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Consider the following reaction:$ A + 3 B \rightarrow 2 C + 2 D$. At some point in the reaction $[B]=0.9986\; M$ and $13.20$ minutes later $[B]=0.9746\;M$. Based on this data, what is the average rate of the reaction during this time in $M/s$?

$\begin{array}{1 1} 1.010 \times 10^{-5}\;M/s \\ 1.010 \times 10^{-3}\;M/s \\ 6.060 \times 10^{-5}\;M/s \\ 60.60 \times 10^{-3}\;M/s \end{array}$
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Answer: $1.010 \times 10^{-5}\;M/s$
For the given reaction, rate of the reaction $ = - \large\frac{1}{3}$$ \large\frac{\Delta B}{\Delta t}$
$\Rightarrow = - \large\frac{1}{3}$$ \large\frac{.0.9746\;M-0.9986\;M}{13.2\times 60 \;s}$$ =1.010 \times 10^{-5}\;M/s$
answered Dec 13, 2013 by sreemathi.v
edited Jul 24, 2014 by balaji.thirumalai

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