Question

The rate constants of some reactions double with every 10 degree rise in temperature. Assume that a reaction takes place at 271 K and 281 K. What must the activation energy be for the rate constant to double as described?

Answer 1

Answer : The activation energy for the reaction is, 119.7 J

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 271 K

`K_2` = rate constant at 281 K =
`2K_1`

`Ea` = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

`T_1` = initial temperature = 271 K

`T_2` = final temperature = 281 K

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

`\log ((2K_1)/(K_1))=(Ea)/(2.303* 8.314J/mole.K)[(1)/(271K)-(1)/(281K)]`

`Ea=119.7J`

Therefore, the activation energy for the reaction is, 119.7 J