Question

f a temperature increase from "18.0 ∘C to 37.0 ∘C" triples the rate constant for a reaction, what is the value of the activation barrier for the reaction?

Answer 1

The value of activation barrier for the reaction is, 43.374 kJ/mol.

Step-by-step 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
`18.0^oC` = k

`K_2` = rate constant at
`37.0^oC` = 3k

`Ea` = activation energy for the reaction = ?

R = gas constant = 8.314 J/mol.K

`T_1` = initial temperature =
`18.0^oC=273+18.0=291 K`

`T_2` = final temperature =
`37.0^oC=273+37.0=310 K`

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

`\log ((3k)/(k))=(Ea)/(2.303* 8.314 J/mol K)[(1)/(291 K)-(1)/(310 K)]`

`Ea=43,374 J/mol=43.374 KJ/mol`

Therefore, the activation energy for the reaction is, 43.374 kJ/mol.