Solved: Temperature and Rate (Section)(a) The activation
Chapter , Problem 54E(choose chapter or problem)
(a) The activation energy for the isomerization of methyl isonitrile (Figure 14.7) is
\(160 \mathrm{~kJ}>\mathrm{mol}\). Calculate the fraction of methyl isonitrile molecules that has an energy of \(160.0 \mathrm{~kJ}\) or greater at \(500 \mathrm{~K}\).
(b) Calculate this fraction for a temperature of \(520 \mathrm{~K}\). What is the ratio of the fraction at \(520 \mathrm{~K}\) to that at \(500 \mathrm{~K}\)?
Equation Transcription:
Text Transcription:
160 kJ>mol
160.0 kJ
500 K
520 K
520 K
500 K
Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.
Becoming a subscriber
Or look for another answer