Activation energy for a chemical reaction

[vladimir69]

hi,
could some one please verify what i have done here or tell me where i went wrong.

suppose the activation energy for a chemical reaction is E*, i have to work out how the rate of the forward reaction depends on temperature T and concentration. (note: i am given the Boltzmann distribution
$$c(E)=c_{0} \exp(\frac{E}{nRT})$$ )
n = number of moles
R = gas constant
E = gravitational potential energy
$$c_{0}$$ = initial concentration
c = c(E) = concentration

i made the following assumptions in my calculations
- the reaction is a unimolecular decomposition reaction ( ie A -> B + C )
- the reaction is first order
- and by analogy E=E* (not so sure about this assumption)

let the rate of the forward reaction for the reaction A -> B + C be
$$r_{f}$$
so
$$r_{f} = -\frac{d[A]}{dt}$$
$$=k[A]$$ where k is the rate constant

therefore
$$r_{f} = kc_{0}\exp(\frac{E*}{nRT})$$

i also need to put some plausible values for E* and comment on the quantitative results - i am not sure how to go about this and not sure what plausible values for E* are

thanks for your time

 

[Clausius2]

hi,
could some one please verify what i have done here or tell me where i went wrong.
.....
- and by analogy E=E* (not so sure about this assumption)

let the rate of the forward reaction for the reaction A -> B + C be
$$r_{f}$$
so
$$r_{f} = -\frac{d[A]}{dt}$$
$$=k[A]$$ where k is the rate constant

therefore
$$r_{f} = kc_{0}\exp(\frac{E*}{nRT})$$

Why E=E*?. And why do you put a 1/n in the exponential factor?
Which are the dimensions of your activation energy needed?. And, is the same "k" in both last equations?.

 

[altered-gravity]

Hi,

Most of the chemical reactions have activation energies around 10-100 kcal/mol.

If you´re assuming that that´s a first order reaction the concentration dependence is set to: $$v=k*[A]^1$$

The temperature dependence is given by the Arrhenius expression:
$$k=C*exp(\frac{-E*}{RT})$$
where C is the preexponential factor, that is different for each reaction.

The preexponential factor can be calculated theoretically aplying models such as collision theory or activated complex theory. You can take it from experimental data too.

So you can write the total expression for the rate of the reaction like this:
$$v=C*exp(\frac{-E*}{RT})*[A]$$

The Boltzman expression is used in collison theory to obtain the distribution of energy of the sample. It gives you (for each value of energy) the amount of particles that have that energy. So introducing kinetic energy (not gravitational) you can get the amount of molecules that are able to collide with enough energy to react.

Sorry if i´m not helpful for you

 

[vladimir69]

hey thanks for the replies
for clausius2, the 1/n comes from the ideal gas equation PV=nRT
n = number of moles
as opposed to the ideal gas equation PV=NkT where N is the number of molecules

and yes k is the same for each expression that uses k

 

[Clausius2]

Hi,


The temperature dependence is given by the Arrhenius expression:
$$k=C*exp(\frac{-E*}{RT})$$
where C is the preexponential factor, that is different for each reaction.


So you can write the total expression for the rate of the reaction like this:
$$v=C*exp(\frac{-E*}{RT})*[A]$$

This is the correct formulation. You have not cleared up enough E* dimensions. And note you have equalled r_f two times with different expresions for k. I absolutely agree with altered_gravity.

Que pasa, De donde eres?. Ya no me acuerdo de nada de esto de la energía de activación.

 

[vladimir69]

clausius2 - i did use 2 different expressions for rf but just so captain obvious didnt come to the rescue i didnt include them. i don't think the expression which altered-gravity kindly provided did what i wanted: which was to get a relationship between rate and temperature and concentration from the Boltzmann distribution. of course i could just pluck the formula from one of my chemistry past exam papers but i think the point of the question was to show how the Boltzmann distribution can be applied to a variety of different situations. hope this makes things clearer for you

 

[altered-gravity]

I´m sorry, i´ve forgotten everything about this subject, perhaps i´m not able to help you. Anyway let´s try, let me see...

Maxwell_Boltzmann distributions of energy (or velocity) are used in molecular kietics models as they give you info about molecular average energy. Let´s try with collision model, it´s the simplest (but not good)

You proposed an unimolecular reaction A->B+C. Lindemann theory proposes this mechanism for unimolecular reactions:

$$A+M \rightleftharpoons A* + M$$
$$A* \rightarrow B+C$$

Theese are two reaction steps with three different reaction rates. Let´s propose a bimolecular reaction:

A+A->B

Here is the Maxwell-Boltzmann velocity distribution:
$$f(v)=4\pi(\frac{m}{2 \pi k T})^\frac{3}{2} v^2 exp(\frac{-m v^2}{2kT})$$

where m is the mass of the molecule
v is the velocity

So the average velocity is:
$$\sqrt{\frac{8kT}{\pi m_A}}$$

From this expression you can deduce the rate of collisions between two systems (A and A in this case). If you want to know how to deduce it, post it. This is the rate:

$$Z=\sqrt{2} \pi (d_A)^2 (\frac{8 k T}{\pi \mu_A})^\frac{1}{2}*[A]^2$$

$$d_A$$ is the effective collisional diameter of the molecule
[A] is the concentration
$$\mu_A$$ is the reduced mass of the system A - A

This is not the rate of reaction still, only collisions upon an energy E* make reaction. So you use the Maxwell-Boltzmann energy distribution (supposed that all the energy is kinetic):

$$\frac{n(E>E*)}{n_0}=exp(\frac{-E*}{RT})$$

n(E>E*) is the number of molecules with equal or more kinetic energy than E*

You multiply by this exponential factor and you get the rate of the reaction:

$$v=\sqrt{2} \pi (d_A)^2 (\frac{8 k T}{\pi \mu_A})^\frac{1}{2}*[A]^2*exp(\frac{-E*}{RT})$$

So the preexponential factor of the Arrhenius expression is:
$$C=\sqrt{2} \pi (d_A)^2 (\frac{8 k T}{\pi \mu_A})^\frac{1}{2}*exp(\frac{-E*}{RT})$$

You´d say: ¿why does preexponential factor depend on temperature? That dependence is very little as it is divided by the reduced mass and multiplied by Boltzmann constant.

Sorry if there are mistakes in expression (I´m new using laTex)

 

[altered-gravity]

Que pasa, De donde eres?. Ya no me acuerdo de nada de esto de la energía de activación.

De Vitoria
Un saludo!

 

[vladimir69]

holy moley
that was a mouth full, let me say though i do appreciate your efforts.
i don't think the answer needs to be that complicated, i just need something similar to my first post - not a heap of equations i have never seen before and will probably never see again (hopefully)
so if it all possible
-few simplifiying assumptions
-from Boltzmann equation and *simple* relationships like rf=-d[A]/dt and rf=k[A] say
-out pops our relationship between reaction rate, temperature and concentration
hopefully you can see what i am trying to get at

thanks again for your time,
vladimir

 

[altered-gravity]

oopss!
Sure you´re right. Forget it, today is not my day!

Well, at least I reviewed my old notes (very old!) Je Je!

 

[Clausius2]

Hi Vladimir. Sorry for not being you very helpful. Anyway, why do you think it will exist a relation between the potential E coming from Boltzmann dist. and Activation Energy?. If I've not read wrong your last post, E is something like an electric potential, isn't it?.

 

[vladimir69]

Hi Vladimir. Sorry for not being you very helpful. Anyway, why do you think it will exist a relation between the potential E coming from Boltzmann dist. and Activation Energy?. If I've not read wrong your last post, E is something like an electric potential, isn't it?.

i am currently working on "similarities" in physics, in particular, applications of the Boltzmann distribution to a variety of situations, one of which being the rate of a chemistry reaction. my understanding has it that the Boltzmann distribution can be applied to very many systems in equilibria of different kinds

 

[timmy1234]

problem

@altered-gravity

if one computes the integral from E to infinity using the Boltzmann equation on gets what you wrote:

$$\frac{n(E>E*)}{n_0}=exp(\frac{-E*}{RT})$$

however, if one takes (which i think in this case one has to) the maxwell-boltzmann distribution given by:

C * sqrt(E) * exp(-E/kt)

the integral from E to infinity is much more messy.

any thoughts on how one still gets exp(-E/kt) as the fraction of molecules having enough energy to react?

 

[Gokul43201]

(I don't think altered gravity is likely to respond)

Doing, the integral, I'm able to recover the correct exponential form under suitable conditions, but I have a troublesome prefactor:

$$f(E>E_a)=\frac{\int _{E_a}^{\infty} \sqrt{E}exp(-E_a/kT)dE}{\int _0^{\infty} \sqrt{E}exp(-E_a/kT)dE} = 2\sqrt{\frac{E_a}{\pi kT}}e^{-E_a/kT} + 1 - erf \left( \sqrt {\frac{E_a}{kT}} \right)$$

For Ea > 2kT, $1 - erf \left( \sqrt {(E_a/kT)} \right) < 0.05$.

So if kT is not big enough to be comparable to Ea, these extra terms can be thrown away, leaving

$$f \approx 2\sqrt{(E_a/\pi kT)}exp(-E_a/kT)$$.