(2 \mathrm{NO}_{2}(g)\) \begin{tabular}{llll} \hline & & Initial Rate \\ Experiment & {\([\mathrm{NO}](M)\)} & {\(\left[\mathrm{O}_{2}\right](M)\)} & $(M / s)$ \\ \hline 1 & \(0.0126\) & \(0.0125\) & \(1.41 \times 10^{-2}\) \\ 2 & \(0.0252\) & \(0.0125\) & \(5.64 \times 10^{-2}\) \\ 3 & \(0.0252\) & \(0.0250\) & \(1.13 \times 10^{-1}\) \\ \hline \end{tabular} (a) What is the rate law for the reaction? (b) What are the units of the rate constant? (c) What is the average value of the rate constant calculated from the three data sets? (d) What is the rate of disappearance of NO when \([\mathrm{NO}]=0.0750 \mathrm{M}\) and $\left[\mathrm{O}_{2}\right]=0.0100 \mathrm{M} ?(\mathrm{e})$ What is the rate of disappearance of \(\mathrm{O}_{2}\) at the concentrations given in part (d)?

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