Obtain a relation log_(10).(k_(2))/(k_(1)) = (E_(a) (T_(2) -T_(1)))/(2.303 R xx T_(1) xx T_(2)) obtaina relationshowingvariationin rateconstant withtemeprature .

Obtain a relation log_(10).(k_(2))/(k_(1)) = (E_(a) (T_(2) -T_(1)))/(2

1.	Obtain a relation log_(10).(k_(2))/(k_(1)) = (E_(a) (T_(2) -T_(1)))/(2.303 R xx T_(1) xx T_(2)) obtaina relationshowingvariationin rateconstant withtemeprature .
Answer» SOLUTION :ByArrhenius equaiton , the rateconstantk of thereactionat a temperatureT is repesented as . `k = A xx e^(-E_(a)//RT)` whereAis a frequncyfactore, R is a gas constantand `E_(a)`is theenergy of activation. By takinglogarithmto the base we, get In = In `(A xx^(-Ea//RT))` = In `A +lne^(-E_(a)//RT) = In A -(E_(a))/(RT)` In e `therefore In k = In A - (E_(a))/(RT)` `therefore2.303 log_(10) k = 2.303 log.(A)/(10) -(E_(a))/(RT)` `therefore log_(10)K = log_(10)A - (E_(a))/(2.303RT)` If `k_(1)` and `k_(2)`are therateconstsntsat temperatures `T_(1)`and `T_(2)`respectively, then `log_(10)k_(1) = log_(10)A -(E_(a))/(2303RT_(1))` `log_(10) k_(2) = log_(10)A-(E)/(2.303RT_(2))` `thereforelog_(10)k_(2) - log_(10)k_(1) = [log_(10)A - (E_(a))/(2.303 RT_(2))] - [log_(10)A-(E_(a))/(2.303RT_(1))]` `log_(10).(k_(2))/(k_(1)) = (E_(a))/(2.303RT) - (E_(a))/(2.303RT)` `= (E_(a))/(2.303R)((1)/(T_(1)) -(1)/(T_(2))) "" log_(10).(k_(2))/(k_(1))=(E_(a)(T_(2) -T_(1)))/(2.303RT_(1) xx T_(2))` Bymeasuringthe rateconstant`k_(1)`and `k_(2)` at TWO DIFFERENTTEMPERATURES `T_(1)` and `T_(2)`the energyof activation `E_(a)`of the reaction can be obtained .