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The reaction $A \rightarrow \text{Products}$ is second order in $A$. Initially $[A]_0 = 1.00 M$ and after $25 \; \text{mins}, \;[A] = 0.25\; M$. What is the rate constant for this reaction?

$\begin{array}{1 1} 0.12\;M^{-1}\text{mins}^{-1} \\ 0.24\;M^{-1}\text{mins}^{-1} \\ 0.36\;M^{-1}\text{mins}^{-1} \\ 0.48\;M^{-1}\text{mins}^{-1} \end{array}$

1 Answer

Answer: $0.12\;M^{-1}\text{mins}^{-1}$
From the integrated rate law for a second order reaction,$\large\frac{1}{[A]}$$ = \large\frac{1}{[A]_0}$$ + kt$
Given $t$, $[A]$ and $[A]_0$ we can simply solve for $k$,
$\Rightarrow \large\frac{1}{0.25\;M}$$ = \large\frac{1}{1\; M}$$ + k\times 25\; \text{mins}$
$\Rightarrow k = \large\frac{3}{25}$$ \;M^{-1}\;\text{mins}^{-1}$$ = 0.12\;M^{-1}\text{mins}^{-1}$
answered Jul 24, 2014 by balaji.thirumalai
edited Jul 24, 2014 by balaji.thirumalai

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