Question

After 60.0 min, 30.0% of a compound has decomposed. What is the half-life of this reaction assuming first-order kinetics?

Answer 1

The integral rate law for a first order reaction is: \[\ln(\frac{[A]_t}{[A]_o}) = -kt\]Using this, we can find the reaction rate coefficient (k), which can be used to determine the half life.
\[k =- \frac {1}{t} * \ln (\frac {[A]_t}{[A]_o}) = -\frac {1}{60 \min} * \ln (\frac {70 %}{100 %}) = 5.94*10^{-3} s^{-1}\]The half life of this particular first order reaction is then:\[t_{1/2} = \frac {\ln 2}{k} = \frac {\ln 2}{5.94*10^{-3} s^{-1}} = 116.6 \min \rightarrow 117 \min\]