Question

If a temperature increase from 19.0 C to 33.0 C triples the rate constant for a reaction, what is the value of the activation barrier for the reaction?

Answer 1

The activation energy barrier is 57.1 kJ/mol.

Explanation:

The Arrhenius equation gives the relation between temperature and reaction rates:

`color(blue)(|bar(ul(color(white)(a/a) k = Ae^(-E_"a"/(RT))color(white)(a/a)|)))" "`

where

`k` = the rate constant
`A` = the pre-exponential factor
`E_"a"` = the activation energy
`R` = the Universal Gas Constant
`T` = the temperature

If we take the logarithms of both sides, we get

`lnk = lnA - E_"a"/(RT)`

Finally, if we have the rates at two different temperatures, we can derive the expression

`color(blue)(|bar(ul(color(white)(a/a) ln(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)color(white)(a/a)|)))" "`

In your problem,

`T_2 =(33.0 + 273.15) K = 306.15 K`
`T_1 = (19.0 + 273.15) K = 292.15 K`
`k_2/k_1 = 3`

Now, let's insert the numbers.

`ln(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)`

`ln3 = E_"a"/("8.314 J"·color(red)(cancel(color(black)("K")))^"-1""mol"^"-1") (1/(292.15 color(red)(cancel(color(black)("K")))) - 1/(306.15 color(red)(cancel(color(black)("K")))))`

`ln3= E_"a"/("8.314 J·mol"^"-1") × 1.565 × 10^"-4"`

`E_a = (ln3 × "8.314 J·mol"^"-1")/(1.565 × 10^"-4") = "57 100 J/mol" = "57.1 kJ/mol"`