Question

The activation energy, Ea. Can be determined from the Arrhenius equation by measuring the rate constants K1 and K2 at 2 different Temperatures T1 and T2

2.303log(k2/k1) = (Ea/R) x ( (T2-T1)/ (T1*T2) ) where T the gas constant is 8.31 mol-1 K-1



Assume that the rate of reaction is ten times faster at 32 than it is at 22. What is the Activation Energy Ea ?

Answer 1

I think this is a Chemistry problem, not a (specifically) Mathematics problem...

Answer 2

I feel its a matter of solving for Ea, where T2 = 32 and T1 = 22, and K1 = 10 X k2,

Answer 3

http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Modeling_Reaction_Kinetics/Temperature_Dependence_of_Reaction_Rates/The_Arrhenius_Law/Arrhenius_Equation

If it helps.

Answer 4

but I cant rearrange it for Ea

Answer 5

Does the link help?

Answer 6

looking at it now, but even then I wouldn't know how to re arrange for Ea lol

Answer 7

You never know till you try. :3

Answer 8

\[2.303\log(k_2/k_1) = (E_a/R) \times \frac{T_2-T_1}{T_1\times T_2}\\
2.303\log(k_2/k_1) \times R= E_a \times \frac{T_2-T_1}{T_1\times T_2}\\
2.303\log(k_2/k_1) \times R\times\frac{T_1\times T_2}{T_2-T_1}= E_a\]