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प्रश्न
The rate constant of a first order reaction are 0.58 S-1 at 313 K and 0.045 S-1 at 293 K. What is the energy of activation for the reaction?
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उत्तर
Given: Rate constant k1 = 0.58 s-1
Rate constant k2 = 0.045 s-1
T1 = 313 K
T2 = 293 K
R = 8.314 J K-1mol-1
To find: Activation energy (Ea)
Formula : `log_10 "k"_2/"k"_1 = "E"_"a"/(2.303"R")[("T"_2 - "T"_1)/("T"_1"T"_2")]`
Calculation : From Formula,
`log10((0.045 "s"^-1)/(0.58 "s"^-1)) = "E"_"a"/(2.303 xx 8.314 "JK"^-1 "mol"^-1)[(293"K" - 313 "K")/(293 "K" xx 313 "K")]`
∴ log 0.0776 = `"E"_"a"/(19.147 "J" "mol"^-1)[(-20)/(293 xx 313)]`
∴ `-1.110 = "E"_"a"/19.147[(-20)/(293 xx 313)]`
∴`"E"_"a" = (-1.110 xx 19.147 xx 293 xx 313)/-20`
= 97455.34 J `"mol"^-1`
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