Question

The activation energy of an uncatalyzed reaction is 70.0 kJ/mol. When a catalyst is added, the activation energy (at 20.0 C) is 42.0 kJ/mol. Theoretically, to what temperature (C) would one have to heat the solution so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction at 20. 0 C? Assume the frequency factor A is constant, and assume the initial concetrations are the same.

Answer 1

Explanation & Calculation steps :

(i) The rate of the uncatalyzed reaction would be equal to rate of the catalyzed reaction when the rate constants of a reaction are equal under both the conditions.

(ii) Let the rate constant for uncatalyzed reaction be represented as k_{(uncatalyzed)} , and that of catalyzed reaction be k_(catalyzed)

(iii) Applying Arrhenius equation,

...........ln k_{(uncatalyzed ) = -}E_a ( uncatalyzed ) / RT _{(uncatalyzed )}   + ln A

&........ln k _{(catalyzed )} .... = - E_a (catalyzed ) / RT _{(catalyzed )}   + ln A

(iv ) when rate constants under the catalyzed and uncatalyzed conditions are equal-

..........................ln k _{(uncatalyzed)}   = ln k _{(catalyzed )}

[ - E_a (uncatalyzed ) / RT_{(uncatalyzed)} ] + ln A = [ - E_a (catalyzed ) / RT_(catalyzed) ] + ln A

therefore , - E_a (uncatalyzed ) / RT _{(uncatalyzed ) =} - E_a (catalyzed ) / RT _{(catalyzed )}

hence ,.....................................E_a (uncatalyzed ) / E_a (catalyzed ) = T_{(uncatalyzed )} / T_(catalyzed)       

(v) Substituting the given values of Ea & temperatures (in Kelvin ) as ,

.............T_(catalyzed) = 293K . T_{(uncatalyzed)}   = ? , E_a (uncatalyzed ) = 70.0 kJ , E_a  ( catalyzed ) = 42.0 kJ

.........T_{(uncatalyzed)}  is calculated = [ ( 70.0 / 42.0 ) x 293 ]

.......................................................= 488.33 K

..............................................or , = ( 488.33 - 273 )^o C

.........................................................= 215.33^o C