# 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? - Chemistry

Sum

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?

#### Solution

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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