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Chapter 12: Problem 61

Write the rate laws for the following elementary reactions. a. \(\mathrm{CH}_{3} \mathrm{NC}(g) \rightarrow \mathrm{CH}_{3} \mathrm{CN}(g)\) b. $\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{NO}_{2}(g)$ c. \(\mathrm{O}_{3}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g)\) d. \(\mathrm{O}_{3}(g)+\mathrm{O}(g) \rightarrow 2 \mathrm{O}_{2}(g)\)

Short Answer

Expert verified
For the given elementary reactions, the rate laws are as follows: a. Rate = k\([\mathrm{CH}_3 \mathrm{NC}]\) b. Rate = k\([\mathrm{O}_3][\mathrm{NO}]\) c. Rate = k\([\mathrm{O}_3]\) d. Rate = k\([\mathrm{O}_3][\mathrm{O}]\)

Step by step solution

01

Reaction a: Convert reactants to products

For reaction a, we are given the elementary reaction: \(\mathrm{CH}_3 \mathrm{NC}(g) \rightarrow \mathrm{CH}_3 \mathrm{CN}(g)\) Since this is an elementary reaction and there is only one reactant, it is a first-order reaction. Therefore, the rate law for this reaction can be written as: Rate = k\([\mathrm{CH}_3 \mathrm{NC}]\)
02

Reaction b: Convert reactants to products

For reaction b, we are given the elementary reaction: \( \mathrm{O_{3}}(g)+\mathrm{NO}(g) \rightarrow \mathrm{O_{2}}(g)+\mathrm{NO_{2}}(g)\) Since this is an elementary reaction with two reactants, it is a second-order reaction (first-order with respect to each reactant). Therefore, the rate law for this reaction can be written as: Rate = k\([\mathrm{O}_3][\mathrm{NO}]\)
03

Reaction c: Convert reactants to products

For reaction c, we are given the elementary reaction: \(\mathrm{O_{3}}(g)\rightarrow \mathrm{O_{2}}(g)+\mathrm{O}(g)\) Since this is an elementary reaction with only one reactant, it is a first-order reaction. Therefore, the rate law for this reaction can be written as: Rate = k\([\mathrm{O}_3]\)
04

Reaction d: Convert reactants to products

For reaction d, we are given the elementary reaction: \(\mathrm{O_{3}}(g)+\mathrm{O}(g) \rightarrow 2 \mathrm{O_{2}}(g)\) Since this is an elementary reaction with two reactants, it is a second-order reaction (first-order with respect to each reactant). Therefore, the rate law for this reaction can be written as: Rate = k\([\mathrm{O}_3][\mathrm{O}]\)

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Most popular questions from this chapter

Provide a conceptual rationale for the differences in the halflives of zero-, first-, and second-order reactions.

At \(40^{\circ} \mathrm{C}, \mathrm{H}_{2} \mathrm{O}_{2}(a q)\) will decompose according to the following reaction: $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(\mathrm{~g}) $$ The following data were collected for the concentration of $\mathrm{H}_{2} \mathrm{O}_{2}$ at various times. $$ \begin{array}{|cc|} \hline \begin{array}{c} \text { Time } \\ (\mathbf{s}) \end{array} & \begin{array}{c} {\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]} \\ (\mathrm{mol} / \mathrm{L}) \end{array} \\ \hline 0 & 1.000 \\ \hline 2.16 \times 10^{4} & 0.500 \\ \hline 4.32 \times 10^{4} & 0.250 \\ \hline \end{array} $$ a. Calculate the average rate of decomposition of $\mathrm{H}_{2} \mathrm{O}_{2}\( between 0 and \)2.16 \times 10^{4} \mathrm{~s}$. Use this rate to calculate the average rate of production of \(\mathrm{O}_{2}(g)\) over the same time period. b. What are these rates for the time period \(2.16 \times 10^{4} \mathrm{~s}\) to \(4.32 \times 10^{4} \mathrm{~s} ?\)

For a first order gas phase reaction \(\mathrm{A} \longrightarrow\) products, \(k=\) \(7.2 \times 10^{-4} \mathrm{s}^{-1}\) at \(660 . \mathrm{K}\) and $k=1.7 \times 10^{-2} \mathrm{s}^{-1}\( at \)720 . \mathrm{K} .$ If the initial pressure of \(\mathrm{A}\) is 536 torr at \(295^{\circ} \mathrm{C},\) how long will it take for the pressure of \(\mathrm{A}\) to decrease to 268 torr?

A first-order reaction is 75.0\(\%\) complete in \(320 .\) s. a. What are the first and second half-lives for this reaction? b. How long does it take for 90.0\(\%\) completion?

The reaction $$ \mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C} $$ is known to be zero order in A and to have a rate constant of $5.0 \times 10^{-2} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\( at \)25^{\circ} \mathrm{C}$ . An experiment was run at \(25^{\circ} \mathrm{C}\) where $[\mathrm{A}]_{0}=1.0 \times 10^{-3} \mathrm{M} .$ a. Write the integrated rate law for this reaction. b. Calculate the half-life for the reaction. c. Calculate the concentration of \(\mathrm{B}\) after $5.0 \times 10^{-3} \mathrm{s}\( has elapsed assuming \)[\mathrm{B}]_{0}=0$
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