Problem 28 - Chemistry and Chemical Reactivity

Question

The compound \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) decomposes in a first-order reaction to elemental Xe with a half-life of 30. min. If you place \(7.50 \mathrm{mg}\) of \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) in a flask, how long must you wait until only 0.25 mg of \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) remains?

Step-by-Step Solution

Answer

The wait time is approximately 147.3 minutes.

Step 1: Identify Known Quantities

We know the initial mass of the compound is 7.50 mg and only 0.25 mg remains. The half-life of the reaction is 30 minutes. This is a first-order reaction.

Step 2: Use First-Order Kinetics Formula

The first-order kinetics formula is \( [A] = [A]_0 \cdot e^{-kt} \), where \([A]\) is the final concentration, \([A]_0\) is the initial concentration, and \(k\) is the rate constant. We need to find the rate constant \(k\) first.

Step 3: Calculate the Rate Constant

For first-order reactions, the rate constant \(k\) can be determined by the formula \( k = \frac{0.693}{\text{half-life}} \). Substituting the given half-life value: \[ k = \frac{0.693}{30 \text{ min}} \approx 0.0231 \text{ min}^{-1} \]

Step 4: Apply the Decay Formula

Now we can apply the first-order decay formula. Substitute the known values: \( 0.25 = 7.50 \cdot e^{-0.0231t} \). We need to solve for \(t\).

Step 5: Solve for Time \(t\)

Rearrange and solve the equation:Divide both sides by 7.50:\[ e^{-0.0231t} = \frac{0.25}{7.50} = 0.0333 \]Take the natural logarithm of both sides:\[ -0.0231t = \ln(0.0333) \approx -3.402 \]Solve for \(t\):\[ t = \frac{-3.402}{-0.0231} \approx 147.3 \text{ minutes} \]

Step 6: Conclusion

The time required for the amount of \( \mathrm{Xe}(\mathrm{CF}_{3})_{2} \) to decrease from 7.50 mg to 0.25 mg under first-order kinetics is approximately 147.3 minutes.

Key Concepts

Kinetics

Kinetics is the study of the rates of chemical reactions and the steps involved in transforming reactants to products. It helps us understand how fast a reaction is occurring and how the concentration of reactants changes over time. In the case of a first-order reaction like the decomposition of \( \mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2} \), the reaction rate depends solely on the concentration of one reactant. This means the rate at which the compound decomposes is directly proportional to its current concentration.

Key characteristics of kinetics in first-order reactions include:

• The rate of reaction decreases over time as the reactant concentration drops.
• The half-life remains constant, meaning the time required to reduce the amount of reactant by half is always the same, regardless of the initial quantity.

Understanding kinetics allows scientists to control and manipulate reactions efficiently, which is crucial in fields like pharmacology and materials science.

Half-Life

The half-life of a reaction is the time taken for half of the reactant to be consumed. For first-order reactions, the half-life is a constant value, making it a critical concept for understanding how quickly a compound diminishes over time.

In the given problem, the half-life of \( \mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2} \) is 30 minutes, which remains consistent as the reaction progresses.

• This property allows us to predict how long it will take for a certain amount of the substance to react.
• It helps in estimating the duration it takes to reach a desired concentration, which is useful in processes like drug dosing and radioactive decay analysis.

With this knowledge, you can also derive the formula to calculate the rate constant, necessary for finding the time required for a specific reduction in concentration.

Rate Constant

The rate constant \( k \) is an essential parameter that defines the speed of a chemical reaction for a given order. For first-order reactions, this constant can be determined using the half-life value. The formula is \[ k = \frac{0.693}{\text{half-life}} \] Plugging in the half-life from the problem, you get: \[ k = \frac{0.693}{30 \text{ min}} \approx 0.0231 \text{ min}^{-1} \]

The rate constant is a unique feature that varies with temperature and catalyst presence but remains constant under given conditions for the same reaction.

• In the problem, this constant helps us find how quickly \( \mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2} \) decomposes over time.
• It's crucial for applying the first-order kinetics formula to solve for time or remaining concentration.

By understanding how to calculate and use the rate constant, you can predict the behavior of chemical reactions in various settings.

Natural Logarithm

The natural logarithm plays a critical role in solving first-order kinetics problems. It is the inverse function of the exponential function and is denoted as \( \ln \).

When you rearrange the first-order decay equation \( [A] = [A]_0 \cdot e^{-kt} \) for time \( t \), you often need to use the natural logarithm to isolate the variable. For example, to solve \[ e^{-0.0231t} = 0.0333 \] you would take the natural logarithm of both sides: \[ -0.0231t = \ln(0.0333) \]

This operation allows you to solve for \( t \) by simplifying the exponential relationship to a linear one.

• Understanding natural logarithms is essential for dealing with exponential decay and growth scenarios in chemistry and physics.
• They simplify the process of solving equations where the variable of interest is in the exponent.

The natural logarithm is a powerful mathematical tool that helps predict how processes change over time in natural, scientific, and financial domains.