Sponsor Area

Chemical Kinetics

Question
CBSEENCH12006322

The rate constant increases 50% when temperature is increased from 298 K to 308 K. THe value of ΔH0 is 15 kJ mol. Calculate the activation energies Eaf and Ear.

Solution

The rate constant at 298 K = k.
When the temperature is raised from 298 k to 308 k.
Increase in rate constant

                      = k×50100 = 0.5 k
By using the equation

        Ink2/k1 = [Eaf/R] 1T1-1T2

We have,

     In(1.5 k/k) = [Eaf/R] 1298k-1308k                     = [Eaf/R] 10298 × 208log (2.303 × 1.5) = Eaf8.314×10298×308log 3.4545 = Eaf8.314×10298×3080.4460 = Eaf8.314×10298×308Eaf = 0.4460×8.314×298×30810     =       34.03 kJ/mol

                rH° = 15 kJ mol-1

but            rH° = Eaf-Ear

              Eaf = (34.03-15) kJ mol       = 19.03 kJ mol-1