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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 $150 \;s$?

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

1 Answer

Answer: 64%
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 $64\%$
answered Jul 24, 2014 by balaji.thirumalai
 

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