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Temperature Problem

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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)(1/301.15-1/352.15)
-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 -

    I 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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