k=Ae−RTEa
ln(K1K2)=REa[T11−T21]
ln(5)=8.314Ea[3001−3151]
1.6094=8.314Ea[300×31515]
Ea=84297J
JEE Main 2020 — Chemistry Physical Chemistry
The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from 27oC to 42oC. Its energy of activation in J/mol is _________ (Take ln5=1.6094;R=8.314Jmol−1)
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