/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 What is the rate law for each of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What is the rate law for each of the following elementary reactions? (a) \(\mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\) (b) \(\mathrm{Cl}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{HCl}(\mathrm{g})+\mathrm{H}(\mathrm{g})\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}(\mathrm{aq}) \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})\)

Short Answer

Expert verified
(a) Rate = k[NO][NO₃]; (b) Rate = k[Cl][H₂]; (c) Rate = k[(CH₃)₃CBr].

Step by step solution

01

Understanding Elementary Reactions

An elementary reaction is a single-step reaction with only one transition state, and its rate law can be directly written using the stoichiometric coefficients of the reactants. The rate of such a reaction is proportional to the concentration of each reactant raised to the power of its coefficient in the balanced equation.
02

Rate Law for Reaction (a)

For the reaction \( \mathrm{NO} + \mathrm{NO}_3 \rightarrow 2 \mathrm{NO}_2 \), the stoichiometric coefficients for the reactants \( \mathrm{NO} \) and \( \mathrm{NO}_3 \) are both 1. Thus, the rate law is given by: \[ \text{Rate} = k[\mathrm{NO}] [\mathrm{NO}_3]\] where \( k \) is the rate constant.
03

Rate Law for Reaction (b)

For the reaction \( \mathrm{Cl} + \mathrm{H}_2 \rightarrow \mathrm{HCl} + \mathrm{H} \), the stoichiometric coefficients for the reactants \( \mathrm{Cl} \) and \( \mathrm{H}_2 \) are both 1. Thus, the rate law is: \[\text{Rate} = k[\mathrm{Cl}] [\mathrm{H}_2]\] where \( k \) is the rate constant.
04

Rate Law for Reaction (c)

For the reaction \( (\mathrm{CH}_3)_3 \mathrm{CBr} \rightarrow (\mathrm{CH}_3)_3 \mathrm{C}^+ + \mathrm{Br}^- \), the reactant \( (\mathrm{CH}_3)_3 \mathrm{CBr} \) has a stoichiometric coefficient of 1. Thus, the rate law is: \[ \text{Rate} = k[(\mathrm{CH}_3)_3 \mathrm{CBr}]\] where \( k \) is the rate constant.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Rate Law
The rate law is a mathematical equation that describes the speed of a chemical reaction. For elementary reactions, this law is straightforward, as it can be directly determined from the stoichiometric coefficients of the reactants in the reaction equation. This is because each elementary reaction represents a single-step process following the specified molecular interactions.

To write a rate law for any elementary reaction, follow these steps:
  • Identify the reactants involved in the step.
  • Determine their stoichiometric coefficients from the balanced equation.
  • Express the rate law using these coefficients as exponents for each reactant concentration.
For instance, in the reaction \( \mathrm{NO} + \mathrm{NO}_3 \rightarrow 2 \mathrm{NO}_2 \), the stoichiometric coefficients for the reactants are 1. Therefore, the rate law is \( \text{Rate} = k[\mathrm{NO}][\mathrm{NO}_3] \), where \( k \) is the rate constant. This tells us the reaction speed depends on the concentration of \( \mathrm{NO} \) and \( \mathrm{NO}_3 \).
Reaction Kinetics
Reaction kinetics is the study of the rates at which chemical processes occur and the factors that affect these rates. It looks at how quickly reactants are converted into products, and the sequence of steps that reactants go through, known as the reaction mechanism.

In elementary reactions, reaction kinetics can be quite simple. Since these processes occur in a single step, the rate-limiting factor typically conforms directly to the concentrations of the reactants as expressed in the rate law. By understanding the kinetics of a reaction, chemists can design processes to optimize production rates or develop strategies to control biological systems.

For example, in the reaction \( \mathrm{Cl} + \mathrm{H}_2 \rightarrow \mathrm{HCl} + \mathrm{H} \), kinetics tells us that the rate is proportional to the concentration of \( \mathrm{Cl} \) and \( \mathrm{H}_2 \), thus the reaction can be influenced by altering these concentrations or temperature.
Stoichiometric Coefficients
Stoichiometric coefficients are numbers placed in front of compounds in a balanced chemical equation. They represent the ratio of moles of a reactant or product involved in the reaction. In the context of an elementary reaction, these coefficients not only help balance the equation but also dictate the form of the rate law.

In the reaction \( (\mathrm{CH}_3)_3 \mathrm{CBr} \rightarrow (\mathrm{CH}_3)_3 \mathrm{C}^+ + \mathrm{Br}^- \), the stoichiometric coefficient is 1 for the reactant \((\mathrm{CH}_3)_3 \mathrm{CBr}\). This informs us that for its rate law, the concentration term of the reactant is raised to the power of 1: \( \text{Rate} = k[(\mathrm{CH}_3)_3 \mathrm{CBr}] \).

Understanding these coefficients is crucial for predicting how changes in concentrations will affect the reaction rate, and thus these coefficients are fundamental for creating efficient reaction plans in both laboratory and industrial settings.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When heated, cyclopropane is converted to propene (see Example 15.5 ). Rate constants for this reaction at \(470^{\circ} \mathrm{C}\) and \(510^{\circ} \mathrm{C}\) are \(k=1.10 \times 10^{-4} \mathrm{s}^{-1}\) and \(k=1.02 \times10^{-3} \mathrm{s}^{-1},\) respectively. Determine the activation energy, \(E_{\mathrm{a}}\) from these data.

Common sugar, sucrose, breaks down in dilute acid solution to form glucose and fructose. Both products have the same formula, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}.\) $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})$$ The rate of this reaction has been studied in acid solution, and the data in the table were obtained. $$\begin{array}{cc}\hline \begin{array}{c}\text { Time } \\\\(\text { min })\end{array} & \begin{array}{c}{\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathbf{0}_{11}\right]} \\\\(\mathrm{mol} / \mathrm{L})\end{array} \\\\\hline 0 & 0.316 \\\39 & 0.274 \\\80 & 0.238 \\\140 & 0.190 \\\210 & 0.146 \\\\\hline\end{array}$$ (a) Plot ln [sucrose] versus time and \(1 /\) [sucrose] versus time. What is the order of the reaction? (b) Write the rate equation for the reaction, and calculate the rate constant, \(k.\) (c) Estimate the concentration of sucrose after 175 min.

The conversion of cyclopropane to propene, described in Example \(15.5,\) occurs with a first-order rate constant of \(5.4 \times 10^{-2} \mathrm{h}^{-1} .\) How long will it take for the concentration of cyclopropane to decrease from an initial concentration \(0.080 \mathrm{mol} / \mathrm{L}\) to \(0.020 \mathrm{mol} / \mathrm{L} ?\)

Ammonium cyanate, NH_NCO, rearranges in water to give urea, (NH\(_{2}\)) \(_{2}\) CO: $$\mathrm{NH}_{4} \mathrm{NCO}(\mathrm{aq}) \longrightarrow\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}(\mathrm{aq})$$ The rate equation for this process is "Rate \(=k\) \(\left[\mathrm{NH}_{4} \mathrm{NCO}\right]^{2}, "\) where \(k=0.0113 \mathrm{L} / \mathrm{mol} \cdot\) min. If the original concentration of \(\mathrm{NH}_{4} \mathrm{NCO}\) in solution is \(0.229 \mathrm{mol} / \mathrm{L}\) how long will it take for the concentration to decrease to \(0.180 \mathrm{mol} / \mathrm{L} ?\)

The decomposition of dinitrogen pentaoxide $$2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ has the following rate equation: \(-\Delta\left[\mathrm{N}_{2} \mathrm{O}_{5}\right] / \Delta t=k\left[\mathrm{N}_{2} \mathrm{O}_{5}\right].\) It has been found experimentally that the decomposition is \(20 \%\) complete in \(6.0 \mathrm{h}\) at \(300 \mathrm{K}\). Calculate the rate constant and the half-life at \(300 \mathrm{K}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.