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A rule of thumb in chemistry is that a reaction doubles in rate for every 10 °C increase in temperature. Considering the impact of a 10 °C increase in global temperature, would the same rule apply to kinetic weather processes – for example, would the average wind speed around the world double?

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    $\begingroup$ Weather is complex and, as ron guessed, I am confident it is not straight and simple as you said. Counter example is NO oxidation by O2. This reaction can be assumed as trimolecular (very rare) and its kinetic constant decrease with temperature. $\endgroup$ – jlandercy May 23 '14 at 21:00
  • $\begingroup$ Weather is a physical not a chemical process so simple rules about chemical reactions are not going to work. Besides the temperature difference at different points on the earth's surface is >100 degrees and we don't see slower wind speeds, for example, in antarctica than in Singapore. $\endgroup$ – matt_black Mar 22 '15 at 11:30
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That is a good rule of thumb. It is based on the Arrhenius equation $$k = A {\operatorname{e}^{\frac{-{E_{\text{act}}}}{RT}}}$$ You'll find that the "temperature doubles every 10 degrees" rule only applies if the following assumptions are valid

  • The reaction must obey the Arrhenius equation
    • it must have an entropic pre-term based on collision probability
    • the reaction must be an activated process (e.g. it must have an activation energy)

The Arrhenius equation works well for simple unimolecular reactions like the ring opening of cyclobutene to buta-1,3-diene.

Now you'll need to ask a weather physicist or a meteorologist if weather can be treated as an activated process. My guess is that the answer is "no", but that's just a guess. It seems to me that there is no minimum energy required (activation energy) for the rain to fall or for the wind to blow.

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  • $\begingroup$ Thanks - I appreciate that different processes may or may not have an "activation energy". Formation of a hurricane may be such a process. $\endgroup$ – iad22agp May 27 '14 at 20:51
  • $\begingroup$ This comment is correct but very incomplete, because the "rule of thumb" only applies to a small subset of reactions that obey the Arrhenius equation. In particular for $k=Ae^{-E/RT}$ and $2k=Ae^{-E/R(T+10)}$, you can solve the equations to get $R \ln{2} = -E_a ((T+10)^{-1}-T^{-1})$. This equation is a constraint on the energy of activation. At 298 K, that means (if I did my math right), the energy of activation has to be 52 kJ/mol, or else the rule doesn't work. $\endgroup$ – Curt F. Mar 21 '15 at 7:20

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