1 L of 2 M CH_3COOH is mixed with 1 L of 3 | vedclass

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Question

(C) Esterification is a second-order reaction.$CH_3COOH + C_2H_5OH \rightarrow CH_3COOC_2H_5 + H_2O$The rate law for this reaction is $r = k[CH_3COOH]

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(C) Esterification is a second-order reaction.$CH_3COOH + C_2H_5OH ightarrow CH_3COOC_2H_5 + H_2O$The rate law for this reaction is $r = k[CH_3COOH][C_2H_5OH]$.When each solution is diluted with an equal volume of water,the total volume doubles,so the concentration of each reactant is reduced to half of its initial value.Let the initial concentrations be $[A]_0$ and $[B]_0$. The initial rate is $r = k[A]_0[B]_0$.After dilution,the new concentrations are $[A]' = \frac{[A]_0}{2}$ and $[B]' = \frac{[B]_0}{2}$.The new rate is $r' = k 	imes (\frac{[A]_0}{2}) 	imes (\frac{[B]_0}{2}) = \frac{1}{4} 	imes k[A]_0[B]_0 = \frac{1}{4}r$.Thus,the rate of reaction becomes $0.25$ times the initial rate.

$1 \ L$ of $2 \ M$ $CH_3COOH$ is mixed with $1 \ L$ of $3 \ M$ $C_2H_5OH$ to form an ester. The rate of the reaction with respect to the initial rate when each solution is diluted with an equal volume of water will be (in times).

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