# A certain first order reaction proceeds through t first order kinetics. The half-life of the reaction is 180 s. What percent of the initial concentration remains after 900s?

$\begin{array}{1 1} 3.12 \% \\ 6.24\;\% \\ 1.56\% \\ 9.36\;\% \end{array}$

From the integrated rate law for a first order reaction, $[A] = [A]_0\; e^{-kt}$.
The half life $t_{1/2} = \large\frac{ ln\;2}{k}$$= \large\frac{0.693}{k}. Given t_{1/2} = 180\;s \rightarrow k = \large\frac{0.693}{180\;s} = 0.00385 s^{-1} The fraction remaining is the concentration divided by the initial concentration = \large\frac{[A]}{[A_0]}$$ = e^{kt} = e^{0.00385 s^{-1} \times 900\; s}$$= 0.0312 = 3.12\%$
edited Jul 24, 2014