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Question
The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
Solution
According to the Arrhenius equation,
`log "k"_2/"k"_1 = ("E"_"a")/(2.303 "R") [1/"T"_1 - 1/"T"_2]`
`"K"_2/"k"_1` = 2,
T1 = 298 K,
T2 = 308 K,
R = 8.314 JK−1 mol−1
∴ `log 2 = "E"_"a"/(2.303 xx 8.314 "JK"^-1 "mol"^-1) [1/(298 "K") - 1/(308 "K")]`
`0.3010 = "E"_"a"/(2.303 xx 8.314 "JK"^-1 "mol")^-1 xx 10/(298 xx 308)`
∴ `"E"_"a" = (0.3010 xx 2.303 xx 8.314 xx 298 xx 308)/10 "J mol"^-1`
= 52897.7 J mol−1
= 52.897 kJ mol−1
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