Prove that a 10-Degree Temperature Increase Doubles the Rate Constant (k), when the Activation Energy is Approximately 50 kJ/mol

I read that increasing the temperature by $10~^\circ\mathrm C$ will double the rate constant ($k$), when the activation energy for the reaction is relatively close to $50~\mathrm{kJ/mol}$.


However, no matter how hard I try to wrap my head around it (through graphical examples that require plotting the natural logarithm of $k$ vs the inverse of temperature), I just can't grasp this so called "rule of thumb". 


As a result, I tried proving this statement through mathematical expressions:


(Note that the activation energy was evaluated for $50\,000$ because it needs to be expressed in $\mathrm{J/mol}$ to be used in Arrhenius' equation)


*Prove that...*



> 
> A rise of temperature of $10\ \mathrm{^\circ C}$ will double the rate constant, when the activation energy for the reaction is approximately $50\ \mathrm{kJ/mol}$.
> 
> 
> 


**Proof starts here -----------------------------**


*For:* $\Delta T = T\_2 - T\_1$ ($T\_2$ is the final temperature, and $T\_1$ the initial temperature)


*When:* $\Delta T = 10 \rightarrow T\_2 - T\_1 = 10$


*It follows that:* The ratio between $k\_2$ (at $T\_2$) and $k\_1$ (at $T\_1$) will equal $2$, owing to the doubling of the rate constant ($k$):


$$
{k\_2\over k\_1} = 2
$$


*Employing Arrhenius' equation:*


$$
{k\_2\over k\_1} = {A\mathrm e^{\left({-E\_\mathrm a\over RT\_2}\right)}\over A\mathrm e^{\left({-E\_\mathrm a\over RT\_1}\right)}}
 = \mathrm e^{\left[{-E\_\mathrm a\over R}\left({1\over T\_2} – {1\over T\_1}\right)\right]}
 = 2
$$


*Or simply:*


$$
{k\_2\over k\_1} = \exp{\!\left({-E\_\mathrm a\over R} \cdot \Delta{1\over T} \right)} = 2
$$


*So:*


$$
\exp{\!\left({-E\_\mathrm a\over R} \cdot \Delta{1\over T} \right)} = 2
$$


**This might be the part where I messed up -------------------------**


$\Delta{1\over T}$ can be set in terms of $\Delta T$:


$$
\Delta{1\over T} = \left({1\over T\_2}-{1\over T\_1}\right)
 = \left({T\_1\over T\_1T\_2}-{T\_2\over T\_1T\_2}\right)
 = \left({T\_1-T\_2\over T\_1T\_2}\right)
 = \left({-\Delta T\over T\_1T\_2}\right)
$$


Since Arrhenius' equation above holds true for $\Delta T = 10$ ...


And since $\Delta{1\over T}$ can be expressed as $-\Delta T \over T\_1T\_2$ ...


Then, the rate constant ($k$) will double when:


$$
\Delta{1\over T} = {-10\over T\_1T\_2}
$$


Substituting into Arrhenius' equation:


$$
\exp{\!\left({-E\_\mathrm a\over R}\cdot{-10\over T\_1T\_2}\right)} = 2
$$


Simplifying:


$$
\begin{align}
\exp{\left({10E\_\mathrm a\over RT\_1T\_2}\right)} &= 2 \\
{10E\_\mathrm a\over RT\_1T\_2} &= \ln{2} \\
{1\over T\_1 T\_2} &= {R\cdot\ln{2}\over 10E\_\mathrm a} \\
T\_1 T\_2 &= {10E\_\mathrm a\over R\cdot\ln{2}}
\end{align}
$$


From $T\_2 - T\_1 = 10$ we can solve for $T\_2$ as $T\_2 = T\_1 + 10$. Substituting above:


$$
\begin{align}
T\_1 (T\_1 + 10) &= {10E\_\mathrm a\over R\cdot\ln{2}} \\
T\_1^2 + 10T\_1 - {10E\_\mathrm a\over R\cdot\ln{2}} &= 0
\end{align}
$$


For $E\_\mathrm a = 50\,000$:


$$
T\_1^2 + 10T\_1 - {500\,000\over R\cdot\ln{2}} = 0
$$


Solving the quadratic equation:


$$
\begin{align}
T\_1 &= 290~\mathrm K \\
T\_2 &= 300~\mathrm K \\
\Delta T &= 10~\mathrm K
\end{align}
$$


I don't think this proof is correct because substituting any value for activation energy ($E\_\mathrm a$) onto the quadratic equation will produce different values for $T\_1$, but that doesn't tell me anything.


What did I do wrong?
Can this be proved any other way? 




**Your method and your mathematics seem perfectly fine, and your calculated result is correct.** You are also quite correct that the result will change depending on the value of $E\_\mathrm a$ that you choose.


The problem you're running into is the assumption that the stated rule of thumb holds *exactly*, regardless of temperature. ***It doesn't.***




---


I would set up the problem as follows. Consider the ratio $\rho$ between the reaction occurring at some Kelvin temperature $T$, and at $T+10$:


$$
\rho = {e^{-E\_\mathrm a\over R\,\left(T+10\right)}\over
 e^{-E\_\mathrm a\over RT}}
 = e^{{E\_\mathrm a\over R}\left({1\over T} - {1\over T+10}\right)}
 = e^{{E\_\mathrm a\over R}\cdot {10\over{T^2+10T}}}
$$


Plotting $\rho$ as a function of $T$ for $E\_\mathrm a = 50~\mathrm{kJ\over mol}$, it becomes immediately clear that the rule of thumb does not hold over a very wide temperature range:


[![Rate ratio versus temp](https://i.stack.imgur.com/MFJaq.png)](https://i.stack.imgur.com/MFJaq.png)


In fact, it only holds strictly for one specific temperature, which is why your math led you to a single temperature as your answer. By my calculations, using a value of $8.3144598~\mathrm{J\over mol\, K}$ for $R$, that temperature is $289.56~K$, **which is equal to your $T\_1=290~K$ to within three significant figures.**


The gray box in the figure marks where the rule holds to within $20\%$ – that is, where $\rho$ falls between $1.6$ and $2.4$. This corresponds to a temperature range of $257~\mathrm K$ to $353~\mathrm K$.


 12