On mixing 1 dm³ of 3 M ethanol with 1 dm³ of 2 M ethanoic acid, an ester is formed.
$C_{2} H_{5} OH + {CH}_{3} COOH ⟶ {CH}_{3} {COOC}_{2} H_{5} + H_{2} O$
If each solution is diluted with an equal volume of water, the decrease in the initial rate would be

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2 times

4 times

0.5 times

0.25 times

answer is B.

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Detailed Solution

$r = k [{CH}_{3} COOH] [C_{2} H_{5} OH] or r = k [\frac{n}{V}⌋ [\frac{n}{V}]$
Increasing volume (V) by twice (due to dilution) will reduce the rate by 4 times.
$i.e. r = 0 .25 k [{CH}_{3} COOH] [C_{2} H_{5} OH]$

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