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Methods for Ester Synthesis

Activation en...

Question

Activation energy $(E_{a})$ and rate constant $(k_{1})$ and $(k_{2})$ of a chemical reaction at two different temperatures $(T_{1})$ and $(T_{2})$ are related by -

A

l n \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})

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B

l n \frac{k_{2}}{k_{1}} = - \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})

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C

l n \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} + \frac{1}{T_{2}})

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D

l n \frac{k_{2}}{k_{1}} = \frac{R}{E_{a}} (\frac{1}{T_{1}} + \frac{1}{T_{2}})

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Solution

The correct option is A $l n \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$
Arrhenius equation:
$k = A e^{\frac{- E_{a}}{R T}}$

Taking natural log on both side,
$l n k = l n (A e^{\frac{- E_{a}}{R T}})$
$l n k = l n A - \frac{E_{a}}{R T}$
Let us consider, for a particular reaction $k_{1}$ and $k_{2}$ are the rate constant at temperature $T_{1}$ and $T_{2}$ .
$∴$
$l n k_{1} = l n A - \frac{E_{a}}{R T_{1}} . . . . . e q n (1)$

$l n k_{2} = l n A - \frac{E_{a}}{R T_{2}} . . . . . e q n (2)$

Subtracting equation (1) from (2) we get,
$l n k_{2} - l n k_{1} = - \frac{E_{a}}{R T_{2}} + \frac{E_{a}}{R T_{1}}$

$l n \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R T_{1}} - \frac{E_{a}}{R T_{2}}$

$l n \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

Option (a) is correct

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