Asked by Anonymous
The respiration of plants is a biochemical reaction and should have a rate determined by the Arrhenius equation, k = Ae-EA/RT. Here, k is the rate, A is a constant, EA is the activation energy for the reaction, R is the gas constant, and T is the temperature in kelvins.
If the rate doubles for a 6 kelvin rise in temperature at around room temperature, what is the activation energy for the reaction?
R is about 8.31 J/K-mol. The value of e is 2.718, but you only have to know that ln(2) is about 0.693 and that you can rewrite the Arrhenius equation as ln(k) = -EA/RT. That means that the reaction rate doubles when the value of EA/RT changes by 0.693, or the value of EA/T changes by 0.693×8.31 J/K-mol = 5.76 J/K-mol.
If the rate doubles for a 6 kelvin rise in temperature at around room temperature, what is the activation energy for the reaction?
R is about 8.31 J/K-mol. The value of e is 2.718, but you only have to know that ln(2) is about 0.693 and that you can rewrite the Arrhenius equation as ln(k) = -EA/RT. That means that the reaction rate doubles when the value of EA/RT changes by 0.693, or the value of EA/T changes by 0.693×8.31 J/K-mol = 5.76 J/K-mol.
Answers
Answered by
Anonymous
120
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.