Part A The activation energy of a certain reaction is 41.6 k J / m o l . At 25 ?

Question

Part A The activation energy of a certain reaction is 41.6kJ/mol . At 25 ?C , the rate constant is 0.0190s?1. At what temperature in degrees Celsius would this reaction go twice as fast?
Part B Given that the initial rate constant is 0.0190s?1 at an initial temperature of 25 ?C , what would the rate constant be at a temperature of 100 ?C for the same reaction described in Part A? please help Part A The activation energy of a certain reaction is 41.6kJ/mol . At 25 ?C , the rate constant is 0.0190s?1. At what temperature in degrees Celsius would this reaction go twice as fast?
Part B Given that the initial rate constant is 0.0190s?1 at an initial temperature of 25 ?C , what would the rate constant be at a temperature of 100 ?C for the same reaction described in Part A? please help

Explanation / Answer

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

=416000/8.314(1/298 -1/373)

ln(k2/k1)=-3.376

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