Question
Asked Sep 22, 2019
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Determine the activation energy for a reaction in which the rate constant triples when the reaction is performed at 518.6 K instead of 300.7 K. Answer in kJ without the unit. 

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Expert Answer

Step 1

Given:

k2 = 3 k1

Temperature T1 = 300.7 K.

Temperature T2 = 518.6 K.

Step 2

Arrhenius equation is by:

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Ea X R 1 1 In T2 T Т.

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Step 3

Calculation for activ...

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k2 n 1 Е, X x R Tz (3k1 In Ea 1 1 X 300.7 K 518.6 K' 8.314 J. K-1. mol-1 Ea n(3) 8.314 J. K-1. mo/-1 X 1.397 x 10-3 K-1 1.098 Ex 1.680 x 10-4 J1.mol 1.098 Ea 1.680 x 10-4 J-1.mol Ea 6535.71 Jmol-1 6.5 KJ. mol1

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Science

Chemistry

Chemical Kinetics

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