Problem: A certain reaction has an activation energy of 37.51 kJ/mol. At what Kelvin temperature will the reaction proceed 5.00 times faster than it did at 349 K?

FREE Expert Solution

We need to use the two-point form of the Arrhenius Equation:


lnk2k1 = -EaR[1T2-1T1]


where:

k1 = rate constant at T

k2 = rate constant at T

Ea = activation energy (in J/mol) 

R = gas constant (8.314 J/mol•K) 

T1 and T2 = temperature (in K).


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Problem Details

A certain reaction has an activation energy of 37.51 kJ/mol. At what Kelvin temperature will the reaction proceed 5.00 times faster than it did at 349 K?

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