[Miffymycat]

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 (Λo OH-) as an exponential decay with a constant half-life [Λo OH-  e-kt] and the rise of X- conductivity as the inverse function of this [0.5 Λo OH- (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, modelling 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. Any thoughts please?

[mjc123]

A number of thoughts in decreasing order of obviousness - hope you are not insulted by the first couple.
1. Are you sure everything is clean and pure? Electrochemical measurements are easily corrupted by small amounts of impurities. (That's why I was never any good at practical electrochemistry!)
2. "Excess RH" - how much excess? Enough to ensure [RH] is effectively constant, giving pseudo-first-order kinetics? Otherwise your conductivity curves will not be exponential at all.
3. Molar conductivity is not constant with concentration, following Kohlrausch's law Λ(c) = Λ₀ - Kc^1/2. But K is small for strong electrolytes.
4. You are ignoring the contribution of the counter ion. The conductivity is the sum of the contributions from all ions present. (You might be including it implicitly, since λ(X-) ≠ 0.5 λ(OH-), but λ(NaX) ≈ 0.5 λ(NaOH).)
5. Having done some quick algebra, I don't think you're right that your initial rates plot (if I have understood it correctly) should have a slope of "k x ratio of ion conductivities". Ignoring Kohlrausch
Λ = λ_M+ + λ_OH-(1-α) + λ_X-α, where the extent of reaction α = 1-e^-kt.
dΛ/dt = (λ_X- - λ_OH-)dα/dt and (dΛ/dt)_t=0 = (λ_X- - λ_OH-)k
Λ_t=∞ = λ_M+ + λ_X-
Thus (dΛ/dt)₀/Λ_∞ = k(λ_X- - λ_OH-)/(λ_M+ + λ_X-); or -k(λ_MOH - λ_MX)/λ_MX or -k(λ_MOH/λ_MX - 1)
Try that and see if it gets you anywhere.

[Miffymycat]

Points 1-4 all OK - thanks for checking
Taking your last line:
"Thus (dΛ/dt)o/Λ∞ = k(λX- - λOH-)/(λM+ + λX-); or -k(λMOH - λMX)/λMX or -k(λMOH/λMX - 1)"
We get
(dΛ/dt)o/Λ∞ = -k(2 - 1) = -k since λ(NaX) ≈ 0.5 λ(NaOH)
So
(dΛ/dt)o = -k Λ∞ which is what one would expect, since [  ]o α Λ∞  and [   ] α Λ.
Plotting values of (dΛ/dt)o vs Λ∞ confirms the value of k is the same from this plot as from a plot of Δ[OH-]o/t vs [OH-]o, where [OH-] values have been obtained from [OH-]t = [OH-]o x {(Λobs-Λ∞)/(Λo-Λ∞).
Thanks for your input!