# Temperature

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Q: A reaction rate has a constant of 1.23 x 10-4/s at 28 degrees C and 0.235 at 79 degrees C. Determine the activation barrier for the reaction.

A: My work thus far:

ln(1.23 x 10-4/0.235)=(Ea/8.314)(1/79-1/28)
-7.551=(Ea/8.314)(0.01265-0.03571)
-7.551=(Ea/8.314)(-0.02305)
327.53=(Ea/8.314)
Ea=2,723.08 J/mol
2,723 J/mol = 0.002723 kJ/mol

My answer is incorrect, and I would like to know where I went wrong and what the correct answer is.

• Temperature -

Thanks for showing your work.
I didn't work it out but your error is you didn't convert T to kelvin.
28 C = 301.15
79 C = 352.15

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