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प्रश्न
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
The rate of reaction becomes four times when the temperature changes from 293 K to 313 K.
Calculate the energy of activation (Ea) of the reaction assuming that it does not change with temperature.
(R = 8.314 J K–1 mol–1)
उत्तर
T1 = 293 K, T2 = 313 K
`log "k"_2/"k"_1 = "E"_"a"/(2.303 "R") [("T"_2 - "T"_1)/("T"_1"T"_2)]`
Ea = `2.303 "R" ("T"_1"T"_2)/("T"_2 - "T"_1) log "k"_2/"k"_1`
`("T"_1"T"_2)/("T"_2 - "T"_1) = (293 "K" xx 313 "K")/(313 "K" - 293 "K")` = 4585.45 K
`log "k"_2/"k"_1 = log 4/1` = log 4 = 0.6021
Ea = 2.303 × 8.314 JK−1 mol−1 × 4585.45 K × 0.6021
= 52863 J mol−1
= 52.8 kJ mol−1
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