Why is the denominator of the ratio possible if the forward rate is 2nd order?

[Settia]

In my textbook there is an example with the following reaction and rate:

C₆H₆ + 3H₂ -> C₆H₁₂

R_f = k_f[C₆H₆][H₂]

Then the the example claims that the ratio of the rate constants is:

k_f / k_r = [C₆H₁₂] / ( [H₂]³ [C₆H₆] )

How is the denominator of that ratio possible if the forward rate is 2nd order?

I tried using latex for the fractions but the subscripts and superscripts were not working correctly in the fractions.

 

[epenguin]

re latex see my sig.

Maybe it is not all that well expressed, but anyway that fraction

[C₆H₁₂] / ( [H₂]³ [C₆H₆] )

refers to the equilibrium concentrations of those substances. That relation is given by thermodynamics, all kinetics has to be consistent with it. Equilibrium happens when forward rate equals back rate R_f = R_r. It follows, and your book surely explains this with examples. E.g. that the back rate kinetic law must be

R_r = k_r[C₆H₁₂]/[H₂]²

It is a challenge to think up a mechanism that would give you that kinetics but it must exist.

 

[Yanick]

Keq's are derived by making the rates equal each other (the definition of an equilibrium) then doing some simple algebra to get a ratio of rate constants on one side and products/reactants on the other. The ratio of the rate constants becomes the new term-Keq-and the ratio of products/reactants can be used to determine concentrations at equilibrium. Rate laws and their orders are determined experimentally, by varying concentrations of reactants and measuring the change in rates.

The stoichiometric coefficients become the exponents in rate laws and since rate laws are used to derive Keq expressions, my guess would be that there is a typo in the book.

 

[Vulgar]

I think this only goes for elementary reactions though,

http://en.wikipedia.org/wiki/Elementary_reaction

 

[epenguin]

Keq's are derived by making the rates equal each other (the definition of an equilibrium) then doing some simple algebra to get a ratio of rate constants on one side and products/reactants on the other. The ratio of the rate constants becomes the new term-Keq-and the ratio of products/reactants can be used to determine concentrations at equilibrium. Rate laws and their orders are determined experimentally, by varying concentrations of reactants and measuring the change in rates.

The stoichiometric coefficients become the exponents in rate laws and since rate laws are used to derive Keq expressions, my guess would be that there is a typo in the book.

They can be. In elementary teaching the equilibrium constant often is introduced this way, via simple kinetics, because that is rather easier to understand than thermodynamics. But thermodynamics, not using rate laws, demonstrates that for that reaction

[C₆H₁₂]_eq/ ( [H₂]³_eq [C₆H₆]_eq )

where the concentrations in brackets are the equilibrium ones - I have made this explicit in the notation - is a constant.

I'll write this

K_eq = [C₆H₁₂]_eq/ ( [H₂]³_eq [C₆H₆]_eq )

This fact is independent of the reaction mechanism. Whereas on the other hand the rate laws are consequence of the reaction mechanism and can be various even for the same reaction. I can imagine a reaction mechanism that generates the law for R_f given by Settia which is quite suggestive actually. But if the mechanism of the reaction is changed - suppose the one with the kinetic law above is not catalysed, one catalysed by a solid metal for instance would give a different rate law but both would have the forward and back rate laws for R_f and R_r such as to give my equation above when R_f = R_r.

Very general - it is still true even if the substances participate in other side reactions, or even if the reaction considered is itself a side reaction!

I think this only goes for elementary reactions though,

http://en.wikipedia.org/wiki/Elementary_reaction

That is just giving an elementary example but what I have said is general. In fact Settia's example is not an elementary reaction.

 

[Vulgar]

If you have an elementary reaction you can put the stoichiometric factor as an exponent in the rate law. You also have reactions where the exponents don't match the stoichiometric factors, and these aren't elementary so kf/kr is not equal to the thermodynamic equilibrium. So I think Settia's example is an elementary reaction

 

[wazani]

In my textbook there is an example with the following reaction and rate:

C₆H₆ + 3H₂ -> C₆H₁₂

R_f = k_f[C₆H₆][H₂]

Then the the example claims that the ratio of the rate constants is:

k_f / k_r = [C₆H₁₂] / ( [H₂]³ [C₆H₆] )

How is the denominator of that ratio possible if the forward rate is 2nd order?

I tried using latex for the fractions but the subscripts and superscripts were not working correctly in the fractions.

Hi
I think that the first formula explains the reaction in the initial concentrations of the reactants of the forwared direction in the beginning of reaction .
The second formula explains the equilibrium constant after the arrival of the system to equilibrium.

 

[epenguin]

I'm not sure if we're saying the same thing. But to make clear I need to give other examples. It would be perfectly plausible that that reaction with solid or other catalyst had rate law

R_f = k_fK₁[C₆H₆][H₂]/(1 + K₁[C₆H₆] + K₂[C₆H₁₂])

But then the law for the reverse reaction must be

R_r = k_rK₂[C₆H₁₂]/[(1 + K₁[C₆H₆] + K₂[C₆H₁₂])[H₂]²]

Or one could conceive a forward reaction law

R_f = k_fK₁[C₆H₆][H₂]³/(1 + K₁[C₆H₆] + K₂[C₆H₁₂])

and the back reaction law would have to be

R_r = k_rK₂[C₆H₁₂]/(1 + K₁[C₆H₆] + K₂[C₆H₁₂])

Always at equilibrium we'd have

[C₆H₆]_eq[H₂]³_eq/[C₆H_12]eq = K_eq.

which equals k_rK₂/k_fK₁ in these two cases. (Oops that's upside down with respect to the first time we did it, but define it which way you like, it's arbitrary.)Student would do well to think what mechanisms could generate the three pairs of rate laws we have considered.

 

[Yanick]

I agree about the elementary reaction bit. I was typing in a hurry and should've thought through my response a bit more.

epenguin, I assume that these concepts you speak of are more advanced than my current knowledge. I was actually quite happy with my understanding of Keq's and rates and such. Goes to show that no matter how much you think you know, you're really just scratching the surface. I assume this is in the realm of Physical Chemistry (seeing as I've not yet been exposed to the expressions of rate that you have posted up)?

 

[epenguin]

They are just kinetics of reaction mechanisms more complicated than elementary - made up in fact of a series of elementary reactions. In Settia's example the kinetics tell us the reaction cannot be elementary. But you cannot expect it to be. I guess an 'elementary reaction' there would be four molecules colliding and producing the product, which does not happen, such a collision probability is so low for a start.

 

[Vulgar]

My reasoning was also completely different than epenguin's. I knew that at equilibrium Rf=Rr, and the expression of Keq. I just assumed that for elementary reactions kf/kr=Keq.

For instance:

3A+2B-->C

Rf=kf[A]^ab
and
Rr=kr[C]^c

When the reaction is elementary, a=3, b=2, c=1, and when you do Rf=Rr, you get the Keq expression. But when it's not elementary, for instance, a=2.5, b=1 and c=2, at equilibrium the reaction has to hold to:

kf/kr=([C]²)/([A]^2.51)

and

Keq=([C])/([A]³²)

I guess that's not mathematically possible for some reason, more than 3 equations in total for instance

 

[epenguin]

You last equilibrium equation Keq=([C])/([A]³²) has to be true for that reaction

You just could not have those rate laws you give:

Rf=kf[A]^ab
and
Rr=kr[C]^c

with a=2.5, b=1 and c=2.

For example if the reverse reaction rate is proportional to C² as in your example there then there has to be a [C] term in the forward rate, likewise if the forward term contains [A]^2.5 the reverse rate term has to have factor [A]^-0.5-1.

Don't worry, if you try to think of reaction mechanisms that generate any kinetic laws, you will find it impossible to come up with any that violate the equilibrium law.

 

[wazani]

Hi
would you please scan the page at your text,that's help us to solve the ambiguity.