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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. (a) $\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CN}^{-}(a q) \longrightarrow \mathrm{HCN}(a q)$ (b) \(\mathrm{CH}_{3} \mathrm{Cl}(\) solv \()+\mathrm{OH}^{-}(\) solv $) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(\( solv \))+\mathrm{Cl}^{-}($ solv \()\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightarrow 2 \mathrm{NO}_{2}\)

Short Answer

Expert verified
(a) The molecularity of the reaction is 2 (bimolecular). The rate law is: \[Rate = k [\mathrm{H}_2 \mathrm{O}] [\mathrm{CN}^-]\] (b) The molecularity of the reaction is 2 (bimolecular). The rate law is: \[Rate = k [\mathrm{CH}_3 \mathrm{Cl}] [\mathrm{OH}^-]\] (c) The molecularity of the reaction is 1 (unimolecular). The rate law is: \[Rate = k [\mathrm{N}_2 \mathrm{O}_4]\]

Step by step solution

01

(a) Molecularity of Reaction 1

In this reaction, there are two reactants: H2O and CN. Hence, the molecularity of this elementary reaction is 2, making it a bimolecular reaction.
02

(a) Rate Law for Reaction 1

The rate law will be the rate constant (k) multiplied by the concentrations of the reactants, each raised to the power of their respective stoichiometric coefficients. Since this is an elementary reaction, the coefficients are equal to the number of molecules of each reactant involved in the reaction. In this case, the rate law becomes: \[Rate = k [\mathrm{H}_2 \mathrm{O}] [\mathrm{CN}^-]\]
03

(b) Molecularity of Reaction 2

In this reaction, there are two reactants: CH3Cl and OH-. Hence, the molecularity of this elementary reaction is 2, making it a bimolecular reaction.
04

(b) Rate Law for Reaction 2

Following the same reasoning as in Reaction 1, the rate law for Reaction 2 will be the rate constant (k) multiplied by the concentrations of the reactants, each raised to the power of their respective stoichiometric coefficients. In this case, the rate law becomes: \[Rate = k [\mathrm{CH}_3 \mathrm{Cl}] [\mathrm{OH}^-]\]
05

(c) Molecularity of Reaction 3

In this reaction, there is only one reactant: N2O4. Hence, the molecularity of this elementary reaction is 1, making it a unimolecular reaction.
06

(c) Rate Law for Reaction 3

Following the same reasoning as in the previous reactions, the rate law for Reaction 3 will be the rate constant (k) multiplied by the concentration of the reactant, raised to the power of its stoichiometric coefficient. In this case, the rate law becomes: \[Rate = k [\mathrm{N}_2 \mathrm{O}_4]\]

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Most popular questions from this chapter

(a) The reaction $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+\( \)\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)$ is first order with in \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)\) and zero-order in \(\mathrm{H}_{2} \mathrm{O}\). At \(300 \mathrm{~K}\) the rate constant equals \(3.30 \times 10^{-2} \mathrm{~min}^{-1} .\) Calculate the half- life at this temperature. \((\mathbf{b})\) If the activation energy for this reaction is \(80.0 \mathrm{~kJ} / \mathrm{mol}\), at what temperature would the reaction rate be doubled?

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 \(\mathrm{NO}, \mathrm{NO}_{2}\), 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? \((\mathbf{c})\) 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.

For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product and disappearance of each reactant: (a) $\mathrm{O}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g)$ (b) $4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$ (c) $2 \mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ (d) $\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2}(g) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{NH}_{3}(g)$

Hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) is a common and troublesome pollutant in industrial wastewaters. One way to remove \(\mathrm{H}_{2} \mathrm{~S}\) is to treat the water with chlorine, in which case the following reaction occurs: $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ The rate of this reaction is first order in each reactant. The rate constant for the disappearance of \(\mathrm{H}_{2} \mathrm{~S}\) at $30{ }^{\circ} \mathrm{C}\( is \)4.0 \times 10^{-2} M^{-1} \mathrm{~s}^{-1}$. If at a given time the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is $2.5 \times 10^{-4} \mathrm{M}\( and that of \)\mathrm{Cl}_{2}\( is \)2.0 \times 10^{-2} \mathrm{M},$ what is the rate of formation of \(\mathrm{H}^{+}\) ?

As shown in Figure 14.23 , the first step in the heterogeneous hydrogenation of ethylene is adsorption of the ethylene molecule on a metal surface. One proposed explanation for the "sticking" of ethylene to a metal surface is the interaction of the electrons in the \(\mathrm{C}-\mathrm{C} \pi\) bond with vacant orbitals on the metal surface. (a) If this notion is correct, would ethane be expected to adsorb to a metal surface, and, if so, how strongly would ethane bind compared to ethylene? (b) Based on its Lewis structure, would you expect ammonia to adsorb to a metal surface using a similar explanation as for ethylene?
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