Arrhenius equation maybe represented as:

l n \frac{A}{k} = \frac{E_{a}}{R T}

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\frac{d l n k}{d T} = \frac{E_{a}}{R T}

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log A = log k + \frac{E_{a}}{2.303 R T}

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log (- \frac{E_{a}}{R T}) = \frac{k}{A}

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Solution

The correct option is A $l n \frac{A}{k} = \frac{E_{a}}{R T}$
Arrenhius Equation is given by,
$k = A e^{\frac{- E_{a}}{R T}}$
Therefore, now taking log on both sides
$l o g k = l o g A - \frac{E_{a}}{R T}$
$\frac{E_{a}}{R T} = l n \frac{A}{k}$

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