# 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

Jun 1, 2017

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

First off, this is an approximation that assumes an activation energy of about $\text{50 kJ/mol}$. 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:

${r}_{1} \left(t\right) = {k}_{1} {\left[A\right]}^{n}$
${r}_{2} \left(t\right) = {k}_{2} {\left[A\right]}^{n}$

where $r$ is the rate in $\text{M/s}$, $k$ is the rate constant in the appropriate units, $\left[A\right]$ is the concentration of $A$ in $\text{M}$, 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:

$\ln \left({k}_{2} / {k}_{1}\right) = - {E}_{a} / R \left[\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right]$

Notice how if we write ${k}_{2} / {k}_{1}$ as a function of the rates, we obtain:

${k}_{2} / {k}_{1} = \left({r}_{2} \left(t\right) \text{/"cancel([A]^n))/(r_1(t)"/} \cancel{{\left[A\right]}^{n}}\right)$

$= \frac{{r}_{2} \left(t\right)}{{r}_{1} \left(t\right)}$,

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 $10$ $\text{K}$ temperature difference near room temperature, and an assumed activation energy of ${E}_{a} = \text{50 kJ/mol}$. 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

${e}^{\ln \left({r}_{2} \text{/} {r}_{1}\right)} = {r}_{2} / {r}_{1} = \textcolor{g r e e n}{2.014} \approx 2$,

and the reaction $\text{10 K}$ hotter ran approximately twice as fast. Alright, so we chose an appropriate activation energy.

Now, consider a change in temperature from ${10}^{\circ} \text{C}$ to ${100}^{\circ} \text{C}$, or a $\text{90 K}$ increase from $\text{283.15 K}$ to $\text{373.15 K}$.

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")]

$\implies \textcolor{b l u e}{{r}_{2} / {r}_{1}} = {e}^{5.1225}$

$=$ $\textcolor{b l u e}{\boldsymbol{\text{167.7 times as fast}}}$

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")]

$\implies \textcolor{b l u e}{{r}_{2} / {r}_{1}} = {e}^{6.2304}$

$=$ $\textcolor{b l u e}{\boldsymbol{\text{508 times as fast}}}$

(Yeah, quite different!)