Question

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If the uncatalyzed reaction takes about 6900 years to occur, about how long will the catalyzed reaction take

Answer 1

The catalyzed reaction will take 2.85 seconds to occur.

Step-by-step explanation:

The activation energy of a reaction is given by:

`k = Ae^{-(E_(a))/(RT)}`

For the reaction without catalyst we have:

`k_(1) = Ae^{-\frac{E_{a_(1)}}{RT}}` (1)

And for the reaction with the catalyst:

`k_(2) = Ae^{-\frac{E_{a_(2)}}{RT}}` (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

`(k_(1))/(k_(2)) = \frac{Ae^{-\frac{E_{a_(1)}}{RT}}}{Ae^{-\frac{E_{a_(2)}}{RT}}}`

`(k_(1))/(k_(2)) = e^{\frac{E_{a_(2)} - E_{a_(1)}}{RT}`

`(k_(1))/(k_(2)) = e^{(59.0 \cdot 10^(3)J/mol - 184 \cdot 10^(3) J/mol)/(8.314 J/Kmol*600 K) = 1.31 \cdot 10^(-11)`

Since the reaction rate is related to the time as follow:

`k = (\Delta [R])/(t)`

And assuming that the initial concentrations ([R]) are the same, we have:

`(k_(1))/(k_(2)) = (\Delta [R]/t_(1))/(\Delta [R]/t_(2))`

`(k_(1))/(k_(2)) = (t_(2))/(t_(1))`

`t_(2) = t_(1)(k_(1))/(k_(2)) = 6900 y*1.31 \cdot 10^(-11) = 9.04 \cdot 10^(-8) y*(365 d)/(1 y)*(24 h)/(1 d)*(3600 s)/(1 h) = 2.85 s`

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!