Arrhenius Equation

[mississippichem]

[math]k=Ae^{\frac{-E_a}{RT}} [/math]

 

-where A is the collision frequency factor

-e is the natural log base

-R is the gas constant [math](8.314510 \times 10^{-3} kJ \ K^{-1} \ mol^{-1})[/math]

-k is the rate constant for the reaction

-Ea is the activation energy

 

Is there a way to calculate the collision frequency factor, A, without just rearranging [assuming activation energy is unknown]?

The collision frequency factor is supposed to be specific for every reaction, so there must be a seperate way to calculate it for it to be of any use.

[Mr Skeptic]

Well you could try to estimate it, or you could just measure it when fitting a curve.

[John Cuthber]

You can measure that rate at more than one temperature then solve the equations.

On the other hand, you can calculate how often molecules bump into eachother from kinetic theory. The second approach gives an upper bound to the reaction rate since it assumes every collision is successful in producing a reaction.

[mississippichem]

or you could just measure it when fitting a curve.

 

Yeah, plotting [math]ln(k)[/math] versus [math]\frac{1}{T}[/math] gives [math]E_a[/math] as the gradient of the resultant line. Then one could solve for [math]A[/math] i guess.

 

You can measure that rate at more than one temperature then solve the equations.

 

Oh, so find [math]k[/math] at two different temperatures, then rearrange to [math]lnk=-(\frac{E_a}{RT})+lnA[/math]. Subtracting the equations from two diiferent temperatures would give:

 

[math]ln \frac{k_2 }{k_1 } = \frac{-E_a }{R} \left[\frac{1}{T_2 } - \frac{1}{T_1 }\right][/math] which conveniently eliminates the [math]A[/math] term. Cool

 

Thanks for your help guys, you set me on the right course.

 

Anyone know a different mathematical treatment to arrive at activation energy?

[cypress]

I recall that there were a couple from my days doing reaction kinetics for reactor designs. Perry's Handbook has a section on this and I think my old textbook does too. I'll have a look on Monday.

[mississippichem]

[math]k= \frac{k_b T}{h} e^\frac{ -\Delta G_a }{RT} [/math]

 

Forgot about the Erying equation, similar to the Arrenhius but gives activation energy in terms of Gibb's energy. Answered my own question, woops.