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A certain reaction has an activation energy of 37.30 kj/mol. At what Kelvin temperature will the reaction proceed 3.00 times then it did at 303 k?

3 answers

Did you make a typo and omit the word FASTER in 3x.... than it did at 303K? Use the Arrhenius equation. No k1 is given but you can call it k1 and k2 becomes 3k1 (in which case k1 cancels) OR you can make up a number for k1, then multiply it by 3 to become k2.
No its not a typo but I did put the wrong activation energy. Its suppose to be a certain reaction has an activation energy of 37.70kj/mol. At what kelvin temperature will the reaction proceed 3.00 times then it did at 303k?

I was working the problem and got stuck on finishing the equation 1/T2=1/303k-1.099/4534.25k?
Yes, I think it is a typo for the sentence doesn't make sense as is. I think it should read, "At what Kelvin temperature will the reaction proceed 3.00 times FASTER THAN it did at 303 k?
ln(k2/k1) = Ea/R(1/T1-1/T2)
ln(3k1/k1) = [37700/8.314](1/303 - 1/T2)