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प्रश्न
Solve
What is the energy of activation of a reaction whose rate constant doubles when the temperature changes from 303 K to 313 K?
उत्तर
Given:
Rate constants; k2 = 2k1,
Temperatures: T1 = 303 K, T2 = 313 K
To find:
Activation energy of the reaction (Ea)
Formula:
`"log"_10 "k"_2/"k"_1 = "E"_"a"/(2.303"R") (("T"_2" - T"_1)/("T"_2"T"_1))`
Calculation:
`"log" (2"k"_1)/"k"_1 = "E"_"a"/(2.303 xx 8.314 "J" "K"^-1 "mol"^-1) ((313 "K" - 303 "K")/(313 "K" xx 303 "K"))`
`"log" 2 = "E"_"a"/(2.303 xx 8.314 "J" "mol"^-1) (10/(313 xx 303))`
0.3010 = `"E"_"a"/(19.147 "J" "mol"^-1) xx 1.0544 xx 10^-4`
Ea = `(0.3010 xx 19.147)/(1.0544 xx 10^-4)` J mol-1
Ea = 54659 J mol-1 = 54.66 kJ mol-1
The energy of activation of the reaction is 54.66 kJ mol-1 .
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