February 25, 2017

Homework Help: Rate Constant - My Work is Incorrect

Posted by Amy on Sunday, February 3, 2013 at 3:26pm.

Hi! I am stuck with this question and I would like to know what I did wrong:

Given that the initial rate constant is 0.0160 s-1 at an initial temperature of 25 degrees C, what would the rate constant be at a temperature of 130 degrees C for the same reaction described in Part A?

I solved for the Ea, and I got 32,900 J.
ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)
ln(k2/0.0160) = (32,900/8.314)(1/298.15 - 1/314.55)
ln(k2/0.0160) = (3,957.180)(0.00174871)
ln(k2/0.0160) = (0.6919986)
(k2/0.0160) = e^(0.6919986)
(k2/0.0160) = 1.997704
k2 = 0.03196

The answer above is not correct.


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