Understanding the Completion of First-Order Reactions

First-order reactions follow a unique characteristic where the rate of reaction depends solely on the concentration of one reactant. This article delves into the fundamental concept of half-life and the rate constant in first-order reactions, providing a detailed explanation of how to calculate the time for a first-order reaction to be 75% complete.

First-Order Reaction Basics

For a first-order reaction, the rate constant k and the half-life t_{0.5} are directly related. The relationship is given by:
t_{0.5} frac{0.693}{k}

Calculating the Rate Constant

Given that t_{0.5} 500, text{sec}, we can find the rate constant k:

k frac{0.693}{500, text{sec}} approx 0.001386, text{sec}^{-1}

Calculating the Time to Complete 75% of the Reaction

To determine the time it takes for a first-order reaction to reach 75% completion, we need to find the time at which 25% of the reactant remains. For a first-order reaction, the relationship between the concentration of reactants and time can be described by the equation:

lnleft(frac{[A]_0}{[A]}right) kt

Since we want to find the time when 25% of the reactant remains:

[A] 0.25, [A]_0

Substituting this into the equation:

lnleft(frac{[A]_0}{0.25, [A]_0}right) kt

This simplifies to:

ln, 4 kt

Calculating ln4:
ln4 approx 1.386

Now, solving for t:

t frac{1.386}{0.001386} approx 1000, text{sec}

Alternative Methods for Solving the Problem

Method 1: Using Half-Life Concept

Let n be the number of half-life periods taken. For a 75% completion:

[A] ,text{left} 100 - 75 25

Since each half-life reduces the reactant by half:

1/2 25

1/2 1/2^2

n 2

Thus, the time taken is:

500, text{sec} times 2 1000, text{sec}

Method 2: Using Integrated Rate Law

Using the integrated rate law for a first-order reaction:

k frac{ln2}{t_{0.5}} frac{ln2}{500, text{sec}} 0.001386, text{sec}^{-1}

The integrated equation for a first-order reaction is:

lnleft(frac{[A]}{[A]_0}right) -kt

Given [A] 0.25, [A]_0, we have:

lnleft(frac{0.25, [A]_0}{[A]_0}right) -kt

This simplifies to:

ln0.25 -0.001386, text{sec}^{-1} t

Solving for t:

t frac{ln0.25}{-0.001386, text{sec}^{-1}} 1000, text{sec}