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A widely used rule-of-thumb for the temperature dependence of a reaction rate is that a ten degree rise in the temperature approximately doubles the rate. This is not generally true, especially when a strong covalent bond must be broken. For a reaction that does show this behavior, what would the activation energy be?

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What activation energy would lead to a doubling in the reaction rate for a 10 degree increase in temperature, using the Arrhenius equation?

You’re right it is not generally true. Let’s set up an equation to solve the problem.

The Arrhenius equation is $$k=A\operatorname{e}^{\frac{-E_\text{a}}{RT}}$$

In this case

$$2=\frac{A\operatorname{e}^{\frac{-E_\text{a}}{R\cdot(T + 10\ \mathrm{K})}}}{A\operatorname{e}^{\frac{-E_\text{a}}{RT}}}$$

or

$$\ln 2 = \frac{ - E_\text{a}}{R\left( T + 10\ \text{K} \right)} - \frac{ - E_\text{a}}{RT}$$

let’s pick a temperature around room temperature, 300 K is a nice even number, but you can re-solve the problem with any temperature you like. $$0.69 = \frac{ - E_\text{a}}{R \cdot 310\ \text{K} } - \frac{ - E_\text{a}}{R \cdot 300\ \text{K} }$$ Using $R=8.31\ \mathrm{J/(K\ mol)}$ will produce an answer in J/mol. Solving we find $$E_\text{a} = 53\,568.49\ \mathrm{J/mol}$$ or, $$53.6\ \mathrm{kJ/mol}~~(12.8\ \mathrm{kcal/mol})$$

So around room temperature a reaction that has an activation energy around 54 kJ/mol (or 13 kcal/mol) would show a doubling in rate for a 10 kelvin increase in temperature.

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  • $\begingroup$ "So around room temperature a reaction that has an activation energy around 54 kJ/m (or 13 kcal/m) would show a doubling in rate for a 10o Kelvin increase in temperature." Does reaction order matter here? $\endgroup$ – Dissenter Dec 28 '14 at 21:44
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    $\begingroup$ @Dissenter No, reaction order doesn't matter. $\endgroup$ – ron Dec 28 '14 at 22:03

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