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

Question

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?

Answer 1

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.

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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:

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 $353~\mathrm K$.