If a temperature of a reaction is raised from 300 to 320K, what factor will the reaction rate increase by [duplicate]








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[Prove that a 10-Degree Temperature Increase Doubles the Rate Constant (k), when the Activation Energy is Approximately 50 kJ/mol](/questions/66233/prove-that-a-10-degree-temperature-increase-doubles-the-rate-constant-k-when)

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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.




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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