Problem 9 - Chemistry in 30 days

Question

For a first-order reaction, the time required for \(99.9 \%\) of the reaction to take place is nearly (a) 10 times that required for half the reaction (b) 100 times that required for two-thirds of the reaction (c) 10 times that required for one-fourth of the reaction (d) 20 times that required for half of the reaction

Step-by-Step Solution

Answer

For 99.9% completion, time required is nearly 10 times that for half the reaction (option a).

Step 1: Understand the First-Order Reaction

A first-order reaction has a rate law of the form \( k[A] \), and the rate of decrease of concentration \( [A] \) of a reactant is proportional to its concentration. The integrated rate equation for a first-order reaction is given by \( [A] = [A]_0 e^{-kt} \), where \( [A]_0 \) is the initial concentration of the reactant.

Step 2: Calculate Time for 99.9% Completion

For 99.9% completion, \( 0.1% \) of the reactant remains, so \( [A] = 0.001[A]_0 \). Substituting in the integrated rate equation gives \( 0.001[A]_0 = [A]_0 e^{-kt} \). This simplifies to \( 0.001 = e^{-kt} \) and taking the natural log of both sides yields \( -6.907755 = -kt \), giving \( t = \frac{6.907755}{k} \).

Step 3: Calculate Time for Half-Reaction

For half of the reaction to complete (t1/2), \([A] = \frac{1}{2}[A]_0 \), so \( [A]_0/2 = [A]_0 e^{-kt} \). Simplifying to \( 1/2 = e^{-kt} \), and taking the natural log gives \( t = \frac{0.693}{k} \).

Step 4: Compare Times for 99.9% and 50% Completion

Comparing \( t = \frac{6.907755}{k} \) for 99.9% completion with \( t_{1/2} = \frac{0.693}{k} \) for half completion, the ratio \( \frac{t_{99.9}}{t_{1/2}} = \frac{6.907755}{0.693} \approx 10 \).

Step 5: Determine the Correct Option

Based on the calculation, the time for 99.9% completion is approximately 10 times the time for half completion. Hence, the correct option is (a).

Key Concepts

Integrated Rate Equation

The integrated rate equation is an essential tool for understanding first-order reactions. This equation describes how the concentration of a reactant changes over time. For a first-order reaction, the equation is written as:

• \([A] = [A]_0 e^{-kt}\)

Here, \([A]\) denotes the concentration of the reactant at any given time \(t\), \([A]_0\) is the initial concentration, \(k\) is the rate constant, and \(e\) is the base of natural logarithms. Understanding this equation helps us predict how quickly a reaction will occur under specific conditions.

When solving problems, inserting known values into the integrated rate equation allows for calculating unknowns, such as the time required for a specific event in the reaction. For example, if we need to determine how long it takes for 99.9% of the reaction to occur, this equation provides the foundation to complete that calculation.

Reaction Kinetics

Reaction kinetics studies the rates at which chemical reactions happen and the factors influencing these rates. In a first-order reaction, the rate depends solely on the concentration of one reactant. The rate is directly proportional to this reactant’s concentration.

First-order reactions have a particular characteristic where their half-life is constant. This is the time it takes for half of the reactant to be consumed, irrespective of its initial concentration. This constancy simplifies calculations and predictions about reaction progress.

Understanding how variables like temperature and concentration affect the rate can help control reactions more efficiently. In our context, using reaction kinetics principles allowed us to identify the correct option in the exercise by understanding the proportional relationship between the time required for different stages of the reaction.

Half-Life in Chemical Reactions

The concept of half-life is a critical aspect of reaction kinetics, providing insight into how quickly a reaction progresses. For a first-order reaction, the half-life is the time it takes for half of the initial amount of the reactant to react or be consumed. It is mathematically represented as:

• \(t_{1/2} = \frac{0.693}{k}\)

Here, \(t_{1/2}\) is the half-life and \(k\) is the rate constant.

The half-life is a constant value in first-order reactions, making it a convenient measure to compare different reactions or stages within the same reaction. By understanding and applying the concept of half-life, one can deduce how long it will take for a certain percentage of reaction completion. In the exercise we tackled, knowing the half-life allowed us to derive that 99.9% completion takes about ten times longer than reaching the half-life point. This is a crucial understanding that helps in predicting the course and the duration needed for particular reactions in various fields, such as pharmacology or environmental science.