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The rate of a reaction at 25.0 oC is 8 times faster than at 20.0 oC. Determine the activation energy (in kJ/mole) of this reaction.?
- Simonizer1218Lv 72 months ago
ln (k2/k1) = -Ea/R (1/T2 - 1/T1)
k1 = 1
k2 = 8
T1 = 20 + 273 = 293
T2 = 25 + 273 = 298
R = 0.008314 kJ/molK
ln (8/1) = -Ea/0.008314 (1/298 - 1/293)
2.08 = -Ea/0.008314 (-0.000057)
Ea = 303 kJ/mol