/*! 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 35 For the reaction \(\mathrm{C}_{2... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 35

For the reaction \(\mathrm{C}_{2} \mathrm{F}_{4} \rightarrow^{1 / 2} \mathrm{C}_{4} \mathrm{F}_{8,}\) a graph of \(1 /\left[\mathrm{C}_{2} \mathrm{F}_{4}\right]\) versus time gives a straight line with a slope of \(+0.04 \mathrm{L} / \mathrm{mol} \cdot \mathrm{s} .\) What is the rate law for this reaction?

Short Answer

Expert verified
Rate law: \( ext{rate} = 0.04 \, {L/mol} \cdot {s} \times [ {C}_2 {F}_4]^2 \).

Step by step solution

01

Identify Reaction Order

The straight line in the graph of \( \frac{1}{[{C}_2 {F}_4]} \) versus time indicates that the reaction follows a second-order rate law.
02

Write the General Rate Law for Second-Order Reactions

The general rate law for a second-order reaction is \( ext{rate} = k [{C}_2 {F}_4]^2 \) where \( k \) is the rate constant.
03

Determine Rate Constant from the Graph Slope

For a second-order reaction, the slope of the graph \( \frac{1}{[{A}]} \) versus time is equal to \( k \). Given the slope is \(+0.04 \, {L/mol} \cdot{s} \), the rate constant \( k \) is \( 0.04 \, {L/mol} \cdot {s} \).
04

Write the Specific Rate Law with the Known Constant

Now, substituting the rate constant into the general rate law, the specific rate law for this reaction becomes \( ext{rate} = 0.04 \, {L/mol} \cdot {s} \times [{C}_2 {F}_4]^2 \).

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 91影视!

Key Concepts

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

Reaction Order
The reaction order is a crucial concept in chemical kinetics, determining how the concentration of reactants affects the rate of reaction. For a reaction involving compound \( \mathrm{C}_2 \mathrm{F}_4 \rightarrow^{1/2} \mathrm{C}_4 \mathrm{F}_8 \), the reaction order can be identified by analyzing how the concentration changes with time.
In practical terms, looking at the graph of \( \frac{1}{[\mathrm{C}_2 \mathrm{F}_4]} \) versus time can reveal the order. If this results in a straight line, it confirms a second-order reaction. This is because a second-order reaction has the form \( \mathrm{rate} = k [A]^2 \).
  • Zero-order reactions: Concentration does not affect the rate.
  • First-order reactions: Rate is directly proportional to concentration.
  • Second-order reactions: Rate is proportional to the square of the concentration.
Understanding the reaction order helps predict how changes in concentration affect the reaction rate. It aids in determining the rate law and further kinetic analysis.
Rate Law
The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. In simple terms, it tells us how the speed of the reaction varies with the concentration of certain compounds involved.
For the reaction \( \mathrm{C}_2 \mathrm{F}_4 \rightarrow^{1/2} \mathrm{C}_4 \mathrm{F}_8 \), the rate law is important to determine because it uses the reaction order as a component. For a second-order reaction like this, the general rate law is \( \mathrm{rate} = k [\mathrm{C}_2 \mathrm{F}_4]^2 \).
  • The rate law is derived from experimental data rather than the stoichiometry of the reaction.
  • It includes the rate constant \( k \), providing crucial data about the reaction's velocity at a certain temperature.
The specific rate law formulated through the reaction order and use of the rate constant enhances our ability to predict how the reaction progresses under various conditions.
Rate Constant
The rate constant, denoted as \( k \), is a vital component in the rate law, responsible for linking the reaction rate with the concentrations of reactants. It's unique to every reaction and remains constant only at a specific temperature.
For the reaction \( \mathrm{C}_2 \mathrm{F}_4 \rightarrow^{1/2} \mathrm{C}_4 \mathrm{F}_8 \), the rate constant is determined using the slope of the graph \( \frac{1}{[\mathrm{C}_2 \mathrm{F}_4]} \) versus time, which is consistent with its second-order nature.
  • For a second-order reaction, the slope of this graph equates to the rate constant \( k \).
  • Given the slope is \( 0.04 \, \mathrm{L}/\mathrm{mol} \cdot \mathrm{s} \), this value serves as the rate constant for the specific reaction.
Understanding and determining the rate constant allows chemists to predict not just the speed but also how sensitive the rate is to concentration changes, thereby providing deeper insights into the reaction dynamics.

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

Formic acid decomposes at \(550^{\circ} \mathrm{C}\) according to the equation $$ \mathrm{HCO}_{2} \mathrm{H}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) $$ The reaction follows first-order kinetics. In an experiment, it is determined that \(75 \%\) of a sample of \(\mathrm{HCO}_{2} \mathrm{H}\) has decomposed in 72 seconds. Determine \(t_{1 / 2}\) for this reaction.

Nitramide, \(\mathrm{NO}_{2} \mathrm{NH}_{2}\), decomposes slowly in aqueous solution according to the following reaction: $$ \mathrm{NO}_{2} \mathrm{NH}_{2}(\mathrm{aq}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) $$ The reaction follows the experimental rate law $$ \text { Rate }=\frac{k\left[\mathrm{NO}_{2} \mathrm{NH}_{2}\right]}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]} $$ (a) What is the apparent order of the reaction in a pH buffered solution? (In a pH buffered solution, the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) is a constant.) (b) Which of the following mechanisms is the most appropriate for the interpretation of this rate law? Explain. (Note that when writing the expression for \(K,\) the equilibrium constant, \(\left.\left[\mathrm{H}_{2} \mathrm{O}\right] \text { is not involved. See Chapter } 15 .\right)\) Mechanism 1 \(\mathrm{NO}_{2} \mathrm{NH}_{2} \stackrel{k_{1}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}\) Mechanism 2 $$ \mathrm{NO}_{2} \mathrm{NH}_{2}+\mathrm{H}_{3} \mathrm{O}^{+} \frac{k_{2}}{\overleftarrow{k_{2}^{\prime}}} \mathrm{NO}_{2} \mathrm{NH}_{3}^{+}+\mathrm{H}_{2} \mathrm{O} $$ \(\mathrm{NO}_{2} \mathrm{NH}_{3}^{+} \stackrel{k_{3}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{3} \mathrm{O}^{+} \quad\) (rate-limiting step) Mechanism 3 \(\mathrm{NO}_{2} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \frac{k_{4}}{k_{4}^{\prime}} \mathrm{NO}_{2} \mathrm{NH}^{-}+\underset{(\text { rapid equilibrium })}{\mathrm{H}_{3} \mathrm{O}^{+}}\) \(\mathrm{NO}_{2} \mathrm{NH}^{-} \stackrel{k_{5}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{OH}^{-} \quad\) (rate-limiting step) \(\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-} \stackrel{k_{6}}{\longrightarrow} 2 \mathrm{H}_{2} \mathrm{O} \quad\) (very fast reaction) (c) Show the relationship between the experimentally observed rate constant, \(k\), and the rate constants in the selected mechanism. (d) Based on the experimental rate law, will the reaction rate increase or decrease if the pH of the solution is increased?

When heated to a high temperature, cyclobutane, \(\mathrm{C}_{4} \mathrm{H}_{8},\) decomposes to ethylene: $$ \mathrm{C}_{4} \mathrm{H}_{8}(\mathrm{g}) \rightarrow 2 \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g}) $$ The activation energy, \(E_{x}\) for this reaction is \(260 \mathrm{kJ} / \mathrm{mol} .\) At \(800 \mathrm{K},\) the rate constant \(k=\) \(0.0315 \mathrm{s}^{-1} .\) Determine the value of \(k\) at \(850 \mathrm{K}\).

What is the rate law for each of the following elementary reactions? (a) \(\mathrm{Cl}(\mathrm{g})+\mathrm{ICl}(\mathrm{g}) \rightarrow \mathrm{I}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) (b) \(\mathrm{O}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{g}) \rightarrow 2 \mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\)

You want to study the hydrolysis of the beautiful green, cobalt-based complex called transdichlorobis-(ethylenediamine)cobalt(III) ion, In this hydrolysis reaction, the green complex ion trans- \(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\) forms the red complex ion \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}\) as a \(\mathrm{Cl}^{-}\) ion is replaced with a water molecule on the \(\mathrm{Co}^{3+}\) ion (en \(=\) \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) trans-\(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) Reactions such as this have been studied extensively, and experiments suggest that the initial, slow step in the reaction is the breaking of the Co-Cl bond to give a five-coordinate intermediate. The intermediate is then attacked rapidly by water. Slow: \(\left.\quad \text { trans-ICo(en) }_{2} \mathrm{Cl}_{2}\right]^{+}(\mathrm{aq}) \rightarrow\) $$ \left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) $$ Fast: \(\quad\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$ \left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}(\mathrm{aq}) $$ (a) Based on the reaction mechanism, what is the predicted rate law? (b) As the reaction proceeds, the color changes from green to red with an intermediate stage where the color is gray. The gray color is reached at the same time, no matter what the concentration of the green starting material (at the same temperature). How does this show the reaction is first-order in the green form? Explain. (c) The activation energy for a reaction can be found by plotting In \(k\) versus \(1 / T .\) However, here we do not need to measure \(k\) directly. Instead, because \(k=-(1 / t) \ln \left([\mathrm{R}] /[\mathrm{R}]_{0}\right),\) the time needed to achieve the gray color is a measure of \(k\). Use the data below to find the activation energy. green $$ \left.\underset{\text { red }}{\operatorname{Co}(\mathrm{en})_{2}}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) $$ The reaction progress is followed by observing the color of the solution. The original solution is green, and the final solution is red, but at some intermediate stage when both the reactant and product are present, the solution is gray. The shape in the middle of the beaker is a vortex that arises because the solutions are being stirred using a magnetic stirring bar in the bottom of the beaker.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation
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.