Rate of a reaction triples when the temperature is increased from ...

[sadsdas]

Rate of a particular reaction triples when the temperature is increased from 25C to 35C. Calculate the activation energy for this reaction?
How to approach the problem when there is no rate constant? What formula to use?

[Yolanda]

anytime you read "activation energy" and "temperature" in the same sentence you should immediately think "Arrhenius equation"
.. k = A x exp(-Ea / RT)

and since you have "rate" not "k" we can do this
.. rate = k x [A]^n.... .  <====.. typical rate equation
so that
.. rate1 = k1 x [A1]^n
.. rate2 = k2 x [A2]^n
and if we assume we start with the same concentration of [A] and just vary "T" and the problem indicates, then [A1]^n = [A2]^n = [A]^n.. so that
.. rate1 / k1 = [A]^n = rate2 / k2
rearranging
.. rate1 / rate2 = k1 / k2
i.e..
.. I can substitute (rate1 / rate2) for (k1 / k2)...  Capice?

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going back to the Arrhenius equation.. since we have 2 conditions
.. k1 = A exp(-Ea / RT1)
.. k2 = A exp(-Ea / RT2)
rearranging
.. k1 / exp(-Ea / RT1) = A
.. k2 / exp(-Ea / RT2) = A
since the both = A, they must = each other
.. k1 / exp(-Ea / RT1) = k2 / exp(-Ea / RT2)
rearranging
.. k1 / k2 = exp(-Ea / RT1) / exp(-Ea / RT2)

now.. since a^b / a^c = a^(b-c)
.. k1 / k2 = exp( (-Ea/RT1) - (-Ea/RT2) )
taking ln of both sides and rearranging a bit..
.. ln(k1 / k2) = (-Ea/R) x (1/T1 - 1/T2)

rearranging one last time and subbing in that rate1 / rate2 for k1 / k2
.. Ea = (-R) x ln(rate1/rate2) / (1/T1 - 1/T2)

and finally... considering rate2 = 3x rate1
.. Ea = (-8.314 J/molK) x ln(1 / 3) / (1/295K - 1/308K) x (1kJ / 1000J) = +63.8 kJ/mol

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do you understand all of this?