Question

Use the following data to determine the half-life of \(^{218} 80\) and the rate constant for the decay of \(^{218} 84\) Po. $$\begin{array}{lllllll} \hline \text { Time } / \mathrm{s} & 0 & 200 & 400 & 600 & 800 & 1000 \\ \text { Moles }_{84}^{218} \mathrm{Po} & 0.250 & 0.110 & 0.057 & 0.025 & 0.012 & 0.005 \\ \hline \end{array}$$

Short answer

The half-life of \(^{218} 84\)Po is approximately 187.47 s, and the rate constant is \(0.0037 \, \text{s}^{-1}\).

Step-by-step solution

Understanding the Decay Process

The decay of a radioactive substance is a first-order reaction. This means that the rate of reaction is directly proportional to the concentration of the substance. The rate constant \( k \) can be found using the formula for first-order kinetics: \[ k = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]} \right) \]where \( [A]_0 \) is the initial concentration, \( [A] \) is the concentration at time \( t \), and \( \ln \) represents the natural logarithm.

Calculating the Rate Constant

Using the given data, select two points: at time \( t = 0 \) s, \( [A]_0 = 0.250 \) moles; at time \( t = 400 \) s, \( [A] = 0.057 \) moles. Plug these into the formula for \( k \):\[ k = \frac{1}{400} \ln \left( \frac{0.250}{0.057} \right) \]Calculating gives \( \ln(4.386) \approx 1.4790 \), so \[ k \approx \frac{1.4790}{400} = 0.0036975 \, \text{s}^{-1} \]

Calculating the Half-life

The half-life \( t_{1/2} \) of a first-order reaction is given by the formula: \[ t_{1/2} = \frac{0.693}{k} \]Substitute \( k = 0.0036975 \, \text{s}^{-1} \):\[ t_{1/2} = \frac{0.693}{0.0036975} \approx 187.47 \, \text{s} \]

Verification Using Additional Data Points

For accuracy, repeat the calculation of \( k \) with different pairs of points from the dataset to ensure consistency. The pattern of radioactive decay should result in the same \( k \) value irrespective of the time points chosen, reinforcing the validity of \( k \) and ensuring any variations in calculations are accounted for.

Key concepts

First-order Kinetics

Radioactive decay often follows a first-order kinetic model. This type of kinetic reaction indicates that the rate at which a substance decays is directly proportional to its current concentration. To understand it easily, consider that as you have more of a radioactive substance, the decay happens faster; less substance results in slower decay.
For first-order kinetics, the equation used is \[ k = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]} \right) \]where:

• \( k \) is the rate constant, indicating how fast the reaction occurs.
• \( t \) is the elapsed time during which the reaction progresses.
• \([A]_0\) is the initial concentration of the substance.
• \([A]\) is the concentration at time \( t \).
• The function \( \ln \) is the natural logarithm.

This formula helps us calculate how fast a radioactive substance decays and allows us to quantify the process under first-order kinetics.

Rate Constant

The rate constant \( k \) is crucial in understanding the speed of a radioactive decay process. In first-order kinetics, the rate constant provides insight into how rapidly the reaction progresses. With the rate constant, we can predict how long it will take for a certain amount of the substance to decay over time.
In our exercise, we determine \( k \) using specific data points: the initial amount of \( ^{218}_{84} \text{Po} \) and its amount after a given period. By applying the formula \[ k = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]} \right) \],we can compute \( k \) between any two data points:

• For instance, at \( t = 0 \) s, \([A]_0 = 0.250\) moles, and at \( t = 400 \) s, \([A] = 0.057\) moles.Calculating the rate constant yields \( k \approx 0.0036975 \, \text{s}^{-1} \).

By calculating \( k \) across various time intervals, we ensure accurate and consistent results, underscoring the reliability of our calculations.

Half-life Calculation

Half-life is a fundamental concept in radioactive decay, representing the time required for half the amount of a radioactive substance to decay. It's a critical measure that helps understand and predict the behavior of radioactive materials over time.
For first-order reactions like radioactive decay, the half-life can be calculated using the formula \[ t_{1/2} = \frac{0.693}{k} \],where \( k \) is the rate constant calculated earlier.

In our problem, using the given rate constant \( k = 0.0036975 \, \text{s}^{-1} \), the half-life is determined as:

• \[ t_{1/2} = \frac{0.693}{0.0036975} \approx 187.47 \, \text{s} \]

Therefore, the half-life represents the decay process's slow and consistent pace, providing essential information about the durability of the radioactive substance and its potential applications or risks in real-world scenarios.