Friday

November 28, 2014

November 28, 2014

Posted by **Amy** on Saturday, February 2, 2013 at 10:16am.

A: My work thus far:

ln(1.23 x 10-4/0.235)=(Ea/8.314)(1/301.15-1/352.15)

-7.551=(Ea/8.314)(0.0048090535)

-7.551=(Ea/8.314)(0.0048090535))

-15,702.808=(Ea/8.314)

Ea= -1,192.24 J/mol

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

- Temperature Problem -
**DrBob222**, Saturday, February 2, 2013 at 3:05pmI believe you have reversed the k1/T1 vs k2/T2. Also I think you are reporting too many significant figures

I would set it up as

ln(k2/k1) = (E/R)(1/T1 - 1/T2)

ln(0.235/1.23E-4) = (Ea/8.314)(1/301.15 - 1/352.15)

My answer is approximately 130,000

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