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Question
A first-order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. (Given : log 2 = 0·3010, log 4 = 0·6021, R = 8·314 JK–1 mol–1)
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Solution
Given
t1/2 = 40 min at temperature (T1) = 300 K
t1/2 = 20 min at temperature (T2) = 320 K
t1/2 = 40 min, t1/2 = 20 min
`k_1 = 0.693/40`
`k_2 = 0.693/20`
According to Arrhenius equation
`log (k_2/k_1) = "E"_"a"/(2.303 " R") [1/"T"_1 - 1/"T"_2]`
`= "E"_"a"/(2.303 " R") [("T"_2 - "T"_1)/("T"_1"T"_2)]`
`log ((0.0693/20)/(0.0693/40)) = "E"_"a"/(2.303 xx 8.314) [(320 - 300)/(300 xx 320)]`
`therefore 0.3010 = "E"_"a"/19.147 [0.0002083]`
Ea = 27664 J/mol
Ea = 27.7 kJ/mol
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