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There is a rule of thumb that says that the rate of many room-temperature chemical reactions approximately doubles when the temperature is increased by 10°C. Calculate the activation energy for a reaction that exactly conforms to this relationship.

The only thing that comes to mind for me it's the Arrhenius equation, but I am not sure if that is the correct way of doing it since I am given so little information. How can I solve this problem ?. Thanks a lot in advance!

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You would use the Arrhenius equation. You are given that the reaction should double in rate when you increase from room temperature (for convenience, I'll say this is 300K) to when the temperature is increased by 10 degrees. So you can equate the Arrhenius equation at $T=310\mathrm{K}$ to double the Arrhenius equation at $T=300\mathrm{K}$ and solve for Ea.

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  • $\begingroup$ A general formula for the activation energy can be derived but I don't think we can achieve a specific value $\endgroup$ – Tan Yong Boon Sep 12 '17 at 7:02
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    $\begingroup$ @TanYongBoon That's not true for specific temperature. $\endgroup$ – Mithoron Sep 12 '17 at 17:50
  • $\begingroup$ @Mithoron There isn't a specific temperature in this case, is there? As in the assumption made here of the increase being from T=300 K to T=310 K is not correct. Because if u take it as T=290 K to T=300 K. The value obtained of the activation energy would differ. $\endgroup$ – Tan Yong Boon Sep 12 '17 at 23:25
  • $\begingroup$ @Tan Yong Boon you are correct in that room temperature does not have an exact meaning and can basically describe anything between 293K and 300K. However, I think for this problem, as it seems to be mainly conceptual rather looking for a specific value (ie about what activation energy will be double in rate with a 10 degree temperature increase) using 300K should be fine and works well because it is a nic round number. $\endgroup$ – Tyberius Sep 13 '17 at 0:24
  • $\begingroup$ Ok then. Since the question asker mentioned "room-temperature chemical reactions", then I guess it should be fine. Just wanted to clarify. $\endgroup$ – Tan Yong Boon Sep 13 '17 at 0:41

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