The half-time of a reaction is defined as the time required to reduce the concentration of the reactant to half of its initial value. It is denoted by the symbol t_{1/2}. Thus,

When x = a/2 , t = t_{1/2}

Putting these in equation 2 mentioned above, we get

$\large k = \frac{2.303}{t_{1/2}} log\frac{a}{a-a/2}$

$\large k = \frac{2.303}{t_{1/2}} log2$

$\large k = \frac{2.303}{t_{1/2}} \times 0.3010$

(∴ log2 = 0.30103)

$\large t_{1/2} = \frac{0.693}{k}$

Since k is a constant for a given reaction at a given temperature and the expression lacks any concentration term so from equation 3 it is evident that half-time of a 1st order reaction is a constant independent of initial concentration of reactant.

This means if we start with 4 mole L^{−1} of a reactant reacting by first−order kinetics and after 20 minute it is reduced to 2 mole L^{−1} will also be 20 minute. That is, after 20 minutes from the start of reaction the concentration of the reactant will be 2 mole L^{−1}, after 40 minutes from the start of reaction of concentration is 1 mole L^{−1} .

After 60 minutes from the start of reaction the concentration of the reactant will be reduced to 0.5 mole L^{−1}. In other words, if during 20 minute 50% of the reaction completes, then in 40 minute 75%, in 60 minute 85.5% of the reaction and on will complete as shown with following plot.

Thus,

Fraction left after n half-lives = (1/2)^{n}

Concentration left after n half lives a_{n} = (1/2)^{n}a_{o}

It is also to be noted that equation 3 helps to calculate t1/2 or k with the knowledge of k or t_{1/2}