[nebula]

Ok. This was a question in a recent year 12 chem test:

Nitrous acid ionises in water according to the following reaction.

HNO₂(aq) + H₂O(l) <--> H₂O⁺(aq) + NO₂^-(aq)

450ml of water is added to a 50ml solution of nitrous acid at constant temperature. Which one of the following answers best describes the system once it has returned to equilibrium?

The multi-choice answers had three categories and choices of higher or lower - concentration of NO₂^-, amount of NO₂^- and final reaction rates as compared to initial equilibrium rates.

I determined that the concentration would be lower but the amount would increase.

As for the final reaction rates, my teacher said that they'd be lower because the concentration of nitrous acid is decreasing. However I thought that reaction rates would only decrease if both reactants weren't water. I mean that decreasing the concentration of nitrous acid will have no effect because the molecules are constantly surrounded by water either way.

So, am I right?

[Borek]

This is a little bit chaotic and hard to address without knowing the original wording of the question, but let's try.

I determined that the concentration would be lower but the amount would increase.

I fail to see how - for a given, known volume - concentration can go down and amount go up.

As for the final reaction rates, my teacher said that they'd be lower because the concentration of nitrous acid is decreasing. However I thought that reaction rates would only decrease if both reactants weren't water. I mean that decreasing the concentration of nitrous acid will have no effect because the molecules are constantly surrounded by water either way.

What is reaction rate proportional to?

[nebula]

That is the original wording of the question. As water is added, the concentration will go down. However, water is also a reactant in the equation and will cause a net forward reaction (because it is an equilibrium reaction). So, there is a net increase in the amount of NO₂^- but the concentration will decrease because more water is being added.

Reaction rate (I think) refers to amount of successful collisions per second. My teacher believed that decreasing the concentration of nitrous acid would reduce the number of successful collisions. However, aren't the water molecules ubiquitous in the container. As such, shouldn't the rate of reaction (number of collisions per second) be the same as before water was added.

[Borek]

OK, you are comparing 50 mL with 500 mL, that explains the difference.

Assuming this is second order reaction its speed is given by

v = k[H₂O][HNO₂]

Unless we are talking about some highly concentrated acid, water concentration can be assumed constant. From the moment molecule is surrounded completely by water molecules number of collisions per molecule of acid doesn't change much.

[nebula]

Thanks. So, rate of reaction does decrease. Oh well.