Half-Life (Second Order Reaction) Equation, Derivation & Rearrangements

Core idea

Overview

The half-life ($t_{1/2}$) of a reaction is the time required for the concentration of a reactant to decrease to half its initial value. For a second-order reaction, unlike a first-order reaction, the half-life is not constant but depends on the initial concentration of the reactant ($[A]_0$) and the rate constant ($k$). This equation is derived from the integrated rate law for a second-order reaction and is crucial for characterizing the speed and concentration dependence of such reactions.

When to use: Apply this equation when you are dealing with a reaction confirmed to be second-order and need to determine how long it takes for half of the reactant to be consumed, or to find the rate constant or initial concentration given the other variables.

Why it matters: Understanding the half-life of second-order reactions is important in fields like environmental chemistry (e.g., degradation of pollutants), pharmacology (e.g., drug metabolism), and industrial chemistry (e.g., optimizing reaction times). It helps predict reaction progress and design experiments or processes where concentration changes are critical.

Symbols

Variables

k = Rate Constant, [A]_0 = Initial Concentration of A, $t_{1/2}$ = Half-Life

k

Rate Constant

L mol⁻¹ s⁻¹

[A]_{0}

Initial Concentration of A

mol/L

t_{1/2}

Half-Life

s

Walkthrough

Derivation

Formula: Half-Life (Second Order Reaction)

This formula calculates the half-life for a second-order reaction, showing its dependence on the rate constant and initial concentration.

The reaction is a simple second-order reaction (e.g., A → products or 2A → products).
The temperature is constant.

1

Start with the Integrated Rate Law for a Second-Order Reaction:

This equation describes how the concentration of reactant A ([A]t) changes over time (t) for a second-order reaction, where [A]₀ is the initial concentration and k is the rate constant.

\frac{1}{[ A ] _{t}} - \frac{1}{[ A ] _{0}} = k t

2

Define Half-Life Condition:

By definition, at the half-life ( $t_{1/2}$ ), the concentration of the reactant ([A]t) has decreased to half of its initial concentration ([A]₀).

A t t = t_{1/2}, [A]_{t} = \frac{[ A ] _{0}}{2}

3

Substitute Half-Life Conditions into Integrated Rate Law:

Replace $t$ with $t_{1/2}$ and $[A]_{t}$ with $[A]_{0} /2$ in the integrated rate law.

\frac{1}{[ A ] _{0} /2} - \frac{1}{[ A ] _{0}} = k t_{1/2}

4

Simplify the Expression:

Simplify the first term on the left side.

\frac{2}{[ A ] _{0}} - \frac{1}{[ A ] _{0}} = k t_{1/2}

5

Combine Terms on the Left Side:

Subtract the fractions on the left side.

\frac{1}{[ A ] _{0}} = k t_{1/2}

6

Rearrange to Solve for Half-Life:

Divide both sides by $k [A]_{0}$ to isolate $t_{1/2}$ , yielding the formula for the half-life of a second-order reaction.

t_{1/2} = \frac{1}{k [ A ] _{0}}

Result

t_{1/2} = \frac{1}{k [ A ] _{0}}

Source: OCR A-level Chemistry A Textbook - Reaction Rates (6.1.2)

Free formulas

Rearrangements

Solve for $k$

Half-Life (Second Order Reaction): Make k the subject

k = \frac{1}{t _{1/2} [ A ] _{0}}

To make $k$ (rate constant) the subject of the second-order half-life formula, multiply both sides by $k$ and then divide by $t_{1/2}$ .

Difficulty: 2/5

Solve for $[A]_{0}$

Half-Life (Second Order Reaction): Make [A]₀ the subject

[A]_{0} = \frac{1}{k t _{1/2}}

To make $[A]_{0}$ (initial concentration) the subject, multiply both sides by $[A]_{0}$ and then divide by $t_{1/2}$ .

Difficulty: 2/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Visual intuition

Graph

The graph forms a hyperbola where the half-life decreases rapidly as the rate constant increases, approaching the horizontal axis as an asymptote. For a chemistry student, this shape illustrates that reactions with large rate constants proceed very quickly with short half-lives, while small rate constants result in significantly longer durations for the reactant concentration to decrease by half. The most important feature of this curve is that the half-life never reaches zero, meaning that regardless of how large the rate constant becomes, the time required for the reaction to progress always remains a positive value.

Graph type: hyperbolic

Why it behaves this way

Intuition

Imagine a curve of reactant concentration versus time that drops sharply initially and then levels off; the half-life is the horizontal distance to reach half the initial concentration, and this distance becomes shorter

t_{1/2}

The time required for the concentration of a reactant to decrease to half its initial value.

A shorter half-life indicates a faster reaction, as it takes less time for half of the reactant to be consumed.

k

The rate constant, a proportionality constant that relates the rate of reaction to the concentrations of reactants.

A larger value of 'k' signifies an intrinsically faster reaction at a given temperature, reflecting the efficiency of molecular collisions leading to product formation.

[A]_{0}

The initial concentration of reactant A at the beginning of the reaction (time t=0).

For a second-order reaction, a higher initial concentration leads to a shorter half-life because the reaction rate is highly dependent on concentration, causing it to consume half the reactant more quickly when starting

Signs and relationships

1/(k[A]_0): The half-life is inversely proportional to both the rate constant (k) and the initial concentration ([A]0). This means that a faster intrinsic reaction (larger k) or a higher starting concentration (larger [A]0)

Free study cues

Insight

Canonical usage

The units of the rate constant k must be the reciprocal of the product of the concentration units and the time units used for the half-life.

Common confusion

Using the first-order half-life logic where concentration is ignored, or using the wrong units for k (such as s^-1), which leads to a dimensionally incorrect result.

Unit systems

$t_{1/2}$ s · Commonly reported in minutes (min) or hours (h) for slower reactions; must match the time unit in k.

$k$ dm^3 mol^-1 s^-1 · Equivalent to M^-1 s^-1. The units of k for a second-order reaction are concentration^-1 time^-1.

$[A]_{0}$ mol dm^-3 · Initial molar concentration of the reactant. Also written as M.

One free problem

Practice Problem

A second-order reaction has a rate constant (k) of 0.05 L mol⁻¹ s⁻¹. If the initial concentration of the reactant ([A]₀) is 0.2 mol/L, calculate the half-life ( $t_{1/2}$ ) of the reaction.

Rate Constant0.05 L mol⁻¹ s⁻¹

Initial Concentration of A0.2 mol/L

Solve for: $t_{h} a l f$

Hint: Ensure units are consistent before calculation.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

Determining how quickly a pollutant in a wastewater treatment plant degrades via a second-order reaction.

Study smarter

Tips

Ensure the reaction is indeed second-order overall or second-order with respect to the specific reactant.
Pay close attention to the units of the rate constant (k), which for a second-order reaction are typically L mol⁻¹ s⁻¹ or M⁻¹ s⁻¹.
The initial concentration ( $[A]_{0}$ ) must be in molarity (mol/L).
Remember that for second-order reactions, $t_{1/2}$ increases as $[A]_{0}$ decreases, meaning it takes longer for half the reactant to disappear at lower initial concentrations.

Avoid these traps

Common Mistakes

Confusing the second-order half-life formula with the first-order half-life formula ( $t_{1/2} = ln (2) / k$ ).
Incorrect units for k or $[A]_{0}$ , leading to incorrect units for $t_{1/2}$ .
Algebraic errors when rearranging the formula to solve for k or $[A]_{0}$ .

Keep going

Related Formulas

Common questions

Frequently Asked Questions

This formula calculates the half-life for a second-order reaction, showing its dependence on the rate constant and initial concentration.

Apply this equation when you are dealing with a reaction confirmed to be second-order and need to determine how long it takes for half of the reactant to be consumed, or to find the rate constant or initial concentration given the other variables.

Understanding the half-life of second-order reactions is important in fields like environmental chemistry (e.g., degradation of pollutants), pharmacology (e.g., drug metabolism), and industrial chemistry (e.g., optimizing reaction times). It helps predict reaction progress and design experiments or processes where concentration changes are critical.

Confusing the second-order half-life formula with the first-order half-life formula ($t_{1/2} = \ln(2)/k$). Incorrect units for k or $[A]_0$, leading to incorrect units for $t_{1/2}$. Algebraic errors when rearranging the formula to solve for k or $[A]_0$.