Haloalkane hydrolysis - stoichiometric vs rate equation

[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it is usual for the student to find the rate expression by measuring the change in [HX] by successive titration of withdrawn samples against standard base over time, or via pH sensor, and many textbooks/references say this.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?!

 

[Ygggdrasil]

A few notes:

a) If water and the haloalkane are present in equimolar amounts than the reaction may not follow a first order rate law. For example, the reaction rate could be given by rate = k[RX][H2O], but since [H2O] >> [RX], [H2O] can be treated as a constant that can be combined with the rate constant to get a new rate constant k' = k[H2O], such that rate = k'[RX].

b) It might be useful to make an ICE chart for the reaction, and find the amounts of each substance presence when [RX] changes by an amount y. This should show you the correct relationship between [RX], [HX], and [ROH]. Since y is small compared to [H2O], [H2O] stays essentially constant throughout the reaction.

 

[Miffymycat]

Thanks again - useful input. I will continue to wrestle with it!

BTW, I could not manage to insert the Powerpoint png directly into the text of my question and had to attach it - how did you do it?!

 

[Ygggdrasil]

I uploaded the image to a hosting server (I used tinypic.com) then used the img tags to insert the picture.

If you have any more questions after you give the problem some more thought, just let us know.

 

[LudusRex]

I would like to clarify some points regarding the stoichiometric and rate equations for the hydrolysis of haloalkanes.

Firstly, your sketch is a valid representation of the concentration vs time profiles for all four substances. The rate of formation of the products, ROH and HX, will be different from the rate of consumption of RX. This is because the rate expression is based on the concentration of RX, and not the products. Additionally, the linear graph for H2O will have the same slope as the average slope for RX and HX, as they are all formed at the same rate.

Regarding your second question, it is correct to say that the molar ratios at any time t will not be as shown in the stoichiometric equation. This is because the stoichiometric equation represents the overall change at the end of the reaction, but not the individual changes at different time points. The reactants and products are not changing in an equimolar or 1:1 ratio at any given time, but the overall stoichiometry will still hold true at the end of the reaction.

Lastly, for your third question, it is important to note that the rate expression is based on the slowest step in the reaction mechanism. In this case, it is the hydrolysis of RX. Therefore, measuring the change in [HX] over time will not give the correct order of the reaction. To determine the correct order, it is necessary to measure the change in [RX] over time, as it is the concentration of RX that affects the rate of the reaction.

In conclusion, while the stoichiometric equation and rate expression may not fully represent the individual changes at different time points, they still accurately reflect the overall change at the end of the reaction. It is also important to consider the slowest step in the reaction mechanism when determining the rate expression for a reaction.