The rate constant of a certain reaction at 100k is 3.465×10^-9 the activation energy of this reaction is 2.303 .The half life of this reaction at 509k in seconds is
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From the formulae:
k = Ae^(-Ea/RT)
logk = logA – 2.303Ea/RT...........(1)
Given eqn. is:
logk = 5.4 - 212/T …..................(2)
Compariing both the eqns. we get,
2.303Ea/R = 212
Ea = > 212*25/(3*2.303) => 767.11 J
Hope it will help you
k = Ae^(-Ea/RT)
logk = logA – 2.303Ea/RT...........(1)
Given eqn. is:
logk = 5.4 - 212/T …..................(2)
Compariing both the eqns. we get,
2.303Ea/R = 212
Ea = > 212*25/(3*2.303) => 767.11 J
Hope it will help you
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