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Chapter 14: Problem 67

What is the molecularity of each of the following elementary reactions? Write the rate law for each. \(\begin{array}{l}{\text { (a) } \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{Cl}(g)} \\ {\text { (b) } \mathrm{OCl}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HOCl}(a q)+\mathrm{OH}^{-}(a q)} \\\ {\text { (c) } \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NOCl}_{2}(g)}\end{array}\)

Short Answer

Expert verified
The molecularity and rate law for each elementary reaction are: (a) Molecularity: 1 (unimolecular), Rate Law: \(Rate = k[\mathrm{Cl}_{2}]\) (b) Molecularity: 2 (bimolecular), Rate Law: \(Rate = k[\mathrm{OCl}^-][\mathrm{H}_{2}\mathrm{O}]\) (c) Molecularity: 2 (bimolecular), Rate Law: \(Rate = k[\mathrm{NO}][\mathrm{Cl}_{2}]\)

Step by step solution

01

Reaction (a) - Molecularity and Rate Law

For the first reaction: \(\mathrm{Cl}_{2}(g) \longrightarrow 2\mathrm{Cl}(g)\) This is an elementary reaction, which means, we can directly obtain the molecularity by counting the number of molecules involved. In this case, there is only one molecule of Cl2 involved. Molecularity: 1 (unimolecular) Now, we can write the rate law for this reaction. For an elementary reaction, the rate law is proportional to the concentration of the reactants. So, Rate = k[Cl2], where k is the rate constant.
02

Reaction (b) - Molecularity and Rate Law

For the second reaction: \(\mathrm{OCl}^{-}(a q)+\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{HOCl}(a q)+\mathrm{OH}^{-}(a q)\) This is also an elementary reaction. Count the number of molecules involved in the reaction. In this case, we have one molecule of OCl^- and one molecule of H2O involved. Molecularity: 2 (bimolecular) Now, let's write the rate law for this reaction. The rate law is proportional to the concentration of the reactants: Rate = k[OCl^-][H2O], where k is the rate constant.
03

Reaction (c) - Molecularity and Rate Law

For the third reaction: \(\mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NOCl}_{2}(g)\) Again, this is an elementary reaction. Counting the number of molecules involved, we get one molecule of NO and one molecule of Cl2. Molecularity: 2 (bimolecular) Now, let's write the rate law for this reaction, which will be proportional to the concentration of the reactants: Rate = k[NO][Cl2], where k is the rate constant. The molecularity and rate law for each elementary reaction are as follows: (a) Molecularity: 1, Rate Law: Rate = k[Cl2] (b) Molecularity: 2, Rate Law: Rate = k[OCl^-][H2O] (c) Molecularity: 2, Rate Law: Rate = k[NO][Cl2]

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Most popular questions from this chapter

(a) What are the units usually used to express the rates of reactions occurring in solution? (b) As the temperature increases, does the reaction rate increase or decrease? (c) As a reaction proceeds, does the instantaneous reaction rate increase or decrease?

The reaction \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2}\) has the rate constant \(k=0.63 M^{-1} \mathrm{s}^{-1}\) . (a) Based on the units for \(k,\) is the reaction first or second order in \(\mathrm{NO}_{2} ?\) ? (b) If the initial concentration of \(\mathrm{NO}_{2}\) is \(0.100 \mathrm{M},\) how would you determine how long it would take for the concentration to decrease to 0.025 \(\mathrm{M}\) ?

The oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is accelerated by \(\mathrm{NO}_{2} .\) The reaction proceeds according to: $$ \begin{array}{l}{\mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g)} \\ {2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)}\end{array}$$ (a) Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of \(S O_{2}\) by \(\mathrm{O}_{2}\) to give \(S O_{3} .(\mathbf{b})\) Do we consider \(N O_{2}\) a catalyst or an intermediate in this reaction? (c) Would you classify NO as a catalyst or as an intermediate? { ( d ) } Is this an example of homogeneous catalysis or heterogeneous catalysis?

Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) or silica \(\left(\mathrm{Si} \mathrm{O}_{2}\right)\) (a) Why is this an effective way of utilizing the catalyst material compared to having powdered metals? ( b) How does the surface area affect the rate of reaction?

The \(\mathrm{NO}_{x}\) waste stream from automobile exhaust includes species such as \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) . Catalysts that convert these species to \(\mathrm{N}_{2}\) are desirable to reduce air pollution. (a) Draw the Lewis dot and VSEPR structures of NO, NO \(_{2}\) and \(\mathrm{N}_{2} .(\mathbf{b})\) Using a resource such as Table \(8.3,\) look up the energies of the bonds in these molecules. In what region of the electromagnetic spectrum are these energies? (c) Design a spectroscopic experiment to monitor the conversion of \(\mathrm{NO}_{x}\) into \(\mathrm{N}_{2},\) describing what wavelengths of light need to be monitored as a function of time.
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