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Chapter 13: Problem 129

Why can't an elementary step in a mechanism have a rate law that is zero order in a reactant?

Short Answer

Expert verified
Answer: An elementary step in a reaction mechanism cannot have a rate law that is zero order in a reactant because it contradicts the fundamental idea that the rate of an elementary step is directly influenced by the concentrations of the reactants involved. In an elementary step, reactants collide directly to form products, so the reaction rate should be intrinsically linked to their concentrations. A zero-order dependence would imply that the reactant with zero order would not contribute to the reaction rate at all, even though it is involved in the elementary step. This is not consistent with the concept of direct collisions and the influence of reactant concentrations on the rate of an elementary step.

Step by step solution

01

Understanding elementary steps

In a reaction mechanism, an elementary step represents a single molecular event. These steps combine in a consecutive or concurrent manner to form the overall reaction. The rate of an elementary step can be influenced by the concentration of the reactants involved and the rate constant of the step.
02

Rate laws and order of reaction

A rate law expresses the relationship between the rate of a reaction and the concentration of the reactants. The order of reaction for a specific reactant represents the power to which its concentration is raised in the rate law and determines the sensitivity of the reaction rate to changes in that reactant's concentration. A zero-order reaction means the rate of the reaction is independent of the concentration of that reactant.
03

Elementary steps and stoichiometry

The rate law for an elementary step is derived from its stoichiometry, as it involves a direct collision of the reactants. For example, consider the following elementary step: A + B -> C. Its rate law can be expressed as: rate = k[A]^[m]*[B]^[n] where k is the rate constant, [A] and [B] are the concentrations of the reactants A and B, and m and n represent the orders of reaction for A and B, respectively.
04

Contradiction of zero-order dependence in an elementary step

Now, let's consider a reactant with a zero-order dependence in an elementary step, which means m=0 or n=0 in the rate law. If m=0, the rate law simplifies to rate = k*[B]^n, and similarly for n=0, rate = k*[A]^m. This means that the rate of the reaction is completely independent of the concentration of one of the reactants. In an elementary reaction, the reactants collide directly to form products, so the rate of the reaction should be intrinsically linked to the concentrations of the reactants. A zero-order dependence in the rate law for an elementary step contradicts this, as the reactant with a zero-order would not contribute to the reaction rate at all, even though it is involved in the elementary step. Therefore, an elementary step in a mechanism cannot have a rate law that is zero order in a reactant because this would contradict the fundamental idea that the rate of an elementary step is directly influenced by the concentrations of the reactants involved. The rate should depend on the collision frequency of the reactants, which is directly proportional to their concentrations.

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Most popular questions from this chapter

Nitric oxide (NO) is a gaseous free radical that plays many biological roles, including regulating neurotransmission and the human immune system. One of its many reactions involves the peroxynitrite ion (ONOO'): $$\mathrm{NO}(g)+\mathrm{ONOO}^{-(a q) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{2}^{-}(a q)}$$ a. Use the following data to determine the rate law and rate constant of the reaction at the experimental temperature at which these data were generated. $$\begin{array}{cccc}\text { Experiment } & \text { [NO]o (M) } & \text { [ONOO }\left.^{-}\right]_{0}(M) & \text { Rate }(M / \mathrm{s}) \\\\\hline 1 & 1.25 \times 10^{-4} & 1.25 \times 10^{-4} & 2.03 \times 10^{-11} \\\\\hline 2 & 1.25 \times 10^{-4} & 0.625 \times 10^{-4} & 1.02 \times 10^{-11} \\\\\hline 3 & 0.625 \times 10^{-4} & 2.50 \times 10^{-4} & 2.03 \times 10^{-11} \\\\\hline 4 & 0.625 \times 10^{-4} & 3.75 \times 10^{-4} & 3.05 \times 10^{-11} \\\\\hline\end{array}$$ b. Draw the Lewis structure of peroxynitrite ion (including all resonance forms) and assign formal charges. Note which form is preferred. c. Use the average bond energies in Appendix Table A4.1 to estimate the value of \(\Delta H_{\mathrm{rxn}}^{\circ}\) using the preferred structure from part (b).

The rate of a chemical reaction is too slow to measure at room temperature. We could either raise the temperature or add a catalyst. Which would be a better solution for making an accurate determination of the rate constant?

The rate of the reaction $$\mathrm{NO}_{2}(g)+\mathrm{CO}(g) \rightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g)$$ was determined in three experiments at \(225^{\circ} \mathrm{C} .\) The results are given in the following table: $$\begin{array}{cccc}\text { Experiment } & \left[\mathrm{NO}_{2}\right]_{0}(\mathrm{M}) & [\mathrm{CO}]_{0}(\mathrm{M}) & \begin{array}{c}\text { Initial Rate } \\\\(M / \mathrm{s})\end{array} \\\\\hline 1 & 0.263 & 0.826 & 1.44 \times 10^{-5} \\\\\hline 2 & 0.263 & 0.413 & 1.44 \times 10^{-5} \\\\\hline 3 & 0.526 & 0.413 & 5.76 \times 10^{-5} \\\\\hline\end{array}$$ a. Determine the rate law for the reaction. b. Calculate the value of the rate constant at \(225^{\circ} \mathrm{C}\) c. Calculate the rate of appearance of \(\mathrm{CO}_{2}\) when \(\left[\mathrm{NO}_{2}\right]=[\mathrm{CO}]=0.500 \mathrm{M}\)

Nitryl chloride, \(\mathrm{NO}_{2} \mathrm{Cl}\), is a reactive chlorine- containing species sometimes found in marine sediments in industrial areas. In the gas phase it decomposes to \(\mathrm{NO}_{2}\) and \(\mathrm{Cl}_{2}\) : $$2 \mathrm{NO}_{2} \mathrm{Cl}(g) \rightarrow 2 \mathrm{NO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ Under laboratory conditions, the rate of formation of \(\mathrm{NO}_{2}(g)\) is \(5.7 \times 10^{-6} M \cdot \mathrm{s}^{-1}\) a. What is the rate of formation of \(\mathrm{Cl}_{2}(g) ?\) b. What is the rate of consumption of \(\mathrm{NO}_{2} \mathrm{Cl}(g)\) ?

Hydroperoxyl radicals react rapidly with ozone to produce oxygen and OH radicals: $$\mathrm{HO}_{2}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{OH}(g)+2 \mathrm{O}_{2}(g)$$ The rate of this reaction was studied in the presence of a large excess of ozone. Determine the pseudo-first-order rate constant and the second-order rate constant for the reaction from the following data: $$\begin{array}{cll} \text { Time (ms) } & {\left[\mathrm{HO}_{2}\right](\mathrm{M})} & {\left[\mathrm{O}_{3}\right](\mathrm{M})} \\ \hline 0 & 3.2 \times 10^{-5} & 1.0 \times 10^{-3} \\ \hline 10 & 2.9 \times 10^{-5} & 1.0 \times 10^{-3} \\ \hline 20 & 2.6 \times 10^{-6} & 1.0 \times 10^{-3} \\ \hline 30 & 2.4 \times 10^{-6} & 1.0 \times 10^{-2} \\ \hline 80 & 1.4 \times 10^{-6} & 1.0 \times 10^{-3} \\ \hline \end{array}$$
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