Question

A catalyst lowers the activation energy of a reaction from 20 kj/mol to 10 kj/mol the temperature at which the uncatalyzed reaction will have the same as that of the catalyzed at 27℃ is

Answer 1

If it is from 20 to 10 the answer will be as follows :-

You need to apply the Arhenius equation

The formula of Arhenius equation is K=Ae^Ea/rt

Suppose if the two rates are equal to each other

E^ea1/rt1=E^ea2/rt2

Ea/t1=Ea2/t2

20/T=10/27

after you solve for t you will  get your answer .

Hope it helped you.

Answer 2

Explanation:

Since, the given reaction is same so rate constant will also be the same. Formula for activation energy is as follows.

                           k = A$e^{-Ea/RT}$

Temperature for catalyzed reaction is $27^{o}C$ or (27 + 273)K equals 300 k. As rate constant is the same. So, formula to calculate temperature for uncatalyzed reaction will be as follows.

         $\frac{E_{a_{1}}}{T_{1}}$ = $\frac{E_{a_{2}}}{T_{2}}$

         $\frac{20 kJ/mol}{T_{1}}$ = $\frac{10 kJ/mol}{300 K}$

                                $T_{1}$ = 600 K

Hence, we can conclude that temperature of the uncatalyzed reaction is 600 K.