Question

At \(573 \mathrm{K},\) gaseous \(\mathrm{NO}_{2}(\mathrm{g})\) decomposes, forming \(\mathrm{NO}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g}) .\) If a vessel containing \(\mathrm{NO}_{2}(\mathrm{g})\) has an initial concentration of \(1.9 \times 10^{-2} \mathrm{mol} / \mathrm{L}_{y}\) how long will it take for \(75 \%\) of the \(\mathrm{NO}_{2}(\mathrm{g})\) to decompose? The decomposition of \(\mathrm{NO}_{2}(\mathrm{g})\) is secondorder in the reactant and the rate constant for this reaction, at \(573 \mathrm{K},\) is \(1.1 \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\).

Short answer

The time required is 143.55 seconds.

Step-by-step solution

Determine the Final Concentration

Since we want 75% of the NO2 to decompose, only 25% of it should remain. Initial concentration, \([NO_2]_0 = 1.9 \times 10^{-2} \, \text{mol/L}\), so the remaining concentration \([NO_2]\) is \( \frac{1}{4}[NO_2]_0 = \frac{1}{4}\times 1.9 \times 10^{-2} = 0.475 \times 10^{-2} \, \text{mol/L}\).

Write the Second Order Kinetics Formula

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

Substitute Known Values

Substitute the initial concentration \([NO_2]_0\), the remaining concentration \([NO_2]\), and the rate constant \(k\) into the second-order reaction formula:\[\frac{1}{0.475 \times 10^{-2}} = \frac{1}{1.9 \times 10^{-2}} + (1.1)(t)\]

Solve for Time \(t\)

First calculate \(\frac{1}{[A]_0}\) and \(\frac{1}{[A]}\):\(\frac{1}{1.9 \times 10^{-2}} = 52.63 \, \text{L/mol}\) \(\frac{1}{0.475 \times 10^{-2}} = 210.53 \, \text{L/mol}\)Substitute these values in:\[210.53 = 52.63 + 1.1t\]Solving for \(t\), subtract 52.63 from both sides:\[157.90 = 1.1t\]\(t = \frac{157.90}{1.1} = 143.55\, \text{seconds}\)

Key concepts

Understanding Second-Order Reactions

In the world of chemical kinetics, reactions are often categorized by their order, which refers to the power to which the concentration of a reactant is raised in the rate law. A second-order reaction involves either one reactant whose concentration squared is concerning the rate, or it can involve two different reactants. This means that the rate of reaction is proportional to the square of the concentration of one reactant or the product of the concentrations of two reactants.
For the decomposition of \(NO_2(g) \)into \(NO(g) \) and \(O_2(g)\), it is a second-order reaction involving the concentration of \(NO_2\) only. Understanding second-order kinetics is crucial because it tells us how concentration and reaction rate are interconnected, and typically, as the concentration decreases, the reaction slows down.

Decoding the Rate Constant

The rate constant, denoted as \(k\), is a crucial part of the rate law equation in chemical kinetics. For second-order reactions, its units are often \(L/mol \, s\). This constant is specific to a particular reaction and is influenced by conditions such as temperature.
In this exercise, the rate constant for the decomposition of \(NO_2\) at \(573 \, K \) is given as \(1.1 \, L / mol \, s\). It provides insight into the reaction's speed: larger rate constants indicate a faster reaction. The constant here allows us to connect changes in reactant concentration over time, allowing us to predict how fast 75% of \(NO_2\) decomposes.

Integrated Rate Law for Second-Order Reactions

The integrated rate law for second-order reactions provides a mathematical way to connect the concentration of a reactant at any time with its initial concentration. The formula we use is:\[\frac{1}{[A]} = \frac{1}{[A]_0} + kt\]
where \([A]\) is the concentration at time \ t\, \([A]_0\) is the initial concentration, and \(k\) is the rate constant.
This equation helps calculate how long it takes for a certain amount of the reactant to be consumed, based on the initial concentration and the rate constant. By rearranging the equation, you can solve for \(t\), the time, providing valuable information on reaction progress.

Monitoring Concentration Change Over Time

Tracking how concentrations change over time is key to understanding and predicting reaction behavior. In this problem, 75% of \(NO_2\) decomposes, meaning only 25% of the original concentration remains. If we started with \([NO_2]_0 = 1.9 \times 10^{-2} \, ext{mol/L}\)\, then the concentration left is \(0.475 \times 10^{-2} \, ext{mol/L}\).
By using the integrated rate law and the provided rate constant, we solved for the time \(t\) needed to reach this remaining concentration. This highlights how understanding and applying the principles of chemical kinetics can predict the duration of a reaction under defined conditions.

• It involves understanding the rate equations and integrating them with respect to time.
• This approach provides a quantitative way to compare how different factors like temperature or catalysts can impact reaction speed.