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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

A certain reaction has the following general form: \(\mathrm{aA} \longrightarrow \mathrm{bB}\) At a particular temperature and \([\mathrm{A}]_{0}=2.00 \times 10^{-2} M\), concentration versus time data were collected for this reaction, and a plot of \(\ln [\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(-2.97 \times 10^{-2} \mathrm{~min}^{-1}\). a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of \(\mathrm{A}\) to decrease to \(2.50 \times 10^{-3} M ?\)

The decomposition of iodoethane in the gas phase proceeds according to the following equation: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{HI}(g) $$ At \(660 . \mathrm{K}, k=7.2 \times 10^{-4} \mathrm{~s}^{-1}\); at 720. \(\mathrm{K}, k=1.7 \times 10^{-2} \mathrm{~s}^{-1}\). What is the value of the rate constant for this first-order decomposition at \(325^{\circ} \mathrm{C} ?\) If the initial pressure of iodoethane is 894 torr at \(245^{\circ} \mathrm{C}\), what is the pressure of iodoethane after three half-lives?

Hydrogen reacts explosively with oxygen. However, a mixture of \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) can exist indefinitely at room temperature. Explain why \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) do not react under these conditions.

What are the units for each of the following if the concentrations are expressed in moles per liter and the time in seconds? a. rate of a chemical reaction b. rate constant for a zero-order rate law c. rate constant for a first-order rate law d. rate constant for a second-order rate law e. rate constant for a third-order rate law

The reaction $$ 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g) $$ was studied at \(-10^{\circ} \mathrm{C}\). The following results were obtained where $$ \text { Rate }=-\frac{\Delta\left[\mathrm{Cl}_{2}\right]}{\Delta t} $$
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