anonymous
  • anonymous
A certain reaction has an activation energy of 46.39 kJ/mol. At what Kelvin temperature will the reaction proceed 5.50 times faster than it did at 299 K?
Chemistry
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SOLVED
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schrodinger
  • schrodinger
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Photon336
  • Photon336
What's the pre exponential factor for this reaction?
Cuanchi
  • Cuanchi
http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Modeling_Reaction_Kinetics/Temperature_Dependence_of_Reaction_Rates/The_Arrhenius_Law/The_Arrhenius_Law%3A_Activation_Energies you can use this formula a derivation of Arrhenius you dont need the A factor is the same at both temperatures. Then k1/k2 = 1/5 and T1= 299K ;T2=?, R=8.314 J/molK, Ea=46.39 x 10^3 J/mol ln(k1/k2)=(1/T2−1/T1)Ea/R \[\ln(\frac{ k _{1} }{ k _{2} })= (\frac{ 1 }{ T _{2} }-\frac{ 1 }{ T _{1} })\frac{ Ea }{ R }\]
Photon336
  • Photon336
@Cuanchi So we don't need A in this problem because it's the same for the particular reaction were studying? so we basically isolate A and then rewrite the Equation to solve for temperature. His question had 5.5 times faster so does this mean K2 = 5.5K1 and then substitute that into the expression Ln(5.5K1/K1) = Ln(5.5) = (rest of equation)? I mean temperature would affect our K.

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Cuanchi
  • Cuanchi
Yes your logic is correct, but if you choose k2=5.5, in the formula Ln(k1/k2)=Ln(1/5.5)!!! just be careful, if you choose k2= 5.5, T2= ?, k1=1, T1= 299K; the activation energy is given in kJ and the R= 8.314 J/molK is in joules, you have to convert the Ea to joules (x10^3). It will not affect the answer which one you assign 1 or 2 so far you keep the pairs k1,T1 and k2,T2 according to the formula. If you mix it up you will get a kelvin temperature with a negative sign (doesn't exist!!) I got ~ 328 K
Photon336
  • Photon336
Great explanation! oh... yeah... I just saw that! I assume that the incorrect Ln value, would affect our T2 value. once you get that you can probably just rearrange and solve for T2

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