The rate constants for a first order reaction are found to be 2.76*10-5 s^-1 at 25 degrees celsius and 6.65*10^-4 s^-1 at 50 degrees celsius, respectively. A) Calculate the activation energy Ea in kilojoules per mole. B) Calculate the rate constant at 75 degrees celsius
a)
Given:
T1 = 25 oC
=(25+273)K
= 298 K
T2 = 50 oC
=(50+273)K
= 323 K
K1 = 2.76*10^-5 s-1
K2 = 6.65*10^-4 s-1
use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(6.65*10^-4/2.76*10^-5) = ( Ea/8.314)*(1/298 - 1/323)
3.182 = (Ea/8.314)*(2.597*10^-4)
Ea = 101856 J/mol
Ea = 101.8556 KJ/mol
Answer: 101.9 KJ/mol
b)
Given:
T1 = 25 oC
=(25+273)K
= 298 K
T2 = 75 oC
=(75+273)K
= 348 K
K1 = 2.76*10^-5 s-1
Ea = 101.8556 KJ/mol
= 101856 J/mol
use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(K2/2.76*10^-5) = (101856/8.314)*(1/298 - 1/348.0)
ln(K2/2.76*10^-5) = 12251*(4.821*10^-4)
K2 = 1.014*10^-2 s-1
Answer: 1.01*10^-2 s-1
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