100 text{ cm}^3 of 1 text{ M } CH_3COOH was m | vedclass

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(D) The rate law for the esterification reaction is $Rate = k[CH_3COOH][CH_3OH]$.In the first case,the total volume is $100 	ext{ cm}^3 + 100 	ext{

Vedclass · Accepted Answer

$0.25$

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(D) The rate law for the esterification reaction is $Rate = k[CH_3COOH][CH_3OH]$.In the first case,the total volume is $100 	ext{ cm}^3 + 100 	ext{ cm}^3 = 200 	ext{ cm}^3$.In the second case,each solution is diluted with an equal volume of water before mixing. Thus,$100 	ext{ cm}^3$ of acid is diluted to $200 	ext{ cm}^3$ and $100 	ext{ cm}^3$ of alcohol is diluted to $200 	ext{ cm}^3$. When these are mixed,the total volume becomes $400 	ext{ cm}^3$.Since the total volume of the mixture is doubled,the concentration of each reactant ($CH_3COOH$ and $CH_3OH$) becomes half of its initial concentration in the mixture.Let the initial concentrations be $[C_1]$ and $[C_2]$. Then $Rate_1 = k[C_1][C_2]$.In the second case,the new concentrations are $[C_1/2]$ and $[C_2/2]$.$Rate_2 = k[C_1/2][C_2/2] = \frac{1}{4} k[C_1][C_2] = 0.25 	imes Rate_1$.Thus,the rate becomes $0.25$ times the initial rate.

$[AB] \ (mol \ dm^{-3})$	Rate of decomposition of $AB \ (mol \ dm^{-3} \ s^{-1})$
$0.20$	$2.75 \times 10^{-8}$
$0.40$	$11.0 \times 10^{-8}$
$0.60$	$24.75 \times 10^{-8}$

$100 \text{ cm}^3$ of $1 \text{ M } CH_3COOH$ was mixed with $100 \text{ cm}^3$ of $2 \text{ M } CH_3OH$ to form an ester. The change in the initial rate if each solution is diluted with an equal volume of water would be: (in $\text{times}$)

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