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Chapter 26: Problem 5

The second-order rate constant for the reaction \\[ \mathrm{O}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{g}) \\] is \(1.26 \times 10^{-15} \mathrm{cm}^{3}\) -molecule \(^{-1} \cdot \mathrm{s}^{-1}\). Determine the value of the rate constant in units of \(\mathrm{dm}^{3} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1}\).

Short Answer

Expert verified
Approximately \( 2.09 \times 10^{-42} \text{ dm}^3 \cdot \text{mol}^{-1} \cdot \text{s}^{-1} \).

Step by step solution

01

Understand the Conversion Required

The given reaction rate constant is in units of \( \mathrm{cm}^{3} \cdot \text{molecule}^{-1} \cdot \mathrm{s}^{-1} \). We need to convert it to units of \( \mathrm{dm}^{3} \cdot \text{mol}^{-1} \cdot \mathrm{s}^{-1} \). This involves converting between volume units and between count units (from molecules to moles).
02

Convert Volume Units

1 dm鲁 is 1000 cm鲁. Thus, to convert volume from cm鲁 to dm鲁, we can use the conversion factor: \( \frac{1\, \text{dm}^3}{1000\, \text{cm}^3} \).
03

Convert Molecules to Moles

Use Avogadro's number, which is \( 6.022 \times 10^{23} \) molecules per mole, to convert the rate constant from the units of molecules to moles. The conversion factor is \( 1 \text{ mol} = 6.022 \times 10^{23} \text{ molecules} \).
04

Apply Conversion Factors

Starting with the given rate constant:\[ 1.26 \times 10^{-15} \text{ cm}^3 \cdot \text{molecule}^{-1} \cdot \text{s}^{-1} \]Convert cm鲁 to dm鲁:\[ 1.26 \times 10^{-15} \frac{\text{dm}^3}{1000 \text{cm}^3} \cdot \text{molecule}^{-1} \cdot \text{s}^{-1} = 1.26 \times 10^{-18} \text{ dm}^3 \cdot \text{molecule}^{-1} \cdot \text{s}^{-1} \]Now, convert molecules to moles:\[ 1.26 \times 10^{-18} \text{ dm}^3 \cdot \frac{1 \text{mol}}{6.022 \times 10^{23} \text{ molecules}} \cdot \text{s}^{-1} \approx 2.09 \times 10^{-42} \text{ dm}^3 \cdot \text{mol}^{-1} \cdot \text{s}^{-1} \]
05

Conclusion

The converted rate constant for the reaction is approximately \( 2.09 \times 10^{-42} \text{ dm}^3 \cdot \text{mol}^{-1} \cdot \text{s}^{-1} \).

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Key Concepts

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

Second-order reactions
In chemical kinetics, reactions are classified based on how the concentration of reactants affects the rate. Second-order reactions are characterized by the rate being proportional to the square of the concentration of one reactant or to the product of the concentrations of two reactants.

For a simple second-order reaction A + B 鈫� Products, the rate can be expressed as:
  • Rate = k[A][B]
In this formula, \( k \) is the second-order rate constant, and [A] and [B] are the concentrations of reactants A and B. Understanding second-order reactions is crucial as it helps predict how changes in concentration affect the reaction rate.
Rate constants
The rate constant, often denoted as \( k \), is a crucial factor in the rate laws of chemical reactions. It links the reaction rate with the concentrations of reactants, and its value can indicate how fast a reaction proceeds.

For a second-order reaction, the rate constant has specific units that reflect its order, typically \( ext{dm}^3 ext{mol}^{-1} ext{s}^{-1} \). These units arise because the rate (concentration over time) and the concentration terms must balance dimensionally.

The magnitude of \( k \) can vary widely depending on the reaction and conditions like temperature, which is why it is essential to convert these constants correctly when using different units.
Unit conversion
Converting units is a critical skill in chemistry, especially when dealing with rate constants. We often need to change the units to match those that are standard or convenient for a specific scenario.

In the original exercise, we converted from \( ext{cm}^3 ext{molecule}^{-1} ext{s}^{-1} \) to \( ext{dm}^3 ext{mol}^{-1} ext{s}^{-1} \). This involved converting:
  • Volume: 1 dm鲁 = 1000 cm鲁, so we apply a factor of \( \frac{1}{1000} \).
  • Particles: 1 mole = \( 6.022 \times 10^{23} \) molecules, using Avogadro's number.
These steps ensure that the rate constant is expressed in units consistent with molar concentrations, which is often more practical in laboratory and industrial settings.
Avogadro's number
Avogadro's number, \( 6.022 \times 10^{23} \), is a fundamental constant in chemistry that relates the number of constituent particles, usually atoms or molecules, in one mole of a substance.

This number allows chemists to translate between atomic and macroscopic scales, making it essential for calculations involving the amount of substance.

In the exercise, Avogadro's number was vital in converting the rate constant from molecules to moles. This conversion is crucial because it facilitates expressing concentration in terms of molarity鈥攁 more manageable measure in chemical equations and reactions.

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Most popular questions from this chapter

The reaction \\[ \mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \\] is first order and has a rate constant of \(2.24 \times 10^{-5} \mathrm{s}^{-1}\) at \(320^{\circ} \mathrm{C}\). Calculate the half-life of the reaction. What fraction of a sample of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g})\) remains after being heated for 5.00 hours at \(320^{\circ} \mathrm{C} ?\) How long will a sample of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g})\) need to be maintained at \(320^{\circ} \mathrm{C}\) to decompose \(92.0 \%\) of the initial amount present?

In this problem, we will derive Equation 26.32 from the rate law (Equation 26.31 ) \\[ -\frac{d[\mathrm{A}]}{d t}=k[\mathrm{A}][\mathrm{B}] \\] Use the reaction stoichiometry of Equation 26.30 to show that \([\mathrm{B}]=[\mathrm{B}]_{0}-[\mathrm{A}]_{0}+[\mathrm{A}]\) Use this result to show that Equation 1 can be written as \\[ -\frac{d[\mathrm{A}]}{d t}=k[\mathrm{A}]\left|[\mathrm{B}]_{0}-[\mathrm{A}]_{0}+[\mathrm{A}]\right| \\] Now separate the variables and then integrate the resulting equation subject to its initial conditions to obtain the desired result, Equation 26.32 \\[ k t=\frac{1}{[\mathbf{A}]_{0}-[\mathbf{B}]_{0}} \ln \frac{[\mathbf{A}][\mathbf{B}]_{0}}{[\mathbf{B}][\mathbf{A}]_{0}} \\]

Consider the base-catalyzed reaction \(\mathrm{OCl}^{-}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \longrightarrow \mathrm{OI}^{-}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\) Use the following initial-rate data to determine the rate law and the corresponding rate constant for the reaction. $$\begin{array}{cccc} {\left[\mathrm{OCl}^{-}\right] / \mathrm{mol} \cdot \mathrm{dm}^{-3}} & {\left[\mathrm{I} \mathrm{I} / \mathrm{mol} \cdot \mathrm{dm}^{-3}\right.} & {\left[\mathrm{OH}^{-}\right] / \mathrm{mol} \cdot \mathrm{dm}^{-3}} & v_{0} / \mathrm{mol} \cdot \mathrm{dm}^{-3} \cdot \mathrm{s}^{-1} \\ \hline 1.62 \times 10^{-3} & 1.62 \times 10^{-3} & 0.52 & 3.06 \times 10^{-4} \\\ 1.62 \times 10^{-3} & 2.88 \times 10^{-3} & 0.52 & 5.44 \times 10^{-4} \\ 2.71 \times 10^{-3} & 1.62 \times 10^{-3} & 0.84 & 3.16 \times 10^{-4} \\ 1.62 \times 10^{-3} & 2.88 \times 10^{-3} & 0.91 & 3.11 \times 10^{-1} \end{array}$$

The equilibrium constant for the reaction. $$ \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftharpoons}} \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$ at \(25^{\circ} \mathrm{C}\) is \(K_{c}=\left[\mathrm{H}_{2} \mathrm{O}\right] /\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=5.49 \times 10^{15} \mathrm{~mol}^{-1} \cdot \mathrm{dm}^{3}\). The time-dependent conductivity of the solution following a temperature jump to a final temperature of \(25^{\circ} \mathrm{C}\) shows a relaxation time of \(\tau=3.7 \times 10^{-5} \mathrm{~s}\). Determine the values of the rate constants \(k_{1}\) and \(k_{-1} .\) At \(25^{\circ} \mathrm{C}\), the density of water is \(\rho=0.997 \mathrm{~g} \cdot \mathrm{cm}^{-3}\).

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), decomposes in water by a first-order kinetic process. A \(0.156\)-mol \(\cdot \mathrm{dm}^{-3}\) solution of \(\mathrm{H}_{2} \mathrm{O}_{2}\) in water has an initial rate of \(1.14 \times\) \(10^{-5} \mathrm{~mol} \cdot \mathrm{dm}^{-3} \cdot \mathrm{s}^{-1}\). Calculate the rate constant for the decomposition reaction and the half-life of the decomposition reaction.
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