Calling kinetics experts: rate law from conductivity isnt possible?

1. Miffymycat

41
Calling kinetics experts: rate law from conductivity isnt possible!!?

Consider the usual primary halogenoalkane aqueous alkaline hydrolysis reaction

RX + OH- --> ROH + X-

We know the rate law is first order in RX and OH-. We could separately represent the drop in OH- conductivity as an exponential decay with a constant half-life (ΛoOH-e-kt) and the rise of X- conductivity as the inverse function of this (0.5ΛoOH-(1-e-kt), taking the conductivity of X- as 0.5x that of OH-.

In practice, using excess RX, the measured (or modelled) solution conductivity during hydrolysis is obviously the sum of the ion conductivities at any point in time. The mixture conductivity drop-off appears to be an exponential-type decay, but attempts to curve fit (albeit only in Excel) show it is not, nor does it fit a recognisable integrated rate law plot. One can therefore not obtain a rate constant or order from this progress curve, which is frustrating - unless I'm mistaken!! {Its not the case for aqueous hydrolysis as this produces ions from neutral molecules rather than an exchange of ions and the graphs work fine}.

Furthermore, taking an initial rates approach and plotting initial (ΔΛ/t) vs Λfinal (over several initial concentrations, rather than a single Λ vs t curve as above) gives a straight line, but whose slope does not appear to be a simple multiple of the calculated k for OH- decay on its own. The stoichiometry is 1:1, so the rate of [OH-] decline = rate of [X-] growth, and I imagined the slope would therefore be k x ratio of ion conductivities ... but it's not. Its a smaller number.

Staff: Mentor

Perhaps I am missing something, but

$$Ae^{-x} + \frac A 2 (1-e^{-x}) = \frac A 2 (1 + e^{-x})$$

doesn't look like something that can be fit to just e-x.

3. Miffymycat

41
Agreed, and seems to support the idea that conductivity data from "ion-exchanges" cant be used to investigate reaction kinetics.

Any ideas on the significance of the slope for the linear plot?

Staff: Mentor

I never said that. You can use the conductivity, you just have to fit it to the right equation.