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Iodine atoms combine to form \(\mathrm{I}_{2}\) in liquid hexane solvent with a rate constant of \(1.5 \times 10^{10} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\). The reaction is second order in I. since the reaction occurs so quickly, the only way to study the reaction is to create iodine atoms almost instanta- neously, usually by photochemical decomposition of \(\mathrm{I}_{2} .\) Suppose a flash of light creates an initial [I] concentration of \(0.0100 \mathrm{M} .\) How long will it take for \(95 \%\) of the newly created iodine atoms to recombine to form \(\mathrm{I}_{2} ?\)

Short Answer

Expert verified
It takes approximately \( 1.39 \times 10^{-8} \) seconds for 95% of the iodine atoms to recombine to form \( \mathrm{I}_{2} \) in the given conditions.

Step by step solution

01

Identify the Reaction Order

Since the reaction is second order in iodine atoms, I, its rate law can be written as rate = k[I]^2, where k is the rate constant, and [I] is the concentration of iodine atoms.
02

Set Up the Integrated Rate Law for a Second-Order Reaction

For a second-order reaction, the integrated rate law is \( \frac{1}{[I]} - \frac{1}{[I]_0} = kt \), where \( [I] \) is the concentration of I at time t, and \( [I]_0 \) is the initial concentration of I.
03

Calculate the Final Concentration of I

If 95% of the iodine atoms recombine, that means only 5% remain unreacted. Calculate the final concentration of iodine atoms, [I], by taking 5% of the initial concentration, \( [I] = 0.05 \times [I]_0 \).
04

Substitute the Known Values into the Integrated Rate Law

Substitute the known values for k, \( [I]_0 \), and \( [I] \) into the integrated rate law equation and solve for t.
05

Solve for the Time, t

Rearrange the integrated rate law to isolate t and calculate the value, giving the time it takes for 95% of the iodine atoms to recombine.

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Key Concepts

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

Rate Law
When studying chemical kinetics, the rate law is a fundamental concept that expresses how the rate of a reaction depends on the concentration of its reactants. Specifically, in our exercise, we examine a second-order reaction, which involves iodine atoms combining in a liquid hexane solvent.

In general terms, the rate law formula for a second-order reaction involving a single reactant, like iodine (I), can be represented as:
\[ \text{rate} = k[I]^2 \]
Here, \(k\) is the rate constant—a measure of how quickly the reaction proceeds—and \([I]\) denotes the reactant iodine atom concentration. For second-order reactions, the reaction rate increases quadratically with the concentration of the reactant, meaning that if the concentration doubles, the rate of the reaction increases by a factor of four. This represents a key characteristic of second-order kinetics.
Integrated Rate Law
While the rate law tells us how the reaction rate depends on concentration, the integrated rate law allows us to understand how concentrations vary over time. For a second-order reaction, the integrated rate law equation is a valuable tool for prediciting the concentration of the reactants at any given time.

The equation is given by:
\[ \frac{1}{[I]} - \frac{1}{[I]_0} = kt \]
where \([I]\) is the concentration of the reactant at time \(t\), \([I]_0\) is the starting concentration, \(k\) is the rate constant, and \(t\) is time. In our exercise, you would use this law to calculate the time required for a certain percentage of the iodine atoms to combine and form \(I_2\).
Reaction Mechanisms
The sequence of steps by which a chemical reaction occurs is known as the reaction mechanism. Each of these steps can involve the breaking and forming of bonds, and they may proceed at different rates. In a reaction mechanism, intermediates—which are not present in the overall balanced equation—may be produced and consumed.

Understanding the mechanism is crucial for explaining why a reaction has its particular rate law. For example, if a reaction has a second-order rate law, the mechanism might involve a step where two reactant molecules collide and react. Reaction mechanisms give us insight into the 'path' the reactants take to become products, which helps chemists to optimize reactions and develop new synthetic pathways.
Photochemical Decomposition
Photochemical decomposition is a process in which a compound is broken down by the influence of light energy. It's a type of photolysis in which light provides the energy required to break chemical bonds. This concept applies to our exercise where iodine molecules (\(I_2\)) are dissociated into iodine atoms (\(I\)) instantaneously through a flash of light—effectively creating the reactants for our second-order reaction.

In the context of chemical kinetics, photochemical reactions are interesting because they can be initiated with a high degree of control over timing and concentration of the reactive species. This control makes such reactions suitable for studying fast processes like the second-order recombination of iodine atoms, as illustrated in the exercise.

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Most popular questions from this chapter

The energy of activation for the decomposition of \(2 \mathrm{~mol}\) of \(\mathrm{HI}\) to \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) in the gas phase is \(185 \mathrm{~kJ}\). The heat of formation of \(\mathrm{HI}(g)\) from \(\mathrm{H}_{2}(g)\) and \(\mathrm{I}_{2}(g)\) is \(-5.65 \mathrm{~kJ} / \mathrm{mol} .\) Find the energy of activation for the reaction of \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) and \(1 \mathrm{~mol}\) of \(\mathrm{I}_{2}\) to form 2 mol of HI in the gas phase.

What are enzymes? What is the active site of an enzyme? What is a substrate?

Explain the difference between homogeneous catalysis and heterogeneous catalysis.

Cyclopropane \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right)\) reacts to form propene \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right)\) in the gas phase. The reaction is first order in cyclopropane and has a rate constant of \(5.87 \times 10^{-4} / \mathrm{s}\) at \(485^{\circ} \mathrm{C}\). If a 2.5 - \(\mathrm{L}\) reaction vessel initially contains 722 torr of cyclopropane at \(485^{\circ} \mathrm{C}\), how long will it take for the partial pressure of cyclopropane to drop to below \(1,00 \times 10^{2}\) torr?

In this chapter, we have seen a number of reactions in which a single reactant forms products. For example, consider the following first-order reaction: \(\mathrm{CH}_{3} \mathrm{NC}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CN}(g)\) However, we also learned that gas-phase reactions occur through collisions. a. One possible explanation for how this reaction occurs is that two molecules of \(\mathrm{CH}_{3} \mathrm{NC}\) collide with each other and form two molecules of the product in a single elementary step. If that were the case, what reaction order would you expect? b. Another possibility is that the reaction occurs through more than one step. For example, a possible mechanism involves one step in which the two \(\mathrm{CH}_{3} \mathrm{NC}\) molecules collide, resulting in the "activation" of one of them. In a second step, the activated molecule goes on to form the product. Write down this mechanism and determine which step must be rate determining in order for the kinetics of the reaction to be first order. Show explicitly how the mechanism predicts first-order kinetics.

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