Question

The activation energy for a reaction is 84.0kJ/mol. Addition of a catalyst lowers the activation energy by 23.0 kJ/mol, while leaving the A-factor unchanged. By what factor ( uncatalyzed catalyzed k k ) does the rate increase at 25 °C?

Answer 1

Answer : The rate constant of the reaction is increased by factor,
`4.93* 10^(10)`

Solution :

According to the Arrhenius equation,

`K=A* e^{(-Ea)/(RT)}`

The expression used with catalyst and without catalyst is,

`(K_2)/(K_1)=\frac{A* e^{(-Ea_2)/(RT)}}{A* e^{(-Ea_1)/(RT)}}`

`(K_2)/(K_1)=e^{(Ea_1-Ea_2)/(RT)}`

where,

`K_2` = rate of reaction with catalyst

`K_1` = rate of reaction without catalyst

`Ea_2` = activation energy with catalyst = 23.0 kJ/mol = 23000 J/mol

`Ea_1` = activation energy without catalyst = 84.0 kJ/mol = 84000 J/mol

R = gas constant = 8.314 J/mol.K

T = temperature =
`25^oC=273+25=298K`

Now put all the given values in this formula, we get

`(K_2)/(K_1)=e^{((84000-23000)J/mol)/(8.314J/mol.K* 298K)}`

`(K_2)/(K_1)=4.93* 10^(10)`

Therefore, the rate constant of the reaction is increased by factor,
`4.93* 10^(10)`