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Chemical Kinetics

Half-Life of Reactions ($t_{1/2}$)

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

By chemca Team • Updated Jan 2026

The Half-Life of a reaction is a key parameter that indicates how fast a reaction proceeds. It is denoted by $t_{1/2}$. At this time, $[A]_t = \frac{[A]_0}{2}$.

1. Zero Order Reaction

Direct Proportionality

For a zero order reaction, the integrated rate equation is $[A]_0 - [A] = kt$.

At $t = t_{1/2}$, substitute $[A] = [A]_0/2$:

$$ [A]_0 - \frac{[A]_0}{2} = k t_{1/2} $$ $$ \frac{[A]_0}{2} = k t_{1/2} $$ $$ t_{1/2} = \frac{[A]_0}{2k} $$

Key Feature: For zero order reactions, $t_{1/2}$ is directly proportional to the initial concentration $[A]_0$. Doubling the concentration doubles the half-life.

2. First Order Reaction

Independent of Concentration

For a first order reaction, the integrated rate equation is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$.

At $t = t_{1/2}$, substitute $[A] = [A]_0/2$:

$$ t_{1/2} = \frac{2.303}{k} \log \frac{[A]_0}{[A]_0/2} $$ $$ t_{1/2} = \frac{2.303}{k} \log 2 $$ $$ t_{1/2} = \frac{0.693}{k} $$

Key Feature: For first order reactions, $t_{1/2}$ is constant and independent of initial concentration.

Application: All radioactive decay processes follow first-order kinetics.

3. Second Order Reaction

Inverse Proportionality

For a simple second order reaction ($2A \to P$), the integrated rate equation is $kt = \frac{1}{[A]} - \frac{1}{[A]_0}$.

At $t = t_{1/2}$, substitute $[A] = [A]_0/2$:

$$ k t_{1/2} = \frac{1}{[A]_0/2} - \frac{1}{[A]_0} = \frac{2}{[A]_0} - \frac{1}{[A]_0} $$ $$ t_{1/2} = \frac{1}{k[A]_0} $$

For second order, half-life is inversely proportional to initial concentration.

4. General Formula for nth Order

The "n-1" Rule

For a reaction of order 'n' (where $n \neq 1$), the half-life is given by:

$$ t_{1/2} \propto \frac{1}{[A]_0^{n-1}} $$

Order (n)	Relation	Nature
0	$t_{1/2} \propto [A]_0$	Directly proportional
1	$t_{1/2} \propto [A]_0^0$	Constant (Independent)
2	$t_{1/2} \propto 1/[A]_0$	Inversely proportional

5. Useful Relations (First Order)

• Time for 75% completion ($t_{75\%}$) = $2 \times t_{1/2}$
• Time for 99.9% completion ($t_{99.9\%}$) $\approx 10 \times t_{1/2}$
• Amount remaining after 'n' half-lives: $[A] = [A]_0 \left(\frac{1}{2}\right)^n$

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