Question

If the temperature coefficient `alpha` for a certain reaction is `2.5`, then if the reaction was run at `283^@ "C"` and `293^@ "C"`, what is the activation energy in `"kJ/mol"`?

Answer 1

I got about:

`E_a = 6.5xx10^3` `"kJ/mol"`

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From Wikipedia, the definition of the temperature coefficient `alpha`, in regards to the rate constant `k` and temperature `T`, is:

`(dk)/(k) = alphadT`

If we integrate this from state 1 to state 2 on the left side, and `T_1` to `T_2` on the right side:

`int_((1))^((2)) 1/k dk = int_(T_1)^(T_2) alphadT`

When we assume that the temperature coefficient stays constant across this small temperature range, we can pull `alpha` out of the integral and obtain (noting that the integral of `1/x` is `lnx`):

`ln(k_2) - ln(k_1)`

`= color(green)(ln(k_2/k_1)) = alpha(T_2 - T_1)`

`= (2.5)(293^@ "C" - 283^@ "C")`

`= color(green)(25)`

where we used the fact that intervals in the celsius temperature scale are equal to intervals on the Kelvin temperature scale.

Thus, we have indirectly calculated `ln(k_2/k_1)`. Recall that this shows up in the form of the Arrhenius equation that demonstrates the temperature dependence of the rate constant:

`ln(k_2/k_1) = -E_a/R[1/T_2 - 1/T_1]`

Therefore, we can now algebraically solve for and calculate the activation energy `E_a`, in `"kJ/mol"`, as:

`color(blue)(E_a) = -(Rln(k_2/k_1))/(1/T_2 - 1/T_1)`

`= -(Rln(k_2/k_1))/((T_1 - T_2)/(T_1T_2))`

`= -Rln(k_2/k_1)[(T_1T_2)/(T_1 - T_2)]`

`= -("0.008314472 kJ/mol"cdot"K")(25)[(("556.15 K")("566.15 K"))/(556.15 - 5"66.15 K")]`

`= color(blue)(6.5_(45)xx10^3)` `color(blue)("kJ/mol")`

where the subscripts indicate digits past the last significant figure.

This means that the reactants have about a `"6500-kJ"` energy barrier in order for the reaction to proceed successfully.