Ask Question
16 May, 13:24

A certain reaction has an activation energy of 51.64 kj/mol. At what temperature Kelvin will the reaction be 5.50 times faster than it did at 343K?

+2
Answers (1)
  1. 16 May, 13:36
    0
    For this, we use the Arrhenius equation to solve the problem. It is expressed as:

    k = Ae^ - (E/RT)

    where k is the reaction rate, A is the pre-exponential factor, E is the activation energy and T is the temperature.

    For a reaction, A is constant at different temperatures. We then isolate the constant and then introduce the second condition to the equation since it will still be equal to the constant. It is then simplified to:

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

    ln (5.50k1/k1) = - 51.64/8.314 (1/T2 - 1/343)

    T2 = 379 K
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A certain reaction has an activation energy of 51.64 kj/mol. At what temperature Kelvin will the reaction be 5.50 times faster than it did ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers