[Miffymycat]

This is a question about stoichiometry and rate expressions.  Consider the hydrolysis of a tertiary haloalkane RX.  The products are the alcohol ROH and hydrogen halide HX.    Usually this is done in alcohol-water solvent, where the nucleophile, H¬2O, is in excess (a solvolysis reaction), allowing the order wrt to RX to be determined with effectively constant [H2O].  In order to make my point (and the graphic) more clearly, consider this reaction done under equimolar conditions in a dipolar aprotic solvent such as propanone, but the same principles/questions apply.   
The stoichiometric equation is RX + H2O  ROH + H2O
The rate expression is Rate = k[RX]
We know that the [   ] vs time graph for a first order process is an exponential decay, and for a zero order a straight line.  I have drawn the following sketch for the reactants and products (solvolytic conditions would show the H2O line at a much higher concentration and as a straight with just a tiny negative slope and therefore awkward to draw on the same axes):

see attached

Several questions arise:
1)   Is my sketch a valid representation of the [   ] vs time profiles for all four substances?  Are the 2 products formed at different rates, in the same way as the reactants?  And does the linear graph for H2O have the same slope as the average slope for RX (and HX as for ROH)?  ? Initially, [RX] is falling faster than [H2O] and vice versa towards the end.  50% conversion of RX has occurred where its curve crosses the ROH curve, since 1mol RX produces 1mol ROH – that seems not to obey conservation of mass - ie total moles of X (or O) atoms at that point do not then seem, to add up to the original amount!!
2)   So is it also correct to say that at any time t, the molar ratios are not as shown in the stoichiometric equation (because it is not depicting an elementary step) – since the reactants are clearly not changing in an equimolar or 1:1 ratio – and that it shows only the overall change at the end?
3)   More interestingly, when studying the reaction kinetics, it is usual for the student to find the rate expression by measuring the change in [HX] by successive titration of withdrawn samples against standard base over time, or via pH sensor, and many text books/references say this.  However, if these graphs are valid, this will not reflect the behaviour of RX and will give the wrong order (zero vs first)!!! Have I missed something?!

[Dan]

consider this reaction done under equimolar conditions

...

1)   Is my sketch a valid representation of the [   ] vs time profiles for all four substances?

No. If the haloalkane and water are in equimolar proportions, their concentrations must always be identical. The concentrations of alcohol and HX must also be identical. I don't understand the logic that got you to this diagram.

[Miffymycat]

Thanks Dan - yes I agree with you, I also felt the concentrations should move together ... but the if the order is zero for water and one for RX then kinetics tells us the concentration-time curves are different, as I have shown?  Any further thoughts Dan/anyone?!

[curiouscat]

Thanks Dan - yes I agree with you, I also felt the concentrations should move together ... but the if the order is zero for water and one for RX then kinetics tells us the concentration-time curves are different, as I have shown? 

No! The differential change in H2O and RX must always be the same as dictated by stoichiometry.

A+B->Products

dA=dB

Rate expressions only tell you how dA and dB respond to [A] and

[Jeremy]

Thanks Dan - yes I agree with you, I also felt the concentrations should move together ... but the if the order is zero for water and one for RX then kinetics tells us the concentration-time curves are different, as I have shown?  Any further thoughts Dan/anyone?!

Just because the reaction is zeroth order with respect to water, it is still first order overall. Thus, whether you plot the concentration of water or RX against time, you'll get a first order curve.

So your linear [H2O] plot is incorrect.

[gritch]

This is a question about stoichiometry and rate expressions.  Consider the hydrolysis of a tertiary haloalkane RX.  The products are the alcohol ROH and hydrogen halide HX.    Usually this is done in alcohol-water solvent, where the nucleophile, H¬2O, is in excess (a solvolysis reaction), allowing the order wrt to RX to be determined with effectively constant [H2O].  In order to make my point (and the graphic) more clearly, consider this reaction done under equimolar conditions in a dipolar aprotic solvent such as propanone, but the same principles/questions apply.

I can't comment excessively on your graphs but I thought I should point out possible complications with the reaction conditions you've described. A tertiary haloalkane can typically go through 3 possible reactions, E2, E1, and Sn1. For the desired alchol product a Sn1 reaction must occur. Sn1 reaction is encouraged by stabilizing both the carbocation intermediate and  leaving group (which is typically why this reaction is a solvolysis reaction). Running this reaction in an aprotic solvent will encourage E2 and E1 reactions that will form alkene's instead of the desired alchol. I would predict as much (if not more) alkene as alchol would be produced. The same principles DO NOT apply.

1)   Is my sketch a valid representation of the [   ] vs time profiles for all four substances?  Are the 2 products formed at different rates, in the same way as the reactants?  And does the linear graph for H2O have the same slope as the average slope for RX (and HX as for ROH)?  ? Initially, [RX] is falling faster than [H2O] and vice versa towards the end.  50% conversion of RX has occurred where its curve crosses the ROH curve, since 1mol RX produces 1mol ROH – that seems not to obey conservation of mass - ie total moles of X (or O) atoms at that point do not then seem, to add up to the original amount!!

Uh... I don't know exactly how you got to you conclusion here but perhaps failing to take into account the alkene product as well as led to the wrong conclusion here.