Question

A certain reaction has an activation energy of 66.41 kJ/mol. At what Kelvin temperature will the reaction proceed 3.00 times faster than it did at 293 K?

Answer 1

Answer : The temperature will be, 392.462 K

Explanation :

According to the Arrhenius equation,

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

or,

`\log ((K_2)/(K_1))=(Ea)/(2.303* R)[(1)/(T_1)-(1)/(T_2)]`

where,

`K_1` = rate constant at
`T_1` =
`K_1`

`K_2` = rate constant at
`T_2` =
`3K_1`

`Ea` = activation energy for the reaction = 66.41 kJ/mole = 66410 J/mole

R = gas constant = 8.314 J/mole.K

`T_1` = initial temperature = 293 K

`T_2` = final temperature = ?

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

`\log ((3K_1)/(K_1))=(66410J/mole)/(2.303* 8.314J/mole.K)[(1)/(293K)-(1)/(T_2)]`

`T_2=392.462K`

Therefore, the temperature will be, 392.462 K