A certain reaction has an activation energy of 51.64 kj/mol. At what temperature Kelvin will the reaction be 5.50 times faster than it did at 343K?

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Azaria Salas

Chemistry

18 September, 11:29

A certain reaction has an activation energy of 51.64 kj/mol. At what temperature Kelvin will the reaction be 5.50 times faster than it did at 343K?

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Brennan Mcdaniel · Answer 1

For this, we use the Arrhenius equation to solve the problem. It is expressed as:

k = Ae^ - (E/RT)

where k is the reaction rate, A is the pre-exponential factor, E is the activation energy and T is the temperature.

For a reaction, A is constant at different temperatures. We then isolate the constant and then introduce the second condition to the equation since it will still be equal to the constant. It is then simplified to:

ln (k2/k1) = - E/R (1/T2 - 1/T1)

ln (5.50k1/k1) = - 51.64/8.314 (1/T2 - 1/343)

T2 = 379 K