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The slope of the graph drawn between ln k (taken on y-axis) verses $\frac{1}{T}$ will have a slope value of $- 2500$ . Then the value of activation energy in cal/mol will be [k = Rate constant]

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5000

543.4

54.72

250

answer is A.

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Detailed Solution

$Slope = \frac{- Ea}{R} 2500 = \frac{E_{a}}{2} = 5000$

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