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 :

Answer:

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

Explanation:

According to the Arrhenius equation,

[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]

or,

[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]

where,

[tex]K_1[/tex] = rate constant at [tex]18.0^oC[/tex] = k

[tex]K_2[/tex] = rate constant at [tex]37.0^oC[/tex]  = 3k

[tex]Ea[/tex] = activation energy for the reaction = ?

R = gas constant = 8.314 J/mol.K

[tex]T_1[/tex] = initial temperature = [tex]18.0^oC=273+18.0=291 K[/tex]

[tex]T_2[/tex] = final temperature = [tex]37.0^oC=273+37.0=310 K[/tex]

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

[tex]\log (\frac{3k}{k})=\frac{Ea}{2.303\times 8.314 J/mol K}[\frac{1}{291 K}-\frac{1}{310 K}][/tex]

[tex]Ea=43,374 J/mol=43.374 KJ/mol[/tex]

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

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