The rate at which popcorn pops was measured at two different temperatures. How many times faster...
The rate at which popcorn pops was measured at two different temperatures. How many times faster does the popcorn pop at 250.0 °C compared with that at 150.0 °C when the activation energy is 53.8 kJ/mol?
Homework Answers
Activation energy = 53.8 kJ/mol
temperature T1 = 150 oC = 423 K
T2 = 250 oC = 523 K
ln (k2 / k1) = Ea / R [1 / T1 - 1/ T2]
ln (k2 / k1) = 53.8 / 8.314 x 10^-3 [1 / 423 - 1/523]
(k2 / k1) = 18.63
rate faster 18.6 times
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