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In the biologically-catalysed oxidation of ethanol, the concentration of ethanol decreases in a first order reaction from $800 mol {dm}^{- 3} to 50 mol {dm}^{- 3}$ in $2 \times 10^{4} s$ The rate constant $(S^{- 1})$ of the reaction is:

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$1, 38 \times 10^{- 4}$

$3.45 \times 10^{- 5}$

$1.00 \times 10^{- 4}$

$5.00 \times 10^{- 5}$

answer is B.

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Detailed Solution

It is known that,

$a t = \ln (\frac{a}{a_{0}})$

So the rate constant value,

$k \times 2 \times 10^{4} = \ln (\frac{800}{50}) \Rightarrow k = 1.38 \times 10^{- 4}$

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