A reaction follows first-order kinetics with t₁/₂= 10 min. Find rate constant.

First-order rate constant with $t_{1/2}=10\ \text{min}$$t$$1/2$$=$$10$$min$
Answer: $k=\dfrac{\ln2}{t_{1/2}}=\dfrac{0.693}{10}=0.0693\ \text{min}^{-1}\approx1.16\times10^{-3}\ \text{s}^{-1}$.
Method: First-order kinetics halves concentration every half-life independent of initial amount. Convert to desired time units by dividing by 60. The integrated law $\ln\frac{[A]_0}{[A]}=kt$ is consistent with this $k$. If asked for 90% decay time, use $t_{0.9}=\ln10/k$.