$2 d m^{3}$ of $4 M$ $C H_{3} C O O H$ is mixed with $2 d m^{3}$ of 4 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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2 times

4 times

0.25 times

0.5 times

answer is C.

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Detailed Solution

$C H_{3} C O O H + C_{2} H_{5} O H \underset{}{\to} C H_{3} C O O C_{2} H_{5} + H_{2} O$

$r = K . {[A c i d]}^{1} {[E t h a n o l]}^{1}$

$r_{1} = K \times (\frac{2 \times 4}{4}) \times (\frac{2 \times 4}{4})$

$r_{2} = K \times {(\frac{2 \times 4}{8})}^{1} \times {(\frac{2 \times 4}{8})}^{1}$

$r_{2} = \frac{r_{1}}{4} = 0.25 \times r_{1}$

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