Question

A first order reaction takes 10 minutes for 25% decomposition. Calculate t_1/2 for the reaction.

(Given : log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021)

Answer 1

The rate constant k for a first-order reaction is given by

\[k = \frac{2 . 303}{t}\log\frac{\left[ R \right]_o}{\left[ R \right]}\]
\[where\]
\[ \left[ R \right]_o = \text{Initial concentration of reactant}\]
\[\left[ R \right] = \text{Final concentration of reactant}\]
\[At t = 10 \min, \]
\[k = \frac{2 . 303}{10}\log\frac{100}{100 - 25} = \frac{2 . 303}{10}\log\frac{4}{3}\]
\[ \Rightarrow k = \frac{2 . 303}{10}\left( \log 4 - \log 3 \right) = \frac{2 . 303}{10}\left( 0 . 6021 - 0 . 4771 \right)\]
\[ \Rightarrow k = 2 . 88 \times {10}^{- 2} \min^{- 1}\]

For a first order reaction, half-life and rate constant are related as

\[t_{1/2} = \frac{0 . 693}{k}\]
\[ \Rightarrow t_{1/2} = \frac{0 . 693}{2 . 88 \times {10}^{- 2}} = 24 . 06 \min\]

Hence, the t_1/2 for the reaction is 24.06 minutes.