Chemistry

The activation energy (E*) for a reaction is 195 KJ. If the k for this reaction is 3.46 x 10-5/min at 35oC, what will the temperature be in OF when k is 1.5 x 10-3/min?

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asked by Jessica
  1. ln(k2/k1) = Ea(1/T1 - 1/T2)/R

    ln(1.5E-3/3.46E-5) = 195000(1/308 - 1/T2)/8.314
    Solve for T2.

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    posted by DrBob222

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