Problem 18 - Chemistry and Chemical Reactivity

Question

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, \(k ?\)

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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 \( \frac{2.56}{2.50} \).

Step 3: Use the First-order Reaction Formula

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

Step 4: Solve for the Rate Constant \( k \)

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

Step 5: Conclusion

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

Key Concepts

Rate Constant

The rate constant, often denoted by the symbol \( k \), is a critical part of understanding a chemical reaction’s 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.