[acidball]

so the slow step for a 3 step mechanism is as follows (it also happens to be step 1):

CO2(g) + H2O(l) --> H2CO3 (aq)

the solution to the problem says in effect that this reaction is essentially

CO2(g) --> CO2(aq), with rate law: Rate = k[CO2]

first of all, how can you make that sort of simplification from the first equation and use that to determine the rate law...why can you essentially ignore H2O(l)? writing rate laws isn't like writing equilibrium expressions as far as i know, where you leave out liquids/solids...isn't that right?

i would think the rate law would simply be:

Rate = k[CO2][H2O]

why isn't that the case? thanks for your *delete me*

[Borek]

i would think the rate law would simply be:

Rate = k[CO2][H2O]

why isn't that the case?

It is the case. However, does the water concentration change? Or is it constant? And if it is constant - can it be incorporated into k?

[realmjumper99]

Normally there is just so much water that it doesn't change the concentration significantly. That is why it is incorporated into the constant,k.

[acidball]

thank you for the thoughtful responses.

so is it fair to say then that rate law expressions are more or less similar to equilibrium expressions, in the sense that you effectively leave out liquids? what about solids?

or is the idea just that anything whose concentration doesn't change appreciably (regardless of whether it's a solid/liquid/gas) just gets incorporated into the rate constant?

[realmjumper99]

correct the second time. It has to do with the change in concentration not the state of the material.

[Yggdrasil]

so is it fair to say then that rate law expressions are more or less similar to equilibrium expressions, in the sense that you effectively leave out liquids? what about solids?

or is the idea just that anything whose concentration doesn't change appreciably (regardless of whether it's a solid/liquid/gas) just gets incorporated into the rate constant?

Anything whose concentration doesn't change appreciably in a reaction can usually be left out of the rate equation.  For example, if you have two gasses reacting in a bimolecular reaction (A + B --> C) and gas B is in a huge excess, you can usually ignore [B ].  This converts a second order reaction into what's called a pseudo first-order reaction since the reaction is bimolecular but the rate law is first order.

For reactions of gasses with solids or liquids, the surface area of the solid/liquid determines the rate of reaction, not the concentration.  Since the surface area does not change significantly throughout the reaction, we can generally ignore solids and liquids in rate equations.

[realmjumper99]

Be careful with how you handle it though. If k is essentially a black box, it is fine to say you throw out the excess. This is incorrect though especially if you must find k because in fact, you simply incorporate it into the rate constant, k. It is not ignored it is simplified.