Problem 28 - Chemistry and Chemical Reactivity

Question

The compound \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) decomposes in a firstorder 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 \(0.25 \mathrm{mg}\) of \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) remains?

Step-by-Step Solution

Answer

It takes 147.2 minutes for the compound to decompose to 0.25 mg.

Step 1: Understand the First-Order Kinetics

The decomposition of \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) follows first-order kinetics. In first-order reactions, the rate of reaction is proportional to the concentration of the reactant. The formula to determine the half-life \(t_{1/2}\) for first-order reactions is given by \(t_{1/2} = \frac{0.693}{k}\), where \(k\) is the rate constant.

Step 2: Calculate the Rate Constant

Using the given half-life of 30 minutes, we can calculate the rate constant \(k\):\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{30} = 0.0231 \, \text{min}^{-1} \]

Step 3: Apply First-Order Integrated Rate Law

The integrated rate law for a first-order reaction is \( \ln \left( \frac{[A]_t}{[A]_0} \right) = -kt \), where \([A]_0\) is the initial concentration and \([A]_t\) is the concentration at time \(t\). Begin by substituting the initial and final concentrations and the rate constant into this formula:\[ \ln \left( \frac{0.25}{7.50} \right) = -0.0231t \]

Step 4: Solve for Time (t)

First, calculate the natural logarithm:\[ \ln \left( \frac{0.25}{7.50} \right) = \ln (0.0333) = -3.4012 \]Now, solve for \(t\):\[ -3.4012 = -0.0231t \]\[ t = \frac{-3.4012}{-0.0231} = 147.2 \, \text{min} \]

Step 5: Verify the Calculation

Check the calculation to ensure accuracy. The given half-life and calculated time should coincide with the proper rate decay expected for first-order reactions. Repeat calculation if necessary to confirm that the time for \(7.50 \, \text{mg}\) to decompose to \(0.25 \, \text{mg}\) is about 147.2 minutes.

Key Concepts

Rate Constant

In chemistry, understanding the rate at which a reaction occurs is crucial. For a first-order reaction, like the one involving the decomposition of \(\mathrm{Xe(CF_3)_2}\), the rate constant \(k\) plays a significant role. It dictates how fast the reactant concentration decreases over time. In first-order reactions, the rate of reaction is directly proportional to the concentration of the single reactant involved.
To determine the rate constant \(k\), we use the half-life \(t_{1/2}\), which is the time taken for the concentration of a reactant to reduce to half its initial value. The formula connecting half-life to rate constant in first-order reactions is \(t_{1/2} = \frac{0.693}{k}\). By inserting the half-life of 30 minutes into this equation, the rate constant \(k\) is calculated to be \(0.0231 \, \text{min}^{-1}\). This value helps predict how quickly the substance will decompose over a given period.

Integrated Rate Law

The integrated rate law is a powerful tool in kinetics, allowing us to relate concentrations of reactants to time elapsed in a reaction. For first-order reactions, the integrated rate law is expressed as \( \ln \left( \frac{[A]_t}{[A]_0} \right) = -kt \), where \([A]_0\) is the initial concentration and \([A]_t\) is the concentration at time \(t\).
This formula helps in finding out how long it will take for a certain amount of reactant to convert into products. Substituting initial concentration \(7.50\, \text{mg}\), final concentration \(0.25\, \text{mg}\), and rate constant \(0.0231\, \text{min}^{-1}\) into the integrated rate law, we set up the equation \[ \ln \left( \frac{0.25}{7.50} \right) = -0.0231t \].
After solving the natural logarithm and rearranging, we find the time \(t\) it takes for \(\mathrm{Xe(CF_3)_2}\) to decompose to the desired concentration.

Half-Life Calculation

Half-life is a central concept when dealing with first-order reactions. It represents the time required for the concentration of the reactant to be reduced by half. In our reaction involving \(\mathrm{Xe(CF_3)_2}\), understanding half-life helps us predict how the reactant concentration decreases over multiple intervals.
For first-order reactions, the half-life is constant and can be calculated using the formula \(t_{1/2} = \frac{0.693}{k}\). Given a rate constant of \(0.0231 \, \text{min}^{-1}\), the half-life is 30 minutes.
This steady half-life allows us to easily predict concentrations at different times, as it does not depend on initial concentration. Calculating multiple half-lives indicates the rapidity of decomposition within the given timeframe, making it essential for understanding reaction kinetics.