In a pseudo first order hydrolysis of ester i | vedclass

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(B) The average rate of reaction is given by the formula:$	ext{Average Rate} = -\frac{\Delta[	ext{Ester}]}{\Delta t} = -\frac{[	ext{Ester}]_{t_2} -

Vedclass · Accepted Answer

$4.67 	imes 10^{-3} \ mol \ L^{-1} \ s^{-1}$

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(B) The average rate of reaction is given by the formula:$	ext{Average Rate} = -\frac{\Delta[	ext{Ester}]}{\Delta t} = -\frac{[	ext{Ester}]_{t_2} - [	ext{Ester}]_{t_1}}{t_2 - t_1}$Given $t_1 = 30 \ s$ and $t_2 = 60 \ s$,the corresponding concentrations are $[	ext{Ester}]_{t_1} = 0.31 \ mol \ L^{-1}$ and $[	ext{Ester}]_{t_2} = 0.17 \ mol \ L^{-1}$.$	ext{Average Rate} = -\frac{0.17 - 0.31}{60 - 30} = -\frac{-0.14}{30} = \frac{0.14}{30} \ mol \ L^{-1} \ s^{-1}$$	ext{Average Rate} = 4.67 	imes 10^{-3} \ mol \ L^{-1} \ s^{-1}$

$t \ (s)$	$0$	$30$	$60$	$90$
$[\text{Ester}] \ (mol \ L^{-1})$	$0.55$	$0.31$	$0.17$	$0.085$

In a pseudo first order hydrolysis of ester in water,the following results were obtained:

$t \ (s)$ $0$ $30$ $60$ $90$

$[\text{Ester}] \ (mol \ L^{-1})$ $0.55$ $0.31$ $0.17$ $0.085$

What will be the average rate of reaction between the time interval $30 \ s$ to $60 \ s$?

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In a pseudo first order hydrolysis of ester in water,the following results were obtained: $t \ (s)$ $0$ $30$ $60$ $90$ $[\text{Ester}] \ (mol \ L^{-1})$ $0.55$ $0.31$ $0.17$ $0.085$ What will be the average rate of reaction between the time interval $30 \ s$ to $60 \ s$?