/*! 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 73 Chlorine dioxide oxidizes iodide... [FREE SOLUTION] | 91Ó°ÊÓ

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Chlorine dioxide oxidizes iodide ion in aqueous solution to iodine; chlorine dioxide is reduced to chlorite ion. $$2 \mathrm{ClO}_{2}(a q)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{ClO}_{2}^{-}(a q)+\mathrm{I}_{2}(a q)$$ The order of the reaction with respect to \(\mathrm{ClO}_{2}\) was determined by starting with a large excess of \(\mathrm{I}^{-}\), so that its concentration was essentially constant. Then $$\text { Rate }=k\left[\mathrm{ClO}_{2}\right]^{m}\left[\mathrm{I}^{-}\right]^{n}=k^{\prime}\left[\mathrm{ClO}_{2}\right]{m}$$ where \(k^{\prime}=k\left[\mathrm{I}^{-}\right]^{n}\). Determine the order with respect to \(\mathrm{ClO}_{2}\) and the rate constant \(k^{\prime}\) by plotting the following data assuming firstand then second-order kinetics. [Data from H. Fukutomi and G. Gordon, J. Am. Chem. Soc., 89,1362 (1967).] \(\begin{array}{cl}\text { Time (s) } & {\left[\text { ClO }_{2}\right] \text { (moll } \text { L) }} \\ 0.00 & 4.77 \times 10^{-4} \\ 1.00 & 4.31 \times 10^{-4} \\ 2.00 & 3.91 \times 10^{-4} \\ 3.00 & 3.53 \times 10^{-4} \\ 5.00 & 2.89 \times 10^{-4} \\ 10.00 & 1.76 \times 10^{-4} \\ 30.00 & 2.4 \times 10^{-5} \\ 50.00 & 3.2 \times 10^{-6}\end{array}\)

Short Answer

Expert verified
The reaction is first-order with respect to \(\mathrm{ClO}_2\), and \(k'\) is determined by the slope of the linear first-order plot.

Step by step solution

01

Determine Reaction Order Assumptions

First, we must assume possible orders of reaction with respect to \(\mathrm{ClO}_2\): a first order \((m=1)\) and then a second order \((m=2)\). We will use these assumptions to linearize the concentration data and see which fits better.
02

Linearize First-Order Kinetics

For a first-order reaction, the rate equation is given by \(\text{Rate} = -\frac{d[\mathrm{ClO}_2]}{dt} = k'[\mathrm{ClO}_2]\). The integrated rate law takes the form \[-\log([\mathrm{ClO}_2]) = k't + \log([\mathrm{ClO}_2]_0)\]where \([\mathrm{ClO}_2]_0\) is the initial concentration. Plot \(-\log([\mathrm{ClO}_2])\) versus time \(t\) to test if the data appears linear, indicating first-order kinetics.
03

Plot First-Order Data

Convert the \([\mathrm{ClO}_2]\) data into \(-\log([\mathrm{ClO}_2])\) and plot against the given time data. The plot appears linear if the reaction is first-order, allowing us to see if the data fits well with the rate law.
04

Analyze First-Order Fit

If the plot in Step 3 is linear, calculate the slope of the line. This slope represents the rate constant \(k'\). Confirm the goodness of fit (linear regression) to determine if the first-order assumption is valid.
05

Linearize Second-Order Kinetics

If the first-order assumption doesn't hold, for a second-order reaction the rate law is \[-\frac{1}{[\mathrm{ClO}_2]} = k't + \frac{1}{[\mathrm{ClO}_2]_0}\]. Plot \(1/[\mathrm{ClO}_2]\) versus time \(t\) to test for a linear relationship, which would indicate second-order kinetics.
06

Plot Second-Order Data

Convert the \([\mathrm{ClO}_2]\) data into \(1/[\mathrm{ClO}_2]\) and plot against the time data. A straight line suggests the reaction follows second-order kinetics. Compare the fit with first-order data.
07

Analyze Second-Order Fit

Calculate the slope from the plot in Step 6 if it's linear. This slope represents the second-order rate constant \(k'\). Again, check the linear regression to assure the best alignment of data points with the assumption.

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is a branch of chemistry that focuses on understanding the speed or rate at which chemical reactions occur. It helps us follow how reactants transform into products over time. By studying kinetics, you can predict the behavior of a reaction under various conditions. This is important for numerous practical applications, such as developing chemical processes, understanding enzyme functions, and even optimizing industrial reactions.
One key aspect of chemical kinetics is determining the reaction rate, which is how fast a reactant is consumed or a product is formed. The rate can be influenced by several factors:
  • Concentration of reactants: Higher concentrations usually increase reaction rates.
  • Temperature: Raising the temperature often speeds up reactions.
  • Catalysts: These substances speed up reactions without being consumed in the process.
Understanding these principles allows chemists to manipulate reactions to achieve desired results more efficiently.
Rate Law
The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. It is usually written in the form:\[ ext{Rate} = k [A]^m [B]^n \]where \(k\) is the rate constant, \([A]\) and \([B]\) are the concentrations of the reactants, and \(m\) and \(n\) are the reaction orders with respect to each reactant.
The reaction order provides insights into how changes in concentration affect the reaction rate. For example:
  • If the reaction is first-order in \[ ext{ClO}_2 \], the rate depends linearly on its concentration.
  • If it's second-order, the rate is proportional to the square of \[ ext{ClO}_2 \]'s concentration.
Rate laws are not always straightforward predictions based on the stoichiometric coefficients of the reactants in the chemical equation; they are determined experimentally. Understanding the rate law aids in predicting how a system will respond to concentration changes.
Integrated Rate Law
The integrated rate law is derived from the rate law and provides a mathematical relationship that describes how concentrations of reactants change over time.
For a first-order reaction, the integrated rate law is given by:\[- ext{log}([ ext{ClO}_2]) = k't + ext{log}([ ext{ClO}_2]_0)\]This equation indicates that plotting \[-\text{log}([ ext{ClO}_2])\] versus time should yield a straight line if the reaction follows first-order kinetics.
For second-order reactions, the integrated rate law is:\[\frac{1}{[ ext{ClO}_2]} = k't + \frac{1}{[ ext{ClO}_2]_0}\]Plotting \[\frac{1}{[ ext{ClO}_2]}\] against time will show a straight line if second-order kinetics are present.
By analyzing these plots, the rate constant and reaction order can be accurately determined, helping in drawing conclusions about the kinetics of a reaction.
Rate Constant
The rate constant \(k\) is a crucial component of the rate law and provides information about the reaction's inherent speed at a particular temperature. It is independent of the concentrations of the reactants, but does depend on other factors such as temperature and the presence of a catalyst.
For different orders of reactions, the units of the rate constant vary:
  • For first-order reactions: \(k\) has units of \[s^{-1}\].
  • For second-order reactions: \(k\) has units of \[M^{-1}s^{-1}\].
This difference in units helps distinguish between reaction orders when analyzing kinetic data.
In practical applications, knowing the rate constant allows chemists to predict how long a reaction will take to reach completion or to attain a certain reactant concentration. The rate constant is particularly useful for comparing the reactivity of different reactions under similar conditions.

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Most popular questions from this chapter

In experiments on the decomposition of azomethane, the following data were obtained: \(\begin{array}{lll} & \text { Initial } & \\ & \begin{array}{l}\text { Concentration } \\ \text { of Azomethane }\end{array} & \text { Initial Rate } \\ \text { Exp. 1 } & 1.13 \times 10^{-2} \mathrm{M} & 2.8 \times 10^{-6} \mathrm{M} / \mathrm{s} \\ \text { Exp. } 2 & 2.26 \times 10^{-2} \mathrm{M} & 5.6 \times 10^{-6} \mathrm{M} / \mathrm{s}\end{array}\) What is the rate law? What is the value of the rate constant?

A second-order decomposition reaction run at \(550^{\circ} \mathrm{C}\) has a rate constant of \(3.1 \times 10^{-2} \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s}) .\) If the initial concentration of the reactant is \(0.10 M\), what is the concentration of this reactant after \(1.0 \times 10^{2} \mathrm{~s} ?\) What is the half-life of this reaction under these conditions?

The decomposition of hydrogen peroxide is a first-order reaction: $$\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g)$$ The half-life of the reaction is \(17.0\) minutes. a. What is the rate constant of the reaction? b. If you had a bottle of \(\mathrm{H}_{2} \mathrm{O}_{2}\), how long would it take for \(86.0 \%\) to decompose? C. If you started the reaction with \(\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=0.100 M\), what would be the hydrogen peroxide concentration after \(15.0\) minutes?

Given the following mechanism for a chemical reaction: $$\begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{I}^{-} & \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{IO}^{-} \\ \mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{IO}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+\mathrm{I}^{-} \end{aligned}$$ a. Write the overall reaction. b. Identify the catalyst and the reaction intermediate. C. With the information given in this problem, can you write the pate law? Fxnlaiv

Cyclobutane, \(\mathrm{C}_{4} \mathrm{H}_{8}\), consisting of molecules in which four carbon atoms form a ring, decomposes, when heated, to give ethylene. The reaction is first order. In an experiment, the initial concentration of cyclobutane was \(0.00150 \mathrm{M}\). After heating at \(450^{\circ} \mathrm{C}\) for \(455 \mathrm{~s}\), this was reduced to \(0.00119 \mathrm{M}\). What was the concentration of cyclobutane after a total of \(750 \mathrm{~s}\) ?

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