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A reaction rate constant is given by ${k = 1.2 × 10}^{14} e^{-25000/RT} {sec}^{-1}$ . It means

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log k versus log 1/T will give a straight line

log k versus T will give a straight line with slope as 25000

log k versus log T will give a straight line with a slope as – 25000

log k versus log 1/T will give a straight line with slope as – 25000

answer is C.

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Detailed Solution

${k = 1.2 × 10}^{14} e^{-25000/RT} {sec}^{-1} or$

${log k = log 1.2 × 10}^{14} - \frac{2500}{R} \frac{1}{T}$

Equation of straight line $s l o p e = - \frac{2500}{R}$

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