Posted by Brun on Monday, October 15, 2012 at 6:58pm.
Yes, Ea goes in with units of J.
You don't need a graph.
ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)
Pick a k1 with T1, then k2 with 300 K for T2 and solve for k2. R is 8.314.
Oh! Thank you! I forgot to pick a second temperature.
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