Temperature dependent equation can be written as:

ln k = ln A - \frac{E_{a}}{R T}

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

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ln k = ln A - e^{R T / E_{a}}

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all of the above

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Solution

The correct option is A $ln k = ln A - \frac{E_{a}}{R T}$
$k = A e^{- \frac{E_{a}}{R T}} ln (k) = ln (A) - \frac{E_{a}}{R T}$

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Similar questions

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

If ln(a+c) , ln(a-c) , ln(a-2b+c) are in A.P, then

l n (a + c), l n (c - a), l n (a - 2 b + c)

ln (\frac{a + b}{3}) = (\frac{ln a + ln b}{2})

\frac{a}{b} + \frac{b}{a}

a (b - c) x^{2} + b (c - a) x y + c (a - b) y^{2} = 0

x - y

\frac{l n (c + a) + l n (a - 2 b + c)}{l n (a - c)}

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