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Define an elementary step and explain why equations for elementary reactions can be used to predict the rate law, but the overall reaction stoichiometry cannot.

Short Answer

Expert verified
Elementary reactions provide direct rate laws from their molecular interactions, unlike overall stoichiometry, which doesn't consider reaction pathways.

Step by step solution

01

Understanding Elementary Steps

An elementary step in a reaction mechanism is a single reaction event that describes a simple molecular process. It cannot be broken down into simpler steps, and it often involves the collision and interaction of species at the molecular level.
02

Rate Law in Elementary Reactions

For an elementary reaction, the rate law can be directly derived from its molecularity, which is the number of reactant molecules involved in the step. If an elementary step involves, for example, two molecules reacting, the rate law is typically second order, involving the concentration of each reactant to the power of one. Thus, for a reaction step like \( A + B \rightarrow C \), the rate law would be \( rate = k[A][B] \).
03

Stoichiometry of Overall Reactions

The stoichiometry of the overall reaction provides a balanced equation for the reaction as a whole but does not give details about the individual steps involved. It reflects the net change in quantities of reactants and products but does not account for intermediates or transition states.
04

Comparing Rate Laws and Stoichiometry

The rate law for a reaction provides the actual rate relationships observed experimentally. For non-elementary (complex) reactions, the stoichiometry of the overall reaction does not directly predict this law because the reaction proceeds through multiple steps with intermediates. In contrast, each elementary step has its own rate law that reflects the immediate interaction of molecules, capturing the kinetic essence of the process.

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Key Concepts

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

Reaction Mechanism
A reaction mechanism is a detailed pathway describing how a chemical reaction occurs at the molecular level. It breaks down the overall reaction into several elementary steps. Each step represents a basic molecular event, typically involving the collision or transformation of molecules.
Understanding reaction mechanisms is crucial because they reveal the stepwise process through which reactants turn into products. This pathway often includes various intermediates, which are species formed in one step and consumed in another.
By analyzing a reaction mechanism, chemists can predict the rate laws of the individual elementary steps and understand the overall kinetics of complex reactions. Knowing these steps helps provide insights into reaction speeds and enables the fine-tuning of conditions to optimize the process.
Rate Law
The rate law is a mathematical expression that relates the rate of a chemical reaction to the concentration of its reactants. For an elementary reaction, the rate law is derived directly from the molecularity, which determines how many molecules collide in a step.
For example, in the elementary step:
  • \[ A + B \rightarrow C \]
  • Rate law: \[ rate = k[A][B] \]
This indicates that the reaction rate depends linearly on the concentration of both species.
It's important to note that for elementary reactions, you can predict the rate law from the stoichiometric coefficients of the reactants in the step. However, this is not true for overall reactions, particularly complex ones.
Molecularity
Molecularity refers to the number of molecules or ions that participate in an elementary step of a chemical reaction. Unlike complex reactions, which may consist of many steps, an elementary reaction is a single collision event.
Common types of molecularity include:
  • Unimolecular: An elementary step involving a single molecule, like isomerization.
  • Bimolecular: A step involving the collision between two molecules or ions.
  • Termolecular: An interaction involving three particles, which is less common due to the low probability of simultaneous collisions.
By understanding molecularity, chemists can deduce the order of the reaction and formulate the rate law directly for elementary steps, which offers insights into the kinetics of the process.
Stoichiometry
Stoichiometry in chemistry deals with the quantitative relationships between reactants and products in a chemical reaction. It provides the balanced equation, which ensures the conservation of atoms according to the law of conservation of mass.
In the context of overall reactions, stoichiometry gives the net conversion of reactants to products but doesn't reveal the underlying mechanism. This is the major distinction since, for complex reactions, stoichiometry doesn't predict the intermediate steps or the precise rate law.
While stoichiometry is crucial for understanding quantities needed and produced, it's the reaction mechanism that reveals the sequence and nature of molecular interactions. Thus, a firm grasp of both stoichiometry and reaction mechanisms is essential for mastering chemical kinetics.

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Most popular questions from this chapter

Nitramide decomposes to water and dinitrogen monoxide. $$ \mathrm{H}_{2} \mathrm{NNO}_{2}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{N}_{2} \mathrm{O}(\mathrm{g}) $$ This reaction was studied by J. N. Brønsted in 1924 as part of research into the fundamental nature of acids and bases. If \(1.00 \mathrm{~L}\) of \(0.440 \mathrm{M}\) nitramide is placed in a reactor at \(20^{\circ} \mathrm{C}\), the following results are expected. (The experiment was actually performed to measure the effect of malate ion on the rate of reaction.) $$ \begin{array}{cc} \mathrm{T} \text { (min) } & P \text { (torr) } \\ \hline 0.00 & 17.54 \\ 0.50 & 18.98 \\ 1.00 & 20.09 \\ 2.00 & 21.65 \\ 4.00 & 23.46 \\ 6.00 & 24.48 \\ 8.00 & 25.14 \\ 10.00 & 25.59 \\ \text { Completion } & 27.54 \end{array} $$ What is the rate law for this reaction? (Hint: The data show the increase in concentration of a product, which differs from the other problems in this book. The problem looks more familiar if you create a decay curve. Note that the initial point represents a small amount of product, and thus a large amount of reactant. The final point represents a large amount of product and no reactant. The problem looks like any other kinetics problem if you first calculate ([final pressure \(-\) current pressure \(]\) to see the data as a decay curve). Graph (final pressure \(-\) current pressure), rather than the current pressure, as a function of time to see the data. If this is not a straight line, you can graph \(\ln (\) final pressure current pressure) and \(1 /\) (final pressure \(-\) current pressure) to determine the order of the reaction. Scientists frequently transform their data to a familiar form. These operations make it easier to interpret the data.

OBJECTIVE. Relate temperature, activation energy, and rate constant through the Arrhenius equation. The decomposition of formic acid (see Exercise 13.49\()\) is measured at several temperatures. The temperature dependence of the first-order rate constant is: $$ \begin{array}{cc} \mathrm{T}(\mathrm{K}) & k\left(\mathrm{~s}^{-1}\right) \\ \hline 800 & 0.00027 \\ 825 & 0.00049 \\ 850 & 0.00086 \\ 875 & 0.00143 \\ 900 & 0.00234 \\ 925 & 0.00372 \end{array} $$ Calculate the activation energy, in kilojoules, and the preexponential term. You may use a graphic method if you have a spreadsheet or graphing calculator, or an algebraic method if you do not.

Explain how to use the method of initial rates to determine the rate law for $$ \mathrm{CH}_{3} \mathrm{Br}+\mathrm{OH}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{OH}+\mathrm{Br}^{-} $$

Define activation energy, and explain how it influences the rate of reaction.

Assume that the chemical reaction is reactants \(\rightarrow\) products. Determine the rate constant and order from the concentration-time dependence. $$ \begin{array}{cc} \text { Time (s) } & \text { [Reactant] (M) } \\ \hline 0 & 0.250 \\ 1 & 0.216 \\ 2 & 0.182 \\ 3 & 0.148 \\ 4 & 0.114 \\ 5 & 0.080 \end{array} $$

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