Problem 19 - General Chemistry: Principles and Modern Applications

Question

The reaction \(A \longrightarrow\) products is first order in A. (a) If \(1.60 \mathrm{g} \mathrm{A}\) is allowed to decompose for 38 min, the mass of A remaining undecomposed is found to be 0.40 g. What is the half-life, \(t_{1 / 2}\), of this reaction? (b) Starting with \(1.60 \mathrm{g} \mathrm{A},\) what is the mass of \(\mathrm{A}\) remaining undecomposed after \(1.00 \mathrm{h} ?\)

Step-by-Step Solution

Answer

The half-life of the reaction is 38 min, and the remaining mass of A after 1 hour is 0.20 g.

Step 1: Calculate the rate constant

In part (a), 1.60 g of A has decomposed to 0.40 g after 38 min, indicating that half of the reactant A has decomposed. The half-life of the reaction is therefore 38 min. Using the equation for the rate constant of a first order reaction, \(k = 0.693 / t_{1/2}\), the rate constant k is calculated by substituting \(t_{1/2}\) with 38 min: \(k = 0.693 / 38 = 0.018 min^{-1}\)

Step 2: Determine the remaining amount of the reactant

In part (b), the remaining mass of A after 1.00 hour (or 60 min) is obtained using the integrated rate law. Rearrange the equation to solve for [A]_t: \(ln[A]_t = -kt + ln[A]_0\), where [A]_0 is the initial concentration (1.60 g), and [A]_t is the concentration at time t. Substituting the known values: \(ln[A]_t = -(0.018 min^{-1} * 60 min) + ln(1.60 g)\). The remaining concentration [A]_t is found by taking the exponential of both sides, since the natural logarithm is involved: \([A]_t = e^{ln[A]_t} = e^{-(0.018 min^{-1} * 60 min) + ln(1.60 g)} = 0.20 g\)

Step 3: Summary

The half-life of the reaction is thus 38 min and the remaining mass of reactant A after 1.00 h is 0.20 g.

Key Concepts

Rate Constant Calculation

In first-order reactions, the rate at which a substance reacts is directly proportional to its concentration. The rate constant, often denoted by the symbol \(k\), provides a measure of how quickly a reaction proceeds. For a first-order reaction, the rate constant can be calculated using the formula: \(k = \frac{0.693}{t_{1/2}}\). This relationship reflects the inherent properties of first-order reactions, linking the decay rate to the half-life, which is the time it takes for half of the reactant to decompose.

The half-life \(t_{1/2}\) is a key figure in determining \(k\). It's essential to note that for first-order reactions, the half-life remains constant throughout the reaction, regardless of the starting concentration. After finding that half of the original quantity of 1.60 g of \(\text{A}\) decomposed in 38 minutes, the half-life \(t_{1/2}\) was determined to be 38 minutes. Plugging this into the equation, we calculate:

• \(k = \frac{0.693}{38} = 0.018 \, \text{min}^{-1}\)

This establishes the reaction's rate constant, serving as a crucial figure when predicting how quickly the remaining quantity will decompose over time.

Half-Life Determination

The concept of half-life is especially significant in first-order reactions, as it represents the uniform period during which half of the reactant is converted into products. For our reaction \(A \rightarrow \text{products}\), half of the initial 1.60 g decomposing into 0.40 g already suggests that one full half-life has passed.

By using the equation for calculating half-life in a first-order process, \(t_{1/2} = \frac{0.693}{k}\), we reaffirm our earlier calculation. Given the rate constant we found, \(k = 0.018 \, \text{min}^{-1}\), substituting \(k\) back into the half-life equation confirms the calculated half-life:

• \(t_{1/2} = \frac{0.693}{0.018} = 38 \, \text{min}\)

Recognizing that the half-life does not change under constant conditions facilitates predictions about future reactant quantities. Understanding this constancy is crucial for various applications, including chemical kinetics and pharmaceutical decay.

Integrated Rate Law

The integrated rate law offers a mathematical means to determine the quantity of reactant remaining at any point in time, an essential tool for first-order reactions. For our scenario, the integrated rate law is expressed as:\[ \ln{[A]_t} = -kt + \ln{[A]_0} \]where \([A]_t\) is the concentration at time \(t\), \(k\) is the rate constant, and \([A]_0\) is the initial concentration. This equation allows us to expertly track the progress of a decomposition over time.

In our example, we started with 1.60 g of \(A\) and wish to know what remains after one full hour (60 minutes). By substituting into the integrated rate law:\[ \ln{[A]_t} = -(0.018 \, \text{min}^{-1} \times 60 \, \text{min}) + \ln(1.60) \]Calculating the right-hand side, we then convert back from the natural logarithmic form:\[ [A]_t = e^{-1.08 + \ln(1.60)} \approx 0.20 \, \text{g} \]This demonstrates how we use the integrated rate law to predict remaining quantities, with careful application of exponential functions. This approach is broadly applicable in understanding not only chemical reactions but also decay processes across different scientific disciplines.