$1 d m^{3}$ of $2 M C H_{3} C O O H$ is mixed with $1 d m^{3}$ of $3 M$ ethanol to form ester. The decrease in the initial rate if each solution is diluted with an equal volume of water would be:

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Solution

The correct option is C $4$ times
$r_{1} = k [C H_{3} C O O H] [C H_{3} C H_{2} O H]$
$r_{1} = k \times 1 \times 1.5 = 1.5 k$

When the solution is diluted with an equal amount of $H_{2} O$ , concentration is reduced to half

$r_{2} = k \times 0.5 \times 0.75 = 0.375 k$

Hence, $\frac{r_{1}}{r_{2}} = 4$ or the rate is reduced by four times

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