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Q.

The slope of the graph drawn between ln k (taken on y-axis) verses 1T 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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a

5000

b

543.4

c

54.72

d

250

answer is A.

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Detailed Solution

Slope=EaR 2500=Ea2=5000

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