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The activation energy of a certain reaction is 40.5Kj/mol.the rate constant is Given 0.0130s^-1 the initial temperature is 20 C , what would the rate constant be at a temperature of 100 C ?

Chemistry
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E=-RT ln (k/A) 40.5 = -0.08206(295.15) ln (0.0130/A) e^(-40.5/(0.08206(293.15))) = 0.0130/A 0.07000234854... approx = A use A to solve for k at 100C (373.15 K): 40.5=-RT ln (k/A) 40.5=-0.08206(373.15) ln (k/30/0.07000234854) 40.5/(-0.08206(373.15))=ln (k/30/0.07000234854) e^(40.5/(-0.08206(373.15)))=k/0.07000234854 k=0.2627397387 pls. double check my calculations :))
ok thank let me do so
i think there's an error

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Other answers:

line 2: 40.5 = -0.08206(293.15) ln (0.0130/A)
im getting a very big answer
last line: 0.01865088557
the rate constant should be at least close to 0.0130s^-1. not geater than 1.

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