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1st Law of Thermodynamics

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Question

According to Arrhenius equation, the rate constant $(k)$ is related to temperature $(T)$ as:

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_{2}} - \frac{1}{T_{1}})

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C

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

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D

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

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Solution

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

$k_{1} = A e^{-^{\frac{E a}{R T_{1}}}}$

$k_{2} = A e^{-^{\frac{E a}{R T_{2}}}}$

$\frac{k_{2}}{k_{1}} = \frac{e^{- E_{a} / R T_{2}}}{e^{-^{E_{a} / R T_{1}}}}$

$\frac{k_{2}}{k_{1}} = e^{- [\frac{E_{a}}{R T_{2}} - \frac{E_{a}}{R T_{1}}]}$

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

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

Option A is correct.

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