# From the integrated rate law for a first order reaction, $[A] = [A]_0\; e^{-kt}$. Suppose rate constant, $k = 2.95 X 10^{-3} s^{-1}$, What percent of $A$ remains after $150 \;s$?

$\begin{array}{1 1} 64\% \\ 32\% \\ 16 \% \\ 48\% \end{array}$

From the integrated rate law for a first order reaction, $[A] = [A]_0\; e^{-kt}$.
The fraction remaining is the concentration divided by the initial concentration $= \large\frac{[A]}{[A_0]}$$= e^{2.95 \times 10^{-3} s^{-1} \times 150\; s}$$ = 0.642$ or $64\%$