Question

The activation energy of a certain reaction is 49.4 kJ/mol. At 20 degrees C, the rate constant is .0130 s^-1. What is the rate constant at 100 degree C?

Answer 1

The rate constant is `"1.00 s"^"-1"`.

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,

`k_1 = "0.0130 s"^"-1"`
`E_"a" = "49.4 kJ/mol" = "49 400 J/mol"`
`R = "8.314 J·K"^"-1""mol"^"-1"`
`T_2 = "100 °C" = "373.15 K"`
`T_1 = "20 °" = "293.15 K"`

Now, let's insert the numbers.

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

`ln(k_2/k_1) = ("49 400" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1")))) (1/(293.15 color(red)(cancel(color(black)("K")))) - 1/(373.15 color(red)(cancel(color(black)("K")))))`

`ln(k_2/k_1) = 5942× 7.313 × 10^"-4" = 4.346`

`(k_2/k_1) = e^4.346 = 77.14`

`k_2 = 77.14k_1 = 77.14 × "0.0130 s"^"-1" = "1.00 s"^"-1"`