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Chapter 14: Problem 66

(a) Can an intermediate appear as a reactant in the first step of a reaction mechanism? (b) On a reaction energy profile diagram, is an intermediate represented as a peak or a valley? (c) If a molecule like \(C l_{2}\) falls apart in an elementary reaction, what is the molecularity of the reaction?

Short Answer

Expert verified
(a) No, an intermediate cannot appear as a reactant in the first step of a reaction mechanism. (b) In a reaction energy profile diagram, an intermediate is represented as a valley. (c) If a molecule like \(Cl_{2}\) falls apart in an elementary reaction, the molecularity of the reaction is unimolecular.

Step by step solution

01

Question (a)

Can an intermediate appear as a reactant in the first step of a reaction mechanism? An intermediate is defined as a species formed during the course of a reaction that is neither a reactant nor a product. In the first step of a reaction mechanism, the species involved are the reactants (starting materials) and will not include any intermediates which are formed in the subsequent steps. So, the answer is no - an intermediate cannot appear as a reactant in the first step of a reaction mechanism.
02

Question (b)

On a reaction energy profile diagram, is an intermediate represented as a peak or a valley? A reaction energy profile diagram depicts the potential energy of a reaction system as it progresses along the reaction pathway. Transition states (a high-energy state where bonds are being broken and formed) are represented as peaks, whereas reactants, products, and intermediates are represented as valleys. Therefore, an intermediate would be represented as a valley on a reaction energy profile diagram.
03

Question (c)

If a molecule like \(Cl_{2}\) falls apart in an elementary reaction, what is the molecularity of the reaction? When a molecule like \(Cl_{2}\) falls apart, it dissociates into two radical species: \[Cl_{2} \rightarrow 2Cl\] This is a single event involving one reactant species, so the molecularity of this reaction is called "unimolecular."

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Most popular questions from this chapter

The decomposition reaction of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride is \(2 \mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) . The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\) . At \(64^{\circ} \mathrm{C}\) the rate constant is \(4.82 \times 10^{-3} \mathrm{s}^{-1}\) (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 M ?(\mathbf{c})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to 0.0480\(M ?(\mathbf{d})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is halved to 0.0120 \(\mathrm{M} ?\)

(a) Most commercial heterogeneous catalysts are extremely finely divided solid materials. Why is particle size important? (b) What role does adsorption play in the action of a heterogeneous catalyst?

The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) is second order in \(\mathrm{NO}\) and first order in \(\mathrm{O}_{2}\) . When \([\mathrm{NO}]=0.040 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M},\) the observed rate of disappearance of \(\mathrm{NO}\) is \(9.3 \times 10^{-5} \mathrm{M} / \mathrm{s}\) . (a) What is the rate of disappearance of \(\mathrm{O}_{2}\) at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of 1.8\(?\)

(a) What is a catalyst? (b) What is the difference between a homogeneous and a heterogeneous catalyst? (c) Do catalysts affect the overall enthalpy change for a reaction, the activation energy, or both?

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?
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