Advertisement Remove all ads

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?

Advertisement Remove all ads

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`

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Advertisement Remove all ads

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×