Question

A general rule of thumb is that a reaction `"10 K"` higher should be twice as fast. Predict how much faster a reaction is when the temperature is changed from `10^@"C"` to `100^@ "C"`?

NOTE: It does depend on the temperature range chosen. This is nontrivial.
- Truong-Son

Answer 1

About `168` times as fast, for your temperature range.

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First off, this is an approximation that assumes an activation energy of about `"50 kJ/mol"`. Let's show how that turns out to give such a ratio of the rates.

We write the general rate law for two trials, each at different temperatures and fixed concentration of `A`, for a one-reactant reaction as:

`r_1(t) = k_1[A]^n`
`r_2(t) = k_2[A]^n`

where `r` is the rate in `"M/s"`, `k` is the rate constant in the appropriate units, `[A]` is the concentration of `A` in `"M"`, and `n` is the order of the reaction with respect to `A` (whatever it may be).

The Arrhenius equation relates the rate constants `k` to each temperature `T`. You may have seen that the log form of the Arrhenius equation is:

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

Notice how if we write `k_2/k_1` as a function of the rates, we obtain:

`k_2/k_1 = (r_2(t)"/"cancel([A]^n))/(r_1(t)"/"cancel([A]^n))`

`= (r_2(t))/(r_1(t))`,

so what you should note is that the ratio of the rate constants is the ratio of the rates. That means `k_2/k_1 = r_2/r_1`.

Now, consider a `10` `"K"` temperature difference near room temperature, and an assumed activation energy of `E_a = "50 kJ/mol"`. Then:

`ln(r_2/r_1) = -("50 kJ/mol" xx "1000 J"/"1 kJ")/("8.314472 J/mol"cdot"K")[1/("298.15 K") - 1/("288.15 K")]`

`= 0.6999`

Therefore, the ratio of the rates is

`e^(ln(r_2"/"r_1)) = r_2/r_1 = color(green)(2.014) ~~ 2`,

and the reaction `"10 K"` hotter ran approximately twice as fast. Alright, so we chose an appropriate activation energy.

Now, consider a change in temperature from `10^@ "C"` to `100^@ "C"`, or a `"90 K"` increase from `"283.15 K"` to `"373.15 K"`.

We solve for `r_2/r_1` to see the change in rate for the hotter reaction.

`ln(r_2/r_1) = -("50 kJ/mol" xx "1000 J"/"1 kJ")/("8.314472 J/mol"cdot"K")[1/("373.15 K") - 1/("283.15 K")]`

`=> color(blue)(r_2/r_1) = e^(5.1225)`

`=` `color(blue)(bb("167.7 times as fast"))`

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Note: here is the same calculation as before, but for a temperature range you are not asked for, to check if the temperature range you are asked for actually matters.

`ln(r_2/r_1) = -("50 kJ/mol" xx "1000 J"/"1 kJ")/("8.314472 J/mol"cdot"K")[1/("343.15 K") - 1/("253.15 K")]`

`=> color(blue)(r_2/r_1) = e^(6.2304)`

`=` `color(blue)(bb("508 times as fast"))`

(Yeah, quite different!)