[Big-Daddy]

If we are given a set of rates of a certain reaction, r₁, r₂, r₃, and the concentrations of the two reactants A and B at the time each rate was measured, c_A,1, c_A,2, c_A,3, c_B,1, c_B,2, c_B,3, how can we calculate the reaction orders for A and B, n(A) and n(B), in terms of these?

I started by noting that

[tex]\frac{r_2}{r_1} = (\frac{c_{A,2}}{c_{A,1}})^{n(A)} \cdot (\frac{c_{B,2}}{c_{B,1}})^{n(B)}[/tex]
and
[tex]\frac{r_3}{r_1} = (\frac{c_{A,3}}{c_{A,1}})^{n(A)} \cdot (\frac{c_{B,3}}{c_{B,1}})^{n(B)}[/tex]

And this went on to yield a fairly arduous-to-type-up solution for n(A) and n(B). However I tried applying it and found that, in several cases, the denominator just ends up being 0. What is the actual solution? Is it possible to solve this properly?

[magician4]

You don't bother with small problems, do you?

Investigations in kinetics usually aren't accomplished by a snapshot of three point type "situations" for a given start-up configuration  (as three points out of a complete graph mostly don't allow for recalculating the underlying function if it came to functions of higher order, as it is the case with most chemical reactions)

to give you an example: if we had a (still quite simple ) reaction 1 A + 1 B  1 C + 1 D  , and if this reaction was second order (which often is the case), it should follow the differential equation

[tex] -  \frac {d[A]}{dt} = - \frac {d[B ]}{dt} = k_2 [A] [B ] [/tex]

resp., if we introduced a reaction progress number [itex] \chi [/itex] (this would be the rate "r" in your question), it should look like this

[tex] - \frac {d \chi}{dt} = k_2 ([A]_0 - \chi) ( [B ]_0 - \chi ) [/tex]



as a result,*⁾ we'd find a function that for all [A]₀ [itex]\neq[/itex]  [B ]₀ could be described as

[tex] \chi (t) = [A]_0 [B ]_0 \cdot \frac {e^{([B ]_0 - [A]_0) k_2 t} - 1  } { [B ]_0 \cdot e^{([B ]_0 - [A]_0) k_2 t} - [A]_0 } [/tex]

whereas (exclusively!) for [A]₀ = [B ]₀ we'd have to use a different mathematical approach,  leading us to

[tex] \chi (t) = \frac { ([A]_0)^2 k_2 t }{ 1 + [A]_0 k_2 t } = \frac { ([B ]_0)^2 k_2 t }{ 1 + [B ]_0 k_2 t } [/tex]

I take it that you'd agree that with simple three-point analysis you'd be completely bursted here, esp if you didn't know that this IS "clean" second order kinetics right from the start, but had to search for this very order instead.

Hence, experiments in kinetics are using a lot of quite clever tricks to facilitate this situation (like: making one of the components a huge excess, so d[B ]/dt becomes approx. zero and thatlike  pseudo - first order kinetics), so reaction orders with respect to single components can be measured separately (and recombined thereafter, on paper, to result the "normal" situation)

regards

Ingo


*⁾
to see how these solutions to the differential equation came about, i would kindly ask you to take a look into the respective literature concerning second order kinetics: else, this might become a huge thread in its own right

[Big-Daddy]

Of course we cannot guess the form of the differential equation solution, so trying to use concentration-time data to find the solution directly is mostly futile. But I'm not looking at concentration-time data but rather rate-time data. e.g. I may have measured that if I start with certain concentrations, the rate is a certain (numerically measured) r1, and if I then have different starting concentrations the rate is r2 instead.

So, we're not trying to determine the form of the solution to the differential equation - I'd just assume my maths is good enough (and if not then I'm sure software could get me numerical answers!) - but rather, using rate-time data from experiment to find the differential rate law, which is of the form r=k[A]^n(A)[B]^n(B).

This topic was motivated by the problems I normally get being very simple, e.g. from trial to trial we always get one reactant's concentration held constant, so that if the other reactant has concentration c1 and c2 leading to respective rates r1 and r2, we can find the order n by

[tex]n = log_{\frac{c_2}{c_1}} (\frac{r_2}{r_1})[/tex]

This comes out of the OP equation too: if we set either of the reactants to having the same concentration across trials 1 and 2, then the dependence of r₂/r₁ on that reactant vanishes.
However, it strikes me that it may/should be possible to find the values of n(B) and n(A) anyway, without needing one to be kept constant to calculate the other. We’ll need 3 data points for 2 reactants (or 2 data points for one reactant), which is what I set up in my original question with r1, r2, r3. But now, the question comes to, what should I do to actually make it work? My solution has so many undefined terms that it’s clearly far from universal.

[curiouscat]

so trying to use concentration-time data to find the solution directly is mostly futile. But I'm not looking at concentration-time data but rather rate-time data.

What's the difference. Isn't one simply the derivative of the other?

*Neglecting error amplification.

[magician4]

let's put it like this:

if the general differential law was -d[itex] \chi [/itex]/dt = r = k[A]^a * [B ]^b , and if you could measure r₁, r₂ and r₃ (and the respective concentrations belonging to)

and if therefore you had like three data sets of the type  r_x([A]_x , [B ]_x)

and hence you could write two ratios like  r_n/r_n+x = ( [A]_n/[A]_n+x )^a * ( [B ]_n/[B ]_n+x )^b
r_n/r_n+z = ( [A]_n/[A]_n+z )^a * ( [B ]_n/[B ]_n+z )^b


only thing you needed to do was to write the double logarithmic form

log ( r_n/r_n+x) = a * [ log ([A]_n/[A]_n+x )] + b * [ log ([B ]_n/[B ]_n+x ) ]
log ( r_n/r_n+z) = a * [ log ([A]_n/[A]_n+z )] + b * [ log ([B ]_n/[B ]_n+z ) ]

to determine a and b (as k drops out of the system this way)


unfortunately, to the best of my knowledge, there is no known way to measure r , i.e. -d[itex] \chi [/itex]/dt directly, so this mathematical route to an easy solution is for all practical purposes impossible to use


regards

Ingo

[Big-Daddy]

What's the difference. Isn't one simply the derivative of the other?

Sure, but this is an experimental problem ... we're starting by assuming that you don't have the function of concentration/time so you can't find the rate just by differentiating. You'd probably get rate by drawing tangents or something. I imagine that if you're a decent drawing hand, and the orders are nice integers roughly between 0 and 3, you can find them even if the rates are as rough as tangents.

log ( r_n/r_n+z) = a * [ log ([A]_n/[A]_n+z )] + b * [ log ([B ]_n/[B ]_n+z ) ]

to determine a and b (as k drops out of the system this way)

Ah, I see. So log to the base of anything would suffice for this equation then? Log rules seem to suggest so...

as you see, two data point sets already would be sufficient to do so

Don't we need a second equation to solve for a and b? For which the third data point would be used...

unfortunately, to the best of my knowledge, there is no known way to measure r , i.e. -d[itex] \chi [/itex]/dt directly, so this mathematical route to an easy solution is for all practical purposes impossible to use

No worries, I guess that makes this a purely mathematical exercise. But it was devoted by my believing that surely we don't actually need to hold one concentration constant to find a and b from the rates, as all my problems do, so solving it is interesting for me.

[magician4]

Ah, I see. So log to the base of anything would suffice for this equation then? Log rules seem to suggest so...

yes, exactly: that's how it's done

Don't we need a second equation to solve for a and b? For which the third data point would be used...

you're right: you would need three data sets r_x([A]_x , [B ]_x) to generate two different ratio expressions rx/rx+y ; rx/rx+z (incl. the respective conc. values belonging to)

two ratio expressions: that's what I meant (but didn't write)

mea culpa: sorry for the confusion

regards

Ingo

hint:
pls. note, that I made respective amendments in my previous post to correct for this error

[Big-Daddy]

Thanks, that clears things up. Instead of trying to take logs to the base of the reactant concentrations as I did (problem arises because log base 1 is almost always undefined), I should have just taken a general base b log and then allowed the solution to drop out.