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Chapter 14: Problem 87

The reaction \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2}\) has the rate constant \(k=0.63 \mathrm{M}^{-1} \mathrm{~s}^{-1}\). Based on the units for \(k\), is the reaction first or second order in \(\mathrm{NO}_{2}\) ? If the initial concentration of \(\mathrm{NO}_{2}\) is \(0.100 \mathrm{M}\), how would you determine how long it would take for the concentration to decrease to \(0.025 \mathrm{M} ?\)

Short Answer

Expert verified
The reaction is second-order in \(\mathrm{NO}_{2}\), based on the given rate constant's units. To find the time it takes for the concentration to decrease from \(0.100 \mathrm{M}\) to \(0.025 \mathrm{M}\), use the integrated second-order rate law: \( \frac{1}{[\mathrm{A(t)}]} - \frac{1}{[\mathrm{A(0)}]} = kt\). Plug in the given concentrations and rate constant, then solve for \(t\), which is approximately \(95.24 \ \mathrm{s}\).

Step by step solution

01

Determine the reaction order from the rate constant's units

The units for the given rate constant \(k\) affects the reaction order. As the given units are \( \mathrm{M}^{-1} \mathrm{~s}^{-1} \), this indicates that the rate law, \( rate = k [\mathrm{NO}_{2}]^{n}\), has the same units as the rate law for a second-order reaction. Thus, the reaction is second-order in \(\mathrm{NO}_{2}\).
02

Write the second-order integrated rate law

For a second-order reaction, we can write the integrated rate law as: \( \frac{1}{[\mathrm{A(t)}]} - \frac{1}{[\mathrm{A(0)}]} = kt\), where \(\mathrm{A(t)}\) is the concentration of \(\mathrm{NO}_{2}\) at time \(t\), \(\mathrm{A(0)}\) is the initial concentration of \(\mathrm{NO}_{2}\), and \(k\) is the rate constant.
03

Use initial and final concentrations to solve for time

We are given the initial concentration of \(\mathrm{NO}_{2}\), \(\mathrm{A(0)} = 0.100 \mathrm{M}\), and we want to know how long it would take for the concentration to decrease to \(\mathrm{A(t)} = 0.025 \mathrm{M}\). Plugging these values into the integrated second-order rate law, we get: \( \frac{1}{0.025 \mathrm{M}} - \frac{1}{0.100 \mathrm{M}} = (0.63 \mathrm{M}^{-1} \mathrm{~s}^{-1})t \) Now, solve for \(t\).
04

Solve for time t

Rearrange the integrated rate law equation to isolate \(t\) : \( t = \frac{\frac{1}{0.025 \mathrm{M}} - \frac{1}{0.100 \mathrm{M}}}{0.63 \mathrm{M}^{-1} \mathrm{~s}^{-1}} \) Calculate \(t\): \( t \approx 95.24 \ \mathrm{s} \) It will take approximately \(95.24 \ \mathrm{s}\) for the concentration of \(\mathrm{NO}_{2}\) to decrease from \(0.100 \mathrm{M}\) to \(0.025 \mathrm{M}\).

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Key Concepts

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

Integrated Rate Law
The integrated rate law is an equation that links the concentration of reactants to time and allows us to calculate how concentrations change as a reaction proceeds. For second-order reactions, the integrated rate law is particularly important. It helps to track the changing concentration over time, providing insights into how long a reaction will take to reach a certain state.

For a second-order reaction, like the one with nitrogen dioxide ( ext{NO}_2), the integrated rate law is expressed as:
  • \( \frac{1}{[\text{A(t)}]} - \frac{1}{[\text{A(0)}]} = kt \)
This formula is derived from the ordinary differential form of the rate law, and it allows you to find the concentration \([\text{A(t)}]\) at any given time \(t\) if you know the initial concentration \([\text{A(0)}]\) and the rate constant \(k\).

In simple terms, it tells us that for a second-order reaction, the change in the reciprocal of the concentration over time is proportional to the product of the rate constant and time. This is very useful when you need to calculate how long it will take for a reaction to proceed to a certain point.
Reaction Rate
The reaction rate of a chemical reaction indicates how fast reactants are converted into products. In essence, it measures the change in concentration of a reactant or product per unit time. This depends on factors like concentration, temperature, and the presence of a catalyst, but it's also directly related to the order of the reaction and the rate constant.

In our context of second-order reactions, the rate can be defined by the equation:
  • \( \text{rate} = k [\text{NO}_2]^2 \)
where \(k\) is the rate constant and \([\text{NO}_2]\) is the concentration of nitrogen dioxide. As the concentration of \(\text{NO}_2\) affects the reaction rate, a higher concentration leads to an increase in the reaction rate. This is due to the fact that in a second-order reaction, the rate of the reaction is proportional to the square of the concentration of the reactant.

Understanding the reaction rate is crucial for predicting how long a process will take, which is essential information for laboratory settings and industrial processes.
Reaction Order
Reaction order refers to the sum of the powers of the concentration terms in the rate law equation. It provides vital information about the relationship between the concentration of reactants and the rate of reaction. In the specific case of nitrogen dioxide ( ext{NO}_2), we have a second-order reaction.

Recognizing the reaction order is straightforward once the units of the rate constant are provided. For our reaction, the rate constant has units \(\text{M}^{-1} \text{s}^{-1}\), characteristic of second-order reactions. This means the rate of reaction is dependent on the square of \([\text{NO}_2]\), written as:
  • \( \text{rate} = k [\text{NO}_2]^2 \)
Knowing the reaction order is essential not only for applying the correct integrated rate law but also for understanding the underlying reaction mechanism. It aids in predicting how a change in concentration will affect the rate, and thus the overall progress of the reaction.

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Most popular questions from this chapter

Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) or silica \(\left(\mathrm{SiO}_{2}\right)\). (a) Why is this an effective way of utilizing the catalyst material? (b) How does the surface area affect the rate of reaction?

Molecular iodine, \(\mathrm{I}_{2}(g)\), dissociates into iodine atoms at \(625 \mathrm{~K}\) with a first-order rate constant of \(0.271 \mathrm{~s}^{-1}\). (a) What is the half-life for this reaction? (b) If you start with \(0.050 \mathrm{M} \mathrm{I}_{2}\) at this temperature, how much will remain after \(5.12\) s assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2}\) ?

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q) & \text { (slow) } \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q) & \text { (fast) } \end{aligned} $$ (a) Write the rate law for each of the elementary reactions of the mechanism. (b) Write the chemical equation for the overall process. (c) Identify the intermediate, if any, in the mechanism. (d) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process.

The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For example, consider the reaction between methane and chlorine to produce methyl chloride and hydrogen chloride: Reaction 1: \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{HCl}(g)\) This reaction is very slow in the absence of light. However, \(\mathrm{Cl}_{2}(g)\) can absorb light to form \(\mathrm{Cl}\) atoms: $$ \text { Reaction 2: } \mathrm{Cl}_{2}(g)+h v \longrightarrow 2 \mathrm{Cl}(g) $$ Once the \(\mathrm{Cl}\) atoms are generated, they can catalyze the reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{Cl}_{2}\), according to the following proposed mechanism: Reaction 3: \(\mathrm{CH}_{4}(g)+\mathrm{Cl}(g) \longrightarrow \mathrm{CH}_{3}(g)+\mathrm{HCl}(g)\) $$ \text { Reaction 4: } \mathrm{CH}_{3}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{Cl}(g) $$ The enthalpy changes and activation energies for these two reactions are tabulated as follows: $$ \begin{array}{lll} \hline \text { Reaction } & \Delta H_{\mathrm{ran}}^{\circ}(\mathrm{k} \mathrm{J} / \mathrm{mol}) & \mathrm{E}_{a}(\mathrm{~kJ} / \mathrm{mol}) \\ \hline 3 & +4 & 17 \\ 4 & -109 & 4 \\ \hline \end{array} $$ (a) By using the bond enthalpy for \(\mathrm{Cl}_{2}\) (Table \(8.4\) ), determine the longest wavelength of light that is energetic enough to cause reaction 2 to occur. \(\ln\) which portion of the electromagnetic spectrum is this light found? (b) By using the data tabulated here, sketch a quantitative energy profile for the catalyzed reaction represented by reactions 3 and 4. (c) By using bond enthalpies, estimate where the reactants, \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g)\), should be placed on your diagram in part (b). Use this result to estimate the value of \(E_{a}\) for the reaction \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow\) \(\mathrm{CH}_{3}(g)+\mathrm{HCl}(g)+\mathrm{Cl}(g) .\) (d) The species \(\mathrm{Cl}(g)\) and \(\mathrm{CH}_{3}(g)\) in reactions 3 and 4 are radicals, that is, atoms or molecules with unpaired electrons. Draw a Lewis structure of \(\mathrm{CH}_{3}\), and verify that it is a radical. (e) The sequence of reactions 3 and 4 comprise a radical chain mechanism. Why do you think this is called a "chain reaction"? Propose a reaction that will terminate the chain reaction.

The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a l c)+\mathrm{OH}^{-}(a l c)\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{I}^{-}(a l c)\), has an activation energy of \(86.8 \mathrm{~kJ} / \mathrm{mol}\) and a frequency factor of \(2.10 \times 10^{11} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). (a) Predict the rate constant for the reaction at \(35^{\circ} \mathrm{C} .\) (b) \(\mathrm{A}\) solution of KOH in ethanol is made up by dissolving \(0.335\) g KOH in ethanol to form \(250.0 \mathrm{~mL}\) of solution. Similarly, \(1.453 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form \(250.0 \mathrm{~mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at \(35^{\circ} \mathrm{C} ?(\mathrm{c})\) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion?
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