The rate constant of a reaction is given by lnk (sec –1 ) = 14.34 – (1.25 × 10 4 )/T What…
Chemistry · JEE Main · NTA Exams — Chemical Kinetics
The rate constant of a reaction is given by
lnk (sec–1) = 14.34 – (1.25 × 104)/T
What will be the energy of activation ?
lnk (sec–1) = 14.34 – (1.25 × 104)/T
What will be the energy of activation ?
- 24.83 Kcal mole
- 12.42 Kcal/mole
- 49.96 Kcal/mole
- None
Answer
(A) 24.83 Kcal mole
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