Question

The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\), at \(70^{\circ} \mathrm{C}\) is \(6.82 \times 10^{-3} \mathrm{~s}^{-1}\). Suppose we start with \(0.0250 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) in a volume of \(2.0 \mathrm{~L}\). (a) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0\) min? (b) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010\) mol? (c) What is the half- life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70^{\circ} \mathrm{C}\) ?

Short answer

(a) 0.00822 mol of N2O5 will remain after 5 minutes, (b) it will take approximately 10.14 minutes for the quantity of N2O5 to drop to 0.010 mol, and (c) the half-life of N2O5 at 70°C is approximately 1.70 minutes.

Step-by-step solution

First-order Integrated Rate Law

For a first-order reaction, we use the Integrated Rate Law, which is given by: \(ln \frac{[A]_0}{[A]} = kt\), where [A]_0 and [A] are the initial and final concentrations, respectively, k is the rate constant, and t is time. We can rearrange this equation to find [A], the final concentration of N2O5 remaining: \([A] = \frac{[A]_0}{e^{kt}}\).

(a) Moles of N2O5 remaining after 5 minutes

We are given: [A]_0 = 0.0250 mol / 2.0 L = 0.0125 M, k = 6.82 x 10^{-3} s^{-1}, and t = 5.0 min = 300 s. Now we can plug these values into the Integrated Rate Law equation to find the final concentration of N2O5: [A] = 0.0125 M / e^(6.82 x 10^{-3} s^{-1} * 300 s) = 0.00411 M. To find the moles of N2O5 remaining, we multiply the concentration by the volume: Moles of N2O5 remaining = 0.00411 M * 2.0 L = 0.00822 mol.

(b) Time for N2O5 to drop to 0.010 mol

We are given: [A]_0 = 0.0125 M, [A] = 0.010 mol / 2.0 L = 0.005 M, and k = 6.82 x 10^{-3} s^{-1}. We can rearrange the Integrated Rate Law equation to find t: \(t = \frac{ln \frac{[A]_0}{[A]}}{k}\). Now we plug the values into the equation: t = ln(0.0125 / 0.005) / (6.82 x 10^{-3} s^{-1}) ≈ 608.4 s. Convert seconds to minutes: Time = 608.4 s / 60 s/min ≈ 10.14 min.

(c) Half-life of N2O5

We can use the half-life formula for a first-order reaction: \(t_{1/2} = \frac{ln 2}{k}\). We are given k = 6.82 x 10^{-3} s^{-1}. Plug this value into the equation: t_{1/2} = ln 2 / (6.82 x 10^{-3} s^{-1}) ≈ 101.7 s. Convert seconds to minutes: Half-life of N2O5 = 101.7 s / 60 s/min ≈ 1.70 min. In summary, (a) 0.00822 mol of N2O5 will remain after 5 minutes, (b) it will take approximately 10.14 minutes for the quantity of N2O5 to drop to 0.010 mol, and (c) the half-life of N2O5 at 70°C is approximately 1.70 minutes.

Key concepts

First-Order Reactions

First-order reactions are chemical processes where the rate at which a reactant converts into products is directly proportional to its concentration. In simple terms, the reaction rate depends only on the concentration of one reactant. This is a common type of reaction in both chemistry labs and the environment.

Characteristics of first-order reactions include:

• The reaction rate decreases over time as the concentration of the reactant decreases.
• A linear relationship between the logarithm of the reactant concentration and time.
• Common in radioactive decay and various chemical reactions involving gases.

Understanding this concept is crucial for solving problems related to reaction kinetics, as it lays the foundation for calculating the remaining concentration of a reactant or predicting how long a reaction will take to reach a certain level of completion.

Rate Constant

The rate constant, often denoted by the symbol \(k\), is a crucial parameter in chemical kinetics that defines the speed of a reaction. For a first-order reaction, the rate constant has units of \;(s^{-1})\, indicating the reaction rate per second. It is called a constant because it remains the same under constant temperature, making it extremely useful for calculations.

Key points about the rate constant:

• The rate constant determines how fast a reaction proceeds. A larger rate constant means a faster reaction.
• The value of \(k\) is temperature dependent—a higher temperature often increases the rate constant.
• In the given exercise, \(k = 6.82 \times 10^{-3} \; s^{-1}\) at \(70^{\circ}C\), helps us calculate the time needed for the reaction to progress to a specific point.

Understanding the rate constant is essential when working with the integrated rate law and calculating reaction times or remaining concentrations.

Integrated Rate Law

The integrated rate law for a first-order reaction provides a mathematical relationship between the concentration of the reactant and time. This formula is useful to determine the concentration of reactants at any given time during the reaction.

The integrated rate law is expressed as:\[ln \left(\frac{[A]_0}{[A]}\right) = kt\]where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), and \(k\) is the rate constant. This equation can be rearranged to solve for different variables based on the available data.

Benefits of using the integrated rate law include:

• Finding the remaining concentration of a reactant after a specified period.
• Determining the time required to reach a certain concentration.
• Calculating the rate constant if concentrations and time are known.

In essence, the integrated rate law acts like a guide for predicting the future or investigating past states of a reaction, making it a powerful tool in kinetic studies.

Half-Life

Half-life is a crucial concept in understanding first-order reactions. It refers to the time required for the concentration of a reactant to decrease to half its initial value. Unlike zero-order or second-order reactions, the half-life for a first-order reaction is constant regardless of the initial concentration.

To calculate the half-life \(t_{1/2}\) for first-order reactions, you can use the formula:\[t_{1/2} = \frac{ln 2}{k}\]where \(ln 2\) is a natural logarithm constant approximately equal to 0.693 and \(k\) is the rate constant. This formula highlights that the half-life is inversely proportional to the rate constant.

A few important notes regarding half-life:

• For first-order reactions, the half-life remains unchanged regardless of how much reactant is present initially.
• It provides a convenient measure for comparing how fast different reactions are, as a smaller half-life indicates a faster reaction.
• In the presented exercise, the calculated half-life of \(N_2O_5\) at \(70^{\circ}C\) is approximately 1.70 minutes, showing a relatively rapid decomposition.

Understanding half-life helps in predicting the dynamics of a reaction and is particularly useful in fields like pharmacology, environmental science, and nuclear physics.