Question

A certain reaction with an activation energy of 165 kJ/mol was run at 455 K and again at 475 K. What was the ratio of `f` at the higher temperature to `f` at the lower temperature?

Answer 1

`(f_2/f_1) = 6.27`

Explanation:

I presume that you are using `f` as a symbol for the rate constant.

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

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

where

`f` = 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

`lnf = 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(f_2/f_1) = E_"a"/R(1/T_1 -1/T_2)color(white)(a/a)|)))" "`

In your problem,

`E_"a" = "165 kJ/mol" = "165 000 J/mol"`
`T_2 = "475 K"`
`T_1 = "455 K"`

Now, let's insert the numbers.

`ln(f_2/f_1) = E_"a"/R(1/T_1 -1/T_2)`

`ln(f_2/f_1) = ("165 000" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1")))) (1/(455 color(red)(cancel(color(black)("K")))) - 1/(475 color(red)(cancel(color(black)("K")))))`

`ln(f_2/f_1) = "19 846" × 9.254 × 10^"-5" = 1.837`

`(f_2/f_1) = e^1.837 = 6.27`