Question

What is the half-life of a compound if 75 percent of a given sample of the compound decomposes in 60 min? Assume first-order kinetics.

Short answer

The half-life of the compound, to the nearest minute, is approximately 21 minutes.

Step-by-step solution

Understand the principles of first-order kinetics

The general formula for first-order decay is given by: \( N = N_0 e^{-kt} \), where \(N\) is the amount remaining, \(N_0\) is the initial amount, \(k\) is the rate constant, and \(t\) is time. For this problem, we need to rearrange this equation to solve for \(k\), which we can then use to find the half-life

Setting up the equation

Given that 75% of the sample decomposes in 60 minutes, meaning that 25% remains. We can substitute this information into the decay formula: \(0.25N_0 = N_0 e^{-k(60)}\)

Solve for the rate constant (k)

After getting rid of the \(N_0\) on both sides and taking the natural logarithm (ln) we obtain: \(-ln(4) = -60k\), rearranging this to solve for \(k\) gives \(k = ln(4) / 60\)

Derive the half-life from the decay constant

The formula connecting the half-life (t1/2) with the rate constant in a first-order reaction is \(t1/2 = ln(2) / k\). Substituting our calculated \(k\) value into this formula will yield the half-life.

Key concepts

Half-life calculation

Understanding the half-life of a compound is crucial in different scientific areas, like chemistry and pharmacology. Half-life is the time it takes for a substance to reduce to half of its initial amount. For first-order reactions, calculating the half-life is straightforward due to the relationship between the decay constant and the natural logarithm. The formula to calculate the half-life is: \[ t_{1/2} = \frac{\ln(2)}{k} \] Here, \( t_{1/2} \) represents the half-life, and \( k \) is the rate constant of the reaction. In the example provided, after calculating \( k \) from the decomposition data, you can substitute \( k \) back into this formula to find how long it will take for the compound to decompose to half of its original amount in a first-order kinetic reaction. It's important to remember that this calculation is specific to first-order reactions.

Rate constant

The rate constant, often symbolized as \( k \), is a critical parameter in kinetics that quantifies the speed of a chemical reaction. In the context of first-order reactions, the rate constant determines how quickly the concentration of a reactant decreases over time.The general formula for a first-order reaction is: \[ N = N_0 e^{-kt} \] Where:

• \( N \) is the amount of substance remaining.
• \( N_0 \) is the initial amount.
• \( t \) is the time elapsed.
• \( k \) is the rate constant.

To find \( k \), you rearrange this formula based on the known quantities such as the time at which a certain percentage of the substance remains. In the exercise, 25% of the compound remains after 60 minutes. By substituting the known values and solving for \( k \), one can understand how swiftly the reaction proceeds.

Exponential decay

Exponential decay describes the process wherein the amount of a substance decreases at a rate proportional to its current value. This is a common characteristic of first-order reactions. The mathematical representation of exponential decay is given by: \[ N = N_0 e^{-kt} \] This formula highlights how the concentration of a reactant diminishes exponentially over time.Key features of exponential decay include:

• The decay rate, which is proportional to the amount of substance remaining.
• The quantity decreases rapidly initially and slows down over time.
• The existence of a constant rate, denoted as \( k \), which dictates the decay speed.

In the decomposition problem, recognizing the exponential nature of decay facilitates the understanding of how 75% of the substance decomposes over 60 minutes. By understanding exponential decay, students can predict future amounts of reactants in similar scenarios.