Better Students Ask More Questions.
The activation energy of a certain reaction is 87 kJ·mol–1. What is the ratio of the...
1 Answer | add yours
Let the rate constant for the reaction at `37^o C` be `k_1`
the rate constant for the reaction at `15^o C` be `k_2`
Here, `T_1=37^o C=(37+273) K=310 K`
`T_2=15^o C=(15+273) K=288 K`
`E_a=87 kJ*mol^-1=87*10^3 J*mol^-1`
Plugging in the values in the above equation:
Therefore, the ratio of the rate constants for this reaction when the temperature is decreased from 37 °C to 15 °C is 13.2.
Posted by llltkl on August 2, 2013 at 1:39 PM (Answer #1)
Join to answer this question
Join a community of thousands of dedicated teachers and students.