# Solve What is the energy of activation of a reaction whose rate constant doubles when the temperature changes from 303 K to 313 K? - Chemistry

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What is the energy of activation of a reaction whose rate constant doubles when the temperature changes from 303 K to 313 K?

#### Solution

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

E= (0.3010 xx 19.147)/(1.0544 xx 10^-4) J mol-1

E= 54659 J mol-1 = 54.66 kJ mol-1

The energy of activation of the reaction is 54.66 kJ mol-1 .

Concept: Temperature Dependence of Reaction Rates
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Chapter 6: Chemical Kinetics - Exercises [Page 137]

#### APPEARS IN

Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 6 Chemical Kinetics
Exercises | Q 4.4 | Page 137
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