A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • one year ago

The rate constant of a reaction is 7.0 x10^-3 s^-1 at 25 degree C, and the activation energy is 33.6 kj/mol. What is k at 75 degree C? I know I have to use ln(k1/k2) = (Ea/R)[(1/T2)-(1/T1)] .. but I'm still confuse. Any help?

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ln (k1/k2) = Ea/R (1/T2-1/T1) R is the gas constant with Joule units: 8.314 J/mol-K You will need to switch your Activation Energy into Joule units: 33600 J and your Temperatures into Kelvin. Once you have done this, plug in and solve for k2. Variables: k1 = 4.7 x 10^(-3) k2 = ? T1 = 289 K T2 = 348 K Ea = 33600 J R = 8.314 J/mol-K

  2. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...


  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.