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March 14th, 2018, 04:34 AM   #1
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Activation energy

I'm suppose to find the activation energy based on this given information:

When double the temperature the rate constant increases with 2.11%. The Arrhenius equation is valid.

I did this:
k1 is the rate constant when T1=T
k2=1.0211k1 is the rate constant when T2=2T

Then I used the Arrhenius for both k-values and divided the two expressions and got

k1/k2=1/1.0211=exp(E/R(1/2T-1/T)=exp(E/R(-1/2T)) but then T is unknown and I don't know how to come up with a solution for E.
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March 14th, 2018, 08:36 AM   #2
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Quote:
Originally Posted by matteamanda View Post
I'm suppose to find the activation energy based on this given information:

When double the temperature the rate constant increases with 2.11%. The Arrhenius equation is valid.

I did this:
k1 is the rate constant when T1=T
k2=1.0211k1 is the rate constant when T2=2T

Then I used the Arrhenius for both k-values and divided the two expressions and got

k1/k2=1/1.0211=exp(E/R(1/2T-1/T)=exp(E/R(-1/2T)) but then T is unknown and I don't know how to come up with a solution for E.
Looked up the equation you referenced in your post ...



$k_1 = Ae^{-\dfrac{E}{RT_1}}$

$k_2 = 1.0211 \cdot k_1 = Ae^{-\dfrac{E}{R \cdot 2T_1}}$

$\dfrac{k_2}{k_1} = 1.0211 = e^{-\dfrac{E}{R \cdot 2T_1} + \dfrac{E}{RT_1}} = e^{\dfrac{E}{R \cdot 2T_1}}$

$\log(1.0211) = \dfrac{E}{R \cdot 2T_1}$

$R \cdot 2T_1 \cdot \log(1.0211) = E$

that's as far as you can go ... obviously values for $R$ and $T_1$ are required to determine a numerical value for $E$.
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March 14th, 2018, 11:14 AM   #3
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Thanks! I thought so, maybe there's something more in the question I don't understand. But thanks a lot for for taking your time to answer
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