What is the value of the rate constant at 17 degrees celsius?
The Ea = 122 kJ/mol
I know I have to use ln k = ln A - Ea/RT.. but I don't know how to calculate it without known ln A and how do you find ln A
Thank you for your help!
8 answers
ln(k2/k1) = (Ea/R)(1/T1 -1/T2)
i have no idea how to apply this formula as we can only use one temperature and this requires the use of two.
Sorry. I have you confused with Gary who posted a similar question.
Is this problem posted exactly as it appears in your homework?
Well, they also gave this information: A reaction has a rate constant of 1.24×10−4 at 25 C and 0.228 at 79 C.
Am I missing something? You have two k values (use one of them) and two T values (use one of them that goes with the k you decided to use) and you have the Ea. So you want k at 17. What's the problem? Plug those values into the equation I wrote and solve for the new k at 17 C.
Im not getting the right value.I get 1.24*10^-4 and that's not correct.
I set this up.
ln(k2/1.24E-4) = (122,000/8.314)(1/298 - 1/290) and solved for k2. I obtained 3.19E-5.
Then just for the fun of it I set this up.
ln(k2/0.228) = (122,000/8.314)(1/352 - 1/290) and solve for k2. I obtained 3.07E-5. Pretty close for such a wide difference in k. The 3.19E-5 should be more accurate (comparing k at 25 with k at 17) instead of comparing k at 79 with k at 17.
ln(k2/1.24E-4) = (122,000/8.314)(1/298 - 1/290) and solved for k2. I obtained 3.19E-5.
Then just for the fun of it I set this up.
ln(k2/0.228) = (122,000/8.314)(1/352 - 1/290) and solve for k2. I obtained 3.07E-5. Pretty close for such a wide difference in k. The 3.19E-5 should be more accurate (comparing k at 25 with k at 17) instead of comparing k at 79 with k at 17.
1 answer