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In general, if the temperature of a reaction is raised from 300k to 320k, the reaction rate will increase by how much?

So the first thing I do is try out the Arrhenius equation:

$k=Ae^{\frac{-E_a}{RT}}$

so I ignore the activation energy and A because they will cancel out to get the following:

$k=e^{\frac{-1}{300}}$

$k=e^{\frac{-1}{320}}$

and I divide the two k's to get 1.000208

but that can't be right.

It seems to me that I do not know how to answer this question and would like some assistance

I do not want to know a proof of anything but rather to know how to execute methods in solving this problem.

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  • $\begingroup$ Your Arrhenius equation lost the negative sign on the exponent. $\endgroup$ – Tyberius Apr 11 '17 at 18:01
  • $\begingroup$ @Tyberius fixed, also not a duplicate, the link that you provided was about the correlation of the commonly accepted idea that if we increase temperature by 10 degrees then we double the K, but in this question rather than doubling, we are trying to calculate the ratio. $\endgroup$ – John Rawls Apr 11 '17 at 18:15
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    $\begingroup$ While that is true, I would argue that the other question effectively shows that this can't be answered in general. You can pick one or maybe a few different activation energies and see what you get, but there isn't a general answer without being given a specific activation energy. If you went by the general rule of thumb, your answer would be 4, since you increase the temperature by 20 degrees, so the rate at 320 would be 4 times the rate at 300 (again, just by that estimation scheme). $\endgroup$ – Tyberius Apr 11 '17 at 18:19
  • $\begingroup$ Agreed with out activation energy it vague and could vary by orders of magnitude. The exponents' use is the activation energy. $\endgroup$ – LinkBerest Apr 11 '17 at 23:23
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The amount of increase is dependent on the activation energy. For a very low activation energy, there will be almost no change (this is what you obtained by removing $E_a$ and $T$). For a high activation energy, the change will be quite large. For example, with an activation energy of $110\mathrm{\frac{kJ}{mol}}$, you the rate will increase by about $16$ times. $$\frac{\exp({\frac{-E_a}{R*320}})}{\exp({\frac{-E_a}{R*300}})}=\frac{\exp({\frac{-111000\text{J}}{8.314*320}})}{\exp({\frac{-111000\text{J}}{8.314*300}})}=\frac{7.6\times10^{-19}}{4.7\times10^{-20}}=16.14$$

So there really isn't an "in general" answer. However, many text books give a crude estimate that the rate will double when the temperature is increased by $10$. This isn't in general the case, as it depends on the temperature range (a change from $300$ to $320$ is more significant that a change from $600$ to $620$) and the the value of the activation energy.

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