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 \(30 .\) minutes. 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?

Short answer

Wait approximately 147 minutes.

Step-by-step solution

Understand the First-Order Reaction

In a first-order reaction, the rate of decomposition depends only on the concentration of the compound. The formula for the remaining amount of a substance in a first-order reaction is \( A(t) = A_0 \times e^{-kt} \), where \( A(t) \) is the amount at time \( t \), \( A_0 \) is the initial amount, and \( k \) is the first-order rate constant.

Determine the Rate Constant

The half-life (\( T_{1/2} \)) for a first-order reaction is related to the rate constant \( k \) by the formula \( T_{1/2} = \frac{0.693}{k} \). Here, \( T_{1/2} = 30 \) minutes. Solving for \( k \), we find \( k = \frac{0.693}{30} \approx 0.0231 \text{ min}^{-1} \).

Set Up the Equation for Remaining Amount

We want to find how long it takes for the compound to decrease from 7.50 mg to 0.25 mg. Using the equation for first-order reactions, set \( A(t) = 0.25 \) mg and \( A_0 = 7.50 \) mg. The equation becomes \( 0.25 = 7.50 \times e^{-0.0231t} \).

Solve for Time (t)

Rearrange the equation to isolate \( e^{-0.0231t} \): \( \frac{0.25}{7.50} = e^{-0.0231t} \). Simplify to find \( 0.0333 = e^{-0.0231t} \). Apply the natural logarithm to both sides: \( \ln(0.0333) = -0.0231t \).

Calculate Time

Solving \( \ln(0.0333) \approx -3.4012 \), we substitute back into the equation: \( -3.4012 = -0.0231t \). Solve for \( t \) by dividing: \( t = \frac{3.4012}{0.0231} \approx 147.3 \) minutes.

Key concepts

Rate Constant

The rate constant (\( k \)) is a crucial component in the study of first-order reactions. It is a measure of the speed at which a chemical reaction occurs. For a first-order reaction, the concentra...lf-life is independent of the initial concentration, which is a unique feature of first-order reactions. Chemists use the half-life to predict when a certain amount of a reactant will have decomposed, making it a valuable tool in chemical analysis.

Natural Logarithm

The natural logarithm (\( \ln \)) is a mathematical function that is essential in calculating rates and times in first-order reactions. It is the inverse function of exponential growth and decay, describing the relationships between time a...ay intuitive. The natural logarithmic function helps us to linearize these equations, making it easier to calculate the unknowns such as time (\( t \)). Mastering the use of logarithms is essential for anyone studying thermodynamics and kinetics.

Chemical Decomposition

Chemical decomposition is a type of chemical reaction where one compound breaks down into two or more simpler substances. In a first-order decomposition reaction, such as the breakdown of \( \text{Xe(CF}_3\text{)}_2 \), the reaction rate ...nel of Xe gas and fluorine compounds. This process is important in a variety of chemical manufacturing and waste treatment processes. Monitoring decomposition rates is critical for predicting the stability and shelf-life of chemical products.