If we plot a graph between log k and 1T by Arrhenius equation, the slope is :

K=A.e−Ea/RTlogK=logA−Ea2.303RTslope=−Ea2.303R

K=A.e−Ea/RTlogK=logA−Ea2.303RTslope=−Ea2.303R

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If we plot a graph between log k and $\frac{1}{T}$ by Arrhenius equation, the slope is :

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$- \frac{E_{a}}{2.303 R}$

$+ \frac{E_{a}}{R}$

$+ \frac{E_{a}}{2.303 R}$

$\frac{- E_{a}}{R}$

answer is C.

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$K = A . e^{- E a / R T}$

$\log K = \log A - \frac{E a}{2.303 R T}$

$s l o p e = \frac{- E a}{2.303 R}$

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