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Chapter 12: Problem 53

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
a. Rate law: \(Rate = k[\mathrm{CH}_{3}\mathrm{NC}]\) b. Rate law: \(Rate = k[\mathrm{O}_{3}][\mathrm{NO}]\) c. Rate law: \(Rate = k[\mathrm{O}_{3}]\) d. Rate law: \(Rate = k[\mathrm{O}_{3}][\mathrm{O}]\)

Step by step solution

01

a. Rate law for $\mathrm{CH}_{3} \mathrm{NC}(g) \rightarrow \mathrm{CH}_{3} \mathrm{CN}(g)

Since this is an elementary reaction involving only one reactant, the rate law can be written as follows: Rate = k[\(\mathrm{CH}_{3} \mathrm{NC}\)] where k is the rate constant, and the concentration of the reactant is raised to the power of 1 because there is only one molecule involved in the reaction.
02

b. Rate law for \(\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{NO}_{2}(g)\)

This reaction involves two reactants, both with a stoichiometric coefficient of 1. The rate law for this elementary reaction is: Rate = k[\(\mathrm{O}_{3}\)][\(\mathrm{NO}\)] Again, the concentrations of each reactant are raised to the power of 1 because both reactants have a stoichiometric coefficient of 1 according to the chemical equation.
03

c. Rate law for \(\mathrm{O}_{3}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g)\)

In this elementary reaction, there is only one reactant, ozone. The stoichiometric coefficient of ozone is 1, so the rate law can be expressed as: Rate = k[\(\mathrm{O}_{3}\)] The concentration of the reactant, ozone, is raised to the power of 1 because there is only one molecule involved in the reaction.
04

d. Rate law for \(\mathrm{O}_{3}(g)+\mathrm{O}(g) \rightarrow 2 \mathrm{O}_{2}(g)\)

This reaction involves two reactants, \(\mathrm{O}_{3}\) and \(\mathrm{O}\), each with stoichiometric coefficients of 1 in the chemical equation. Thus, the rate law for this elementary reaction is: Rate = k[\(\mathrm{O}_{3}\)][\(\mathrm{O}\)] The concentrations of both reactants are raised to the power of 1 because each reactant has a stoichiometric coefficient of 1.

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Most popular questions from this chapter

Consider the hypothetical reaction $$ \mathrm{A}+\mathrm{B}+2 \mathrm{C} \longrightarrow 2 \mathrm{D}+3 \mathrm{E} $$ where the rate law is $$ \text { Rate }=-\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{~A}][\mathrm{B}]^{2} $$ An experiment is carried out where \([\mathrm{A}]_{0}=1.0 \times 10^{-2} M\), \([\mathrm{B}]_{0}=3.0 \mathrm{M}\), and \([\mathrm{C}]_{0}=2.0 \mathrm{M} .\) The reaction is started, and after \(8.0\) seconds, the concentration of \(\mathrm{A}\) is \(3.8 \times 10^{-3} \mathrm{M}\). a. Calculate the value of \(k\) for this reaction. b. Calculate the half-life for this experiment. c. Calculate the concentration of A after \(13.0\) seconds. d. Calculate the concentration of \(\mathrm{C}\) after \(13.0\) seconds.

The central idea of the collision model is that molecules must collide in order to react. Give two reasons why not all collisions of reactant molecules result in product formation.

The rate law for the decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) is $$ \text { Rate }=-\frac{\Delta\left[\mathrm{PH}_{3}\right]}{\Delta t}=k\left[\mathrm{PH}_{3}\right] $$ It takes \(120 .\) s for \(1.00 M \mathrm{PH}_{3}\) to decrease to \(0.250 \mathrm{M}\). How much time is required for \(2.00 \mathrm{M} \mathrm{PH}_{3}\) to decrease to a concentration of \(0.350 \mathrm{M} ?\)

Sulfuryl chloride undergoes first-order decomposition at \(320 .{ }^{\circ} \mathrm{C}\) with a half-life of \(8.75 \mathrm{~h}\). $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ What is the value of the rate constant, \(k\), in \(\mathrm{s}^{-1} ?\) If the initial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is 791 torr and the decomposition occurs in a \(1.25\) -L container, how many molecules of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) remain after \(12.5 \mathrm{~h}\) ?

Consider the general reaction $$ \mathrm{aA}+\mathrm{bB} \longrightarrow \mathrm{cC} $$ and the following average rate data over some time period \(\Delta t\) : $$ \begin{aligned} -\frac{\Delta \mathrm{A}}{\Delta t} &=0.0080 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \\ -\frac{\Delta \mathrm{B}}{\Delta t} &=0.0120 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \\ \frac{\Delta \mathrm{C}}{\Delta t} &=0.0160 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \end{aligned} $$ Determine a set of possible coefficients to balance this general reaction.
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