Chemistry · Ch 6 — Chemical Kinetics

Half life and rate constant of the first order reaction

6.5.4

Half life and rate constant of the first order reaction

The integrated rate law for the first order reaction is

k = \frac{2.303}{t}\,\log_{10}\frac{\mathrm{[A]_0}}{\mathrm{[A]_t}}

where $\mathrm{[A]_0}$ is the initial concentration of the reactant at $t = 0$ . It falls to $\mathrm{[A]_t}$ at time $t$ after the start of the reaction. The time required for $\mathrm{[A]_0}$ to become $\mathrm{[A]_0}/2$ is denoted as $t_{1/2}$ , or

\mathrm{[A]_t} = \mathrm{[A]_0}/2 \;\;\text{ at }\; t = t_{1/2}

Putting this condition in the integrated rate law we write

k = \frac{2.303}{t_{1/2}}\,\log_{10}\frac{\mathrm{[A]_0}}{\mathrm{[A]_0}/2} = \frac{2.303}{t_{1/2}}\,\log_{10} 2 = \frac{2.303}{t_{1/2}} \times 0.3010

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

$t_{1/2} = \frac{0.693}{k} \qquad \text{...(6.10)}$ …

Figure 6.2Exponential decay of reactant concentration versus time for a first order reaction, with equal-width arrows marking that each successive halving of the concentration takes the same time - the constant half life.

Fig. 6.2 — Exponential decay of reactant concentration versus time for a first order reaction, with equal-width arrows marking that each successive halving of the concentration takes the same time - the constant half life.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. The concentration [A] decays exponentially with time. The row of short horizontal arrows steps down the curve at successive halving levels: each arrow spans the same width along the time axis, showing that the time needed to halve the concentration — the half life $t_{1/2}$ — is the same wherever you start on the curve. That is the graphical meaning of Eq. (6.10): the half l …

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