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Chapter 15: Problem 18

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in \(\mathrm{CCl}_{4}\) is a first-order reaction. If \(2.56 \mathrm{mg}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is present initially, and \(2.50 \mathrm{mg}\) is present after 4.26 min at \(55^{\circ} \mathrm{C},\) what is the value of the rate constant, \(k ?\)

Short Answer

Expert verified
The rate constant \( k \) is approximately \( 0.0055 \text{ min}^{-1} \).

Step by step solution

01

Understanding First-Order Reactions

First-order reactions follow the equation \( ext{ln}([A]_0/[A]) = kt \), where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \( t \), and \( k \) is the rate constant. In this scenario, concentrations are replaced by masses since the volumes are not changing.
02

Calculate the Ratio of Concentrations

Given \( 2.56 \text{ mg of } \mathrm{N}_2\mathrm{O}_5 \) initially and \( 2.50 \text{ mg} \) after \( 4.26 \text{ min} \), the ratio \( \frac{[A]_0}{[A]} \) is \( \frac{2.56}{2.50} \).
03

Use the First-order Reaction Formula

Substitute the known values into the equation: \[ \ln\left(\frac{2.56}{2.50}\right) = k \times 4.26.\]
04

Solve for the Rate Constant \( k \)

Calculate \( \ln\left(\frac{2.56}{2.50}\right) \) which equals approximately 0.0234. Then solve for \( k \):\[k = \frac{0.0234}{4.26} \approx 0.0055 \text{ min}^{-1}.\]
05

Conclusion

The rate constant \( k \) for the decomposition of \( \mathrm{N}_2\mathrm{O}_5 \) is approximately \( 0.0055 \text{ min}^{-1} \).

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Key Concepts

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

Rate Constant
The rate constant, often denoted by the symbol \( k \), is a critical part of understanding a chemical reaction鈥檚 speed. For a first-order reaction, the rate constant links the reaction rate with the concentration of a single reactant. This value is unique to each reaction and is significantly influenced by factors such as temperature.
In our given decomposition, \( k \) helps determine how fast \( \mathrm{N}_2\mathrm{O}_5 \) decomposes into its products. Calculating the rate constant involves understanding the initial and remaining quantities of \( \mathrm{N}_2\mathrm{O}_5 \) and the time it takes for that change to occur. Having the rate constant also helps predict how long it will take for a significant amount of reactant to decompose, which is vital for practical applications in chemical kinetics.
Decomposition Reaction
Decomposition reactions are a type of chemical reaction where a single compound breaks down into two or more simpler substances. In the case of \( \mathrm{N}_2\mathrm{O}_5 \), it decomposes to form nitrogen dioxide (\( \mathrm{NO}_2 \)) and oxygen (\( \mathrm{O}_2 \)).
These reactions can be initiated by heat, light, or other energy forms, and they are usually endothermic, meaning they absorb energy from the surroundings. Understanding the decomposition of \( \mathrm{N}_2\mathrm{O}_5 \) helps explain how and why certain reactions yield specific products, which is essential in fields such as environmental chemistry and industrial processes.
Chemical Kinetics
Chemical kinetics is the branch of chemistry that studies the speed at which chemical reactions occur and the factors affecting these speeds. It involves understanding the rate of reactions, the sequence of steps in a reaction mechanism, and how various conditions can change reaction rates.
In the context of the decomposition of \( \mathrm{N}_2\mathrm{O}_5 \), chemical kinetics allows us to pinpoint not just how fast the reaction occurs (through the rate constant) but also how changing concentration, temperature, or pressure might influence this speed. Having insight into these kinetics is vital for optimizing reactions in industrial applications, ensuring safety, and understanding natural processes.
N2O5
\( \mathrm{N}_2\mathrm{O}_5 \), or dinitrogen pentoxide, is a chemical compound composed of nitrogen and oxygen that often serves as a precursor to nitric acid. It is known for being a strong oxidizer, capable of initiating decomposition reactions under the right conditions.
Its role in the atmosphere is particularly noteworthy, as it can contribute to atmospheric pollution and acid rain formation. When analyzing its decomposition, the processes and kinetics involved help us devise strategies to control emissions and mitigate environmental impact.

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Most popular questions from this chapter

At temperatures below \(500 \mathrm{K},\) the reaction between carbon monoxide and nitrogen dioxide $$ \mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) $$ has the following rate equation: Rate \(=k\left[\mathrm{NO}_{2}\right]^{2} .\) Which of the three mechanisms suggested here best agrees with the experimentally observed rate equation? Mechanism 1 \(\quad\) single, elementary step $$\mathrm{NO}_{2}+\mathrm{CO} \longrightarrow \mathrm{CO}_{2}+\mathrm{NO}$$ Mechanism \(2 \quad\) Two steps $$\begin{aligned}&\text { Slow } \quad \mathrm{NO}_{2}+\mathrm{NO}_{2} \longrightarrow \mathrm{NO}_{3}+\mathrm{NO}\\\&\text { Fast } \quad \mathrm{NO}_{3}+\mathrm{CO} \longrightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2}\end{aligned}$$ Mechanism 3 \(\quad\) Two steps $$\begin{aligned}&\text { Slow } \quad \mathrm{NO}_{2} \longrightarrow \mathrm{NO}+\mathrm{O}\\\&\text { Fast } \quad \mathrm{CO}+\mathrm{O} \longrightarrow \mathrm{CO}_{2}\end{aligned}$$

The decomposition of HOF occurs at \(25^{\circ} \mathrm{C}\) $$2 \mathrm{HOF}(\mathrm{g}) \longrightarrow 2 \mathrm{HF}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ Using the data in the table below, determine the rate law and then calculate the rate constant. $$\begin{array}{lc}\hline \begin{array}{l}{[\mathrm{HOF}]} \\\\(\mathrm{mol} / \mathrm{L})\end{array} & \begin{array}{l}\text { Time } \\\\(\mathrm{min})\end{array} \\\\\hline 0.850 & 0 \\\0.810 & 2.00 \\\0.754 & 5.00 \\\0.526 & 20.0 \\\0.243 & 50.0 \\\\\hline\end{array}$$

The decomposition of gaseous dimethyl ether at ordinary pressures is first order. Its half-life is 25.0 min at \(500^{\circ} \mathrm{C}\) $$\mathrm{CH}_{3} \mathrm{OCH}_{3}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$$ (a) Starting with \(8.00 \mathrm{g}\) of dimethyl ether, what mass remains (in grams) after 125 min and after 145 min? (b) Calculate the time in minutes required to decrease \(7.60 \mathrm{ng}\) (nanograms) to 2.25 ng. (c) What fraction of the original dimethyl ether remains after 150 min?

Data for the decomposition of dinitrogen oxide $$2 \mathrm{N}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{N}_{2} (\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ on a gold surface at \(900^{\circ} \mathrm{C}\) are given below. Verify that the reaction is first order by preparing a graph of \(\ln \left[\mathrm{N}_{2} \mathrm{O}\right]\) versus time. Derive the rate constant from the slope of the line in this graph. Using the rate law and value of \(k\), determine the decomposition rate at \(900^{\circ} \mathrm{C}\) when \(\left[\mathrm{N}_{2} \mathrm{O}\right]=\) \(0.035 \mathrm{mol} / \mathrm{L}.\) $$\begin{array}{cc}\hline \begin{array}{c}\text { Time } \\\\\text { (min) }\end{array} & \begin{array}{c}{\left[\mathrm{N}_{2} 0\right]} \\\\(\mathrm{mol} / \mathrm{L})\end{array} \\\\\hline 15.0 & 0.0835 \\\30.0 & 0.0680 \\\80.0 & 0.0350 \\\120.0 & 0.0220 \\\\\hline\end{array}$$

Data in the table were collected at \(540 \mathrm{K}\) for the following reaction: $$\mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NO}(\mathrm{g})$$ (a) Derive the rate equation. (b) Determine the reaction order with respect to each reactant. (c) Calculate the rate constant, giving the correct units for \(k.\) $$\begin{array}{lll}\hline \text { Initial Concentration }(\mathrm{mol} / \mathrm{L}) & {\text { Initial Rate }} \\\\\hline[\mathrm{C} 0] & {\left[\mathrm{NO}_{2}\right]} & (\mathrm{mol} / \mathrm{L} \cdot \mathrm{h}) \\\\\hline 5.1 \times 10^{-4} & 0.35 \times 10^{-4} & 3.4 \times 10^{-8} \\\5.1 \times 10^{-4} & 0.70 \times 10^{-4} & 6.8 \times 10^{-8} \\\5.1 \times 10^{-4} & 0.18 \times 10^{-4} & 1.7 \times 10^{-8} \\\1.0 \times 10^{-3} & 0.35 \times 10^{-4} & 6.8 \times 10^{-8} \\\1.5 \times 10^{-3} & 0.35 \times 10^{-4} & 10.2 \times 10^{-8} \\\\\hline\end{array}$$
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