# If sucrose gets hydrolyzed by water to yield two equivalents of "C"_6"H"_12"O"_6, what is the rate law? How should you describe the molecularity, and why is the concentration of sucrose not needed to describe the rate (provided there is excess water)?

##### 1 Answer
Feb 5, 2016

This reaction is sucrose (${\text{C"_12"H"_22"O}}_{11}$) getting hydrolyzed by water, and it's saying the products are two equivalents of ${\text{C"_6"H"_12"O}}_{6}$ (one of them is glucose and the other is fructose, so it's more proper to not write $2 {\text{C"_6"H"_12"O}}_{6}$).

The idea is that the concentration of water is not going to change significantly with respect to the sucrose concentration.

That's the equivalent of saying:

$r \left(t\right) \approx k \left[\text{H"_2"O"]^0["Sucrose}\right]$

$\to \textcolor{b l u e}{r \left(t\right) \approx k \left[\text{Sucrose}\right]}$

Remember that the rate of overall zero order reactions does not depend on the concentration of the compound. In other words, suppose there was no sucrose. Then:

$r \left(t\right) \approx k {\left[\text{H"_2"O}\right]}^{0} \approx k$

and evidently the concentration is not needed to know the rate.

But water and sucrose are both there. Water is in excess, so its concentration doesn't change much relative to the change in concentration of sucrose.

A mathematical way to say this is:

$\textcolor{b l u e}{- \frac{d \left[\text{Sucrose"])/(dt) " >> " -(d["H"_2"O}\right]}{\mathrm{dt}}}$

or, " the rate of disappearance of sucrose is much faster than that of water."

Since there truly are two reactants and sucrose doesn't hydrolyze automatically on its own with no help, it is not unimolecular, but pseudo-unimolecular.

It is basically unimolecular but, strictly speaking, is not actually unimolecular. It is just assumed to be unimolecular to simplify calculations without compromising accuracy.