Reaction Half-Life Kinetics and Calculations

Definition of Reaction Half-Life

The half-life of a reaction, designated as $t_{1/2}$ , is defined as the time required for the concentration of a given reactant to fall to exactly $\frac{1}{2}$ ( $50\%$ ) of its initial value.
The mathematical expression and properties of reaction half-lives depend directly on the reaction order (first-order, second-order, or zero-order).

First-Order Reaction Half-Life

Mathematical Expression:
- $t_{1/2} = \frac{0.693}{k}$
- $t_{1/2} = \frac{\ln(2)}{k}$
- In these equations, $t_{1/2}$ represents the half-life, and $k$ represents the specific rate constant for the reaction.
- The constant $0.693$ originates from taking the natural logarithm of $2$ (i.e., $\ln(2) \approx 0.693$ ). (Note: Contextual references to $2.693$ represent a slight typo for $\ln(2)$ , but the exact natural logarithm value is $\ln(2)$ ).
Independence from Initial Concentration:
- A defining characteristic of any first-order reaction is that its half-life is entirely independent of the initial reactant concentration ( $[A]_0$ ).
- The half-life equation for a first-order process contains no concentration terms.
Concentration Decay Behavior:
- Because the half-life remains constant, every consecutive half-life interval reduces the reactant concentration by half over the exact same time duration.
- For example, if a first-order reaction has a half-life of $t_{1/2} = 100\,\text{s}$ and begins with an initial concentration of $1.00\,\text{M}$ :
  - At $t = 0\,\text{s}$ : Concentration is $1.00\,\text{M}$ .
  - After $100\,\text{s}$ (1st half-life): Concentration decreases to $0.50\,\text{M}$ .
  - After another $100\,\text{s}$ ( $t = 200\,\text{s}$ , 2nd half-life): Concentration decreases to $0.25\,\text{M}$ .
  - After another $100\,\text{s}$ ( $t = 300\,\text{s}$ , 3rd half-life): Concentration decreases to $0.125\,\text{M}$ .
Graphical Identification:
- First-order reactions can be identified from experimental concentration-versus-time plots by observing whether the time taken to halve the concentration remains constant across successive intervals.

Second-Order Reaction Half-Life

Mathematical Expression:
- $t_{1/2} = \frac{1}{k [A]_0}$
- In this equation, $k$ is the rate constant and $[A]_0$ is the initial concentration of the reactant.
Dependence on Initial Concentration:
- Unlike first-order reactions, the half-life of a second-order reaction is directly dependent on the initial concentration $[A]_0$ .
- Because $[A]_0$ is located in the denominator, the half-life is inversely proportional to the initial concentration.
Concentration Decay Behavior:
- As the reaction proceeds and reactant concentration decreases, subsequent half-lives become progressively longer ( $t_{1/2}$ increases).
- For example, starting with an initial concentration of $1.00\,\text{M}$ :
  - Reducing concentration from $1.00\,\text{M}$ to $0.50\,\text{M}$ takes $100\,\text{s}$ .
  - Reducing concentration from $0.50\,\text{M}$ to $0.25\,\text{M}$ (the next half-life) takes $300\,\text{s}$ .

Zero-Order Reaction Half-Life

Mathematical Expression:
- $t_{1/2} = \frac{[A]_0}{2k}$
- In this equation, $[A]_0$ is the initial concentration in the numerator, and $k$ is the rate constant in the denominator.
Dependence on Initial Concentration:
- The half-life of a zero-order reaction is dependent on the initial reactant concentration.
- Because $[A]_0$ appears in the numerator, the half-life is directly proportional to the initial concentration.
Concentration Decay Behavior:
- As reactant concentration decreases over time, each subsequent half-life becomes progressively shorter ( $t_{1/2}$ decreases).
- For example, starting with an initial concentration of $1.00\,\text{M}$ :
  - Reducing concentration from $1.00\,\text{M}$ to $0.50\,\text{M}$ takes $100\,\text{s}$ .
  - Reducing concentration from $0.50\,\text{M}$ to $0.25\,\text{M}$ takes $50\,\text{s}$ .
  - Further reduction from $0.25\,\text{M}$ to $0.125\,\text{M}$ takes $20\,\text{s}$ .

Summary of Reaction Order Half-Life Progression

Zero-Order: Half-life is proportional to $[A]_0$ ; half-life gets shorter as concentration drops.
First-Order: Half-life is independent of $[A]_0$ ; half-life remains constant regardless of concentration changes.
Second-Order: Half-life is inversely proportional to $[A]_0$ ; half-life gets longer as concentration drops.

Application and Problem-Solving Strategy

Certain problems require combining the reaction half-life equation with the corresponding integrated rate law.
Key Principle: All radioactive decay/decomposition processes follow first-order kinetics.

Step-by-Step Sample Problem

Problem Statement: The half-life of a radioactive phosphorus isotope is $14.3\,\text{days}$ . How long does it take for a sample of this phosphorus isotope to lose $99\%$ of its radioactivity?
Identify Reaction Kinetics:
- Process type: Radioactive isotope decomposition (First-Order Kinetics).
- Given half-life ( $t_{1/2}$ ): $14.3\,\text{days}$ .
- Initial radioactivity amount ( $[A]_0$ ): $100\%$ .
- Radioactivity lost: $99\%$ .
- Remaining radioactivity amount at time $t$ ( $[A]_t$ ): $100\% - 99\% = 1\%$ .
Formulas Required:
1. First-order half-life formula: $t_{1/2} = \frac{\ln(2)}{k}$
2. First-order integrated rate law: $\ln[A]_t = -kt + \ln[A]_0$
Step 1: Determine the Rate Constant ( $k$ ):
- Substitute $t_{1/2} = 14.3\,\text{days}$ into the half-life equation: $14.3\,\text{days} = \frac{\ln(2)}{k}$
- Rearrange to solve for $k$ : $k = \frac{\ln(2)}{14.3\,\text{days}}$
- Calculated rate constant value: $k = 0.04847183081\,\text{day}^{-1}$
Step 2: Calculate the Required Duration ( $t$ ):
- Substitute $k = 0.04847183081\,\text{day}^{-1}$ , $[A]_t = 1$ , and $[A]_0 = 100$ into the integrated rate law: $\ln(1) = -(0.04847183081\,\text{day}^{-1}) \times t + \ln(100)$
- Since $\ln(1) = 0$ : $0 = -(0.04847183081) \times t + \ln(100)$
- Rearrange to isolate time ( $t$ ): $(0.04847183081) \times t = \ln(100)$ $t = \frac{\ln(100)}{0.04847183081}$
- Unrounded calculation: $t = 95.7714351\,\text{days}$
- Final rounded answer: $t \approx 95\,\text{days}$