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

Short Answer

Expert verified
(a) Molecularity: 1 (unimolecular). Rate law: \(Rate = k[Cl_2]\) (b) Molecularity: 2 (bimolecular). Rate law: \(Rate = k[OCl^-][H_2O]\) (c) Molecularity: 2 (bimolecular). Rate law: \(Rate = k[NO][Cl_2]\)

Step by step solution

01

Reaction (a) Molecularity

In this reaction, a Cl鈧 molecule dissociates into two Cl atoms. Only one molecule is involved in the rate-determining step, so the molecularity is 1, also known as a unimolecular reaction.
02

Reaction (a) Rate Law

The rate of the reaction is directly proportional to the concentration of the reactant Cl鈧. Thus, the rate law can be written as: Rate = k[Cl鈧俔, where k is the rate constant.
03

Reaction (b) Molecularity

In this reaction, OCl鈦 reacts with H鈧侽 to form HOCl and OH鈦. Two molecules are involved in the rate-determining step, so the molecularity is 2, also known as a bimolecular reaction.
04

Reaction (b) Rate Law

The rate of the reaction is directly proportional to the concentration of the reactants OCl鈦 and H鈧侽. Thus, the rate law can be written as: Rate = k[OCl鈦籡[H鈧侽], where k is the rate constant.
05

Reaction (c) Molecularity

In this reaction, NO and Cl鈧 react to form NOCl鈧. Two molecules are involved in the rate-determining step, making this another bimolecular reaction.
06

Reaction (c) Rate Law

The rate of the reaction is directly proportional to the concentration of the reactants NO and Cl鈧. Thus, the rate law can be written as: Rate = k[NO][Cl鈧俔, where k is the rate constant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecularity of Reaction
The molecularity of a reaction refers to the number of reactant molecules that collide and participate in a single elementary step. Understanding the molecularity is vital because it provides insights into the reaction mechanism, which is the step-by-step sequence of elementary reactions that make up the overall reaction.

Molecularity is always an integer and can be unimolecular, involving a single molecule, bimolecular, involving two molecules, or termolecular, involving three molecules. It's important to note that molecularity is defined only for elementary reactions - these are reactions that occur in a single step and represent the actual physical process at the molecular level.

For instance, in the example reaction \(\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{Cl}(g)\), we determined the molecularity to be unimolecular because only one molecule of \(\mathrm{Cl}_{2}\) is involved in the rate-determining step.
Rate Law
The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. It is derived empirically, meaning it is based on experimental observations rather than theoretical predictions.

A typical rate law takes the form of \(\text{Rate} = k[\text{Reactant}_1]^{n_1}[\text{Reactant}_2]^{n_2}...\), where \(k\) is the rate constant and \(n_1\), \(n_2\), etc., are the reaction orders with respect to each reactant. These orders can be integers, fractions, or even zero which indicate that the reaction rate is independent of the concentration of that particular reactant.

In the case of the reaction \((b)\), \(\mathrm{OCl}^{-} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HOCl} + \mathrm{OH}^{-}\), the rate law was found to be \(\text{Rate} = k[\mathrm{OCl}^{-}][\mathrm{H}_{2}O]\), suggesting a first order dependence on each reactant.
Unimolecular Reaction
A unimolecular reaction involves a single reactant molecule undergoing a chemical change to transform into products. This type of reaction is characterized by its molecularity of one, indicating that the reaction proceeds when just one reactant molecule is activated or has enough energy to undergo the change.

An example includes the decomposition of \(\mathrm{Cl}_{2}\) into chlorine atoms, as seen in reaction \(\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{Cl}(g)\). Since it is an elementary reaction, the rate at which it occurs depends solely on the concentration of \(\mathrm{Cl}_{2}\), which fits with the rate law \(\text{Rate} = k[\mathrm{Cl}_{2}]\). The simplicity of unimolecular reactions makes them important conceptual tools in understanding chemical kinetics.
Bimolecular Reaction
In contrast to unimolecular reactions, bimolecular reactions involve the collision between two reactant molecules. The molecularity for these reactions is two. The reactions are governed by the kinetic theory of gases, which suggests that the frequency and energy of collisions between molecules greatly influence reaction rates.

Examples of bimolecular reactions include \((b)\) \(\mathrm{OCl}^{-} + \mathrm{H}_{2} \mathrm{O}\) and \((c)\) \(\mathrm{NO} + \mathrm{Cl}_{2}\), both of which require the interaction of two reactant molecules for the reaction to occur. The corresponding rate laws are \(\text{Rate} = k[\mathrm{OCl}^{-}][\mathrm{H}_{2}O]\) and \(\text{Rate} = k[\mathrm{NO}][\mathrm{Cl}_{2}]\), respectively, which indicate that the reaction rate is proportional to the product of the concentrations of both reactants involved in the elementary step.

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Most popular questions from this chapter

The reaction $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). Using the following kinetic data, determine the magnitude of the first-order rate constant: $$ \begin{array}{rl} \hline \text { Time (s) } & \text { Pressure } \mathrm{SO}_{2} \mathrm{Cl}_{2} \text { (atm) } \\ \hline 0 & 1.000 \\ 2,500 & 0.947 \\ 5,000 & 0.895 \\ 7,500 & 0.848 \\ 10,000 & 0.803 \\ \hline \end{array} $$

The gas-phase reaction \(\mathrm{Cl}(g)+\operatorname{HBr}(g) \longrightarrow \mathrm{HCl}(g)+\) \(\mathrm{Br}(g)\) has an overall enthalpy change of \(-66 \mathrm{~kJ}\). The activation energy for the reaction is \(7 \mathrm{~kJ}\). (a) Sketch the energy profile for the reaction, and label \(E_{g}\) and \(\Delta E\). (b) What is the activation energy for the reverse reaction?

Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in NO. Write the rate law. (b) If the rate constant for this reaction at \(1000 \mathrm{~K}\) is \(6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\), what is the reaction rate when \([\mathrm{NO}]=0.035 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathrm{c})\) What is the reaction rate at \(1000 \mathrm{~K}\) when the concentration of \(\mathrm{NO}\) is increased to \(0.10 \mathrm{M}\), while the concentration of \(\mathrm{H}_{2}\) is \(0.010 \mathrm{M}\) ?

The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For example, consider the reaction between methane and chlorine to produce methyl chloride and hydrogen chloride: Reaction 1: \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(\mathrm{g})+\mathrm{HCl}(\mathrm{g})\) This reaction is very slow in the absence of light. However, \(\mathrm{Cl}_{2}(g)\) can absorb light to form \(\mathrm{Cl}\) atoms: $$ \text { Reaction 2: } \mathrm{Cl}_{2}(g)+h v \longrightarrow 2 \mathrm{Cl}(g) $$ Once the \(\mathrm{Cl}\) atoms are generated, they can catalyze the reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{Cl}_{2}\), according to the following proposed mechanism: Reaction 3: \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}(\mathrm{g}) \longrightarrow \mathrm{CH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g})\) Reaction 4: \(\mathrm{CH}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{Cl}(g)\) The enthalpy changes and activation energies for these two reactions are tabulated as follows: $$ \begin{array}{lll} \hline \text { Reaction } & \Delta H_{\mathrm{ran}}^{\circ}(\mathrm{kJ} / \mathrm{mol}) & E_{a}(\mathrm{~kJ} / \mathrm{mol}) \\ \hline 3 & +4 & 17 \\ 4 & -109 & 4 \end{array} $$ (a) By using the bond enthalpy for \(\mathrm{Cl}_{2}\) (Table \(8.4\) ), determine the longest wavelength of light that is energetic enough to cause reaction 2 to occur. In which portion of the electromagnetic spectrum is this light found? (b) By using the data tabulated here, sketch a quantitative energy profile for the catalyzed reaction represented by reactions 3 and 4. (c) By using bond enthalpies, estimate where the reactants, \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g)\), should be placed on your diagram in part (b). Use this result to estimate the value of \(E_{a}\) for the reaction \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow\) \(\mathrm{CH}_{3}(g)+\mathrm{HCl}(g)+\mathrm{Cl}(g) .\) (d) The species \(\mathrm{Cl}(g)\) and \(\mathrm{CH}_{3}(\mathrm{~g})\) in reactions 3 and 4 are radicals, that is, atoms or molecules with unpaired electrons. Draw a Lewis structure of \(\mathrm{CH}_{3}\), and verify that is a radical. (e) The sequence of reactions 3 and 4 comprise a radical chain mechanism. Why do you think this is called a "chain reaction"? Propose a reaction that will terminate the chain reaction.

Consider two reactions. Reaction (1) has a constant halflife, whereas reaction (2) has a half-life that gets longer as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

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