Problem: The activation energy of a certain reaction is 36.9 kJ/mol. At 20°C, the rate constant is 0.0140 s-1. At what temperature in degrees Celsius would this reaction go twice as fast?

FREE Expert Solution

We’re being asked to determine the temperature at which the reaction would go twice as fast. 


We’re given the rate constant at another temperature and the activation energy of the reaction. This means we need to use the two-point form of the Arrhenius Equation:


ln k2k1=-EaR[1T2-1T1]


where:

k1 = rate constant at T1 

k2 = rate constant at T2 

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

The activation energy of a certain reaction is 36.9 kJ/mol. At 20°C, the rate constant is 0.0140 s-1. At what temperature in degrees Celsius would this reaction go twice as fast?

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