$100 {c m}^{3}$ of $1 M$ ${C H}_{3} C O O H$ was mixed with $100 {c m}^{3}$ of $2 M$ ${C H}_{3} O H$ to form an ester. The change in the initial rate if each solution is diluted with equal volume of water would be:

2

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4

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0.5

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0.25

times

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Solution

The correct option is D $0.25$ times
$\begin{matrix} Rate of reaction = k [CH3COOH] [CH3OH] if both are diluted to double their initial volume concentration of both will be halved. ∴ rate of reaction changes by \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25 \end{matrix}$

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$100 c m_{3}$ of 1 M $C H_{3} C O O H$ was mixed with $100 c m_{3}$ of 2M $C H_{3} O H$ to form an ester. The change in the intial rate if each solution is diluted with equal volume of water would be

Q. One litre of 2M acetic acid is mixed with one litre of 3M ethyl alcohol to form ester

C H_{3} C O O H + C_{2} H_{5} O H \Leftrightarrow C H_{3} C O O C_{2} H_{5} + H_{2} O

. The decrease in the initial rate if each solution is diluted by an equal volume of water would be

Q. One litre of 2M acetic acid is mixed with one litre of 2M ethyl alcohol to form ester as

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

The decrease in initial reaction rate if each solution is diluted by an equal volume of water would be

1 dm3 of 2 M CH3 COOH is mixed with 1 dm3 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

Ans 0.25 times

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100 cm3 of 1M CH3COOH was mixed with 100 cm3 of 2M CH3OH to form an ester. The change in the initial rate if each solution is diluted with equal volume of water would be:

The correct option is D 0.25 times Rate of reaction =k [CH3COOH] [CH3OH] if both are diluted to double their initial volume concentration of both will be halved. ∴ rate of reaction changes by 12×12=14=0.25

$100 {c m}^{3}$ of $1 M$ ${C H}_{3} C O O H$ was mixed with $100 {c m}^{3}$ of $2 M$ ${C H}_{3} O H$ to form an ester. The change in the initial rate if each solution is diluted with equal volume of water would be:

The correct option is D $0.25$ times
$\begin{matrix} Rate of reaction = k [CH3COOH] [CH3OH] if both are diluted to double their initial volume concentration of both will be halved. ∴ rate of reaction changes by \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25 \end{matrix}$