# Activation energy... help appreciated?

• Activation energy... help appreciated?

Answer #1 | 23/12 2013 08:29
Ok I think the way you are writing it out is confusing you. Arrhenius' Equation: k = A e^-(Ea / RT), where k is the rate constant, A is a constant that you don't need to worry about right now, e is the symbol for exponential, Ea is activation energy, R is the gas constant, and T is temperature in Kelvin. Now the activation energy of a process is the difference between a set of reactants and the top of the energy peak the reactants have to get over in order to react and form the products. Or to take it one step further it is the difference in energy between the reactants and their transition state between them and the product. If that makes no sense to you just ignore it. So if we just set up that equation. We are presuming that the Ea for the different temperatures is constant. So: Change in k = k(2) - k(1) = A e^-(Ea / R * T(2)) - A e^-(Ea / R * T(1)) Now a little bit of algebra: ln(k(2)) - ln(k(1)) = ln A - (Ea / R * T(2)) - ln A + (Ea / R * T(1)), note that (Ea / R * T(1)) has become positive on the right because it is minus a negative. Now there is ln A - ln A on the right side. Which means these can cancel. Multiply by R to move it over to the right. And factorise to get Ea out of the fractions. What you should now have is: ( ln(k(2)) - ln(k(1)) ) * R = Ea * ( (1 / T(1) ) - (1 / T(2) ) ), if you are unsure of how I did this I would have a go at it yourself on paper to prove to yourself this is correct. Then divide through by that bunch of brackets on the right: (ln (k(2) / k(1)) * R) / ( (1 / T(1) ) - (1 / T(2) ) ) = Ea again on the left I have changed the form of the ln(k) terms. This is simply the fact that: ln (A) - ln (B) = ln (A/B). Again if you are unsure of this try it for yourself with some numbers. Otherwise it is simply: ( ( ln( k(2) ) - ln( k(1) ) ) * R ) / ( (1 / T(1) ) - (1 / T(2) ) ) = Ea. Few, lots of brackets later: Now plug the numbers you have in for k(1), T(1), k(2), T(2) and calculate Ea. I don't know if you have the answer handy but I get 79.76 kJ / mol. The whole point is to get rid of the thing you don't know in this case A, so you can calculate Ea. If you do this on paper it will probably be a LOT easier to see. On computer using brackets and writing formulae is not ideal.
Positive: 53 %
Answer #2 | 23/12 2013 08:29
what is all this ln(kr (2) = Ea (1 - 1) (kr (1) R ( T(1) T(2) ) kr (1) = 4.0 x 10-4 dm3 mol-1 s-1 at 297 K kr(2) = 15.5 x10-4 dm3 mol-1 s-1 at 310K ??????? ******** the Arrhenius equation is ..k = Ao exp(-Ea/RT) so for 2 conditions.. 1 and 2 .. k1 = Ao exp(-Ea/RT1) .. k2 = Ao exp(-Ea/RT2) rearranging both .. k1 / exp(-Ea/RT1) = Ao .. k2 / exp(-Ea/RT2) = Ao since they both = Ao, they must = each other .. k1 / exp(-Ea/RT1) = k2 / exp(-Ea/RT2) rearranging .. k1 / k2 = exp(-Ea/RT1) / exp(-Ea/RT2) since a^b / a^c = a^(b-c) .. k1 / k2 = exp( (-Ea/RT1) - (-Ea/RT2) ) simplifying .. k1 / k2 = exp( (Ea/RT2) - (Ea/RT1) ) natural log both sides .. ln(k1 / k2) = (Ea/RT2) - (Ea/RT1) rearranging .. ln(k1 / k2) = (Ea/R) x (1/T2 - 1/T1).... . <--- look familiar? rearranging one last time .. Ea = R x ln(k1 / k2) / (1/T2 - 1/T1) make sure your units are consistent when YOU finish the problem ... R = 8.314 J/molK ... T must be in K
Positive: 47 %

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