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Asked Sep 29, 2019
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A certain reaction has an activation energy of 59.88 kJ/mol. At what Kelvin temperature will the reaction proceed 5.00 times
faster than it did at 313 K?
T =
K
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A certain reaction has an activation energy of 59.88 kJ/mol. At what Kelvin temperature will the reaction proceed 5.00 times faster than it did at 313 K? T = K

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Expert Answer

Step 1

The expression for the ratio of two rate constants at the given temperature is given by Arrhenius equation labelled in the equation (1); k1 is the initial rate constant at temperature T1, k2 is the final rate constant at temperature T2, Ea is the activation energy and R is the universal gas constant.

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Е, 1 RT T 1 a In . () 2

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Step 2

The activation energy of the given reaction is 59.88 kJ/mol which is equal to 59.88 × 103 J/mol because 1 kJ/mol ...

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5 59.88 x 10' J/mol In 1 1 1 8.314 J/mol K313 K T 1 1.61 7202.310.00319- Т, 0.00319-1.61 T2 1 7202.31 =0.00297 1 T2 0.00297 = 336 K

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Chemical Kinetics

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