[robzy]

Hey guys,

So here is a question in our chemistry exam on rates:

Continuous line is the first experiment. Dotted line is the second experiment.

Circle one or more changes that could have been made to the experiment
(individually) to change the continuous line to the dotted line.

(A) Increase in surface area
(B) Increase in concentration
(C) Increase in temprature
(D) Somethingorother

http://members.optusnet.com.au/robzy/rates.jpg

I (as well as quite a few others) said the answer was B.... but according to my chem teacher the answer was A and B.

We argued that it couldnt be A, for no other reason that the higher line implies there are more particles. We argued that if A was the answer the vertical-axis should be "exposed particles" or "particles involved in collisions".

He responded by saying, "Well, there is no other way to show a change in surface area on the graph".

Grr.... Just because there is no other way to show a change it does not make that one corrent.

Anyways, me and some friends are really hoping we could come across someone with a bit more "weight" than us to help us argue our point. Could you possibly help us out?

Its annoying because normally this guy is amazing. He's a great teacher and often very logical, this time however it doesnt feel so.

Thanks a lot,
Rob.

[Borek]

IMHO he is right.

The thing is - in the real life you have to work with the data as it is given to you. It often means you have to work with incomplete information or with misrepresented information, thus you have to be flexible. That's such case. Imagine you are presented with these two plots and you are told "you see, there was increase of concentration". Will you accept this answer? If you do, you are overlooking something - perhaps the second experiment was done with the same concentration, but someone was not aware of the fact that size matters?

Sorry if it is not the answer you waited for And sorry that this lesson was learnt the hard way.

[robzy]

Sorry if it is not the answer you waited for And sorry that this lesson was learnt the hard way.

Its cool, do you mind if i argue (debate? ) it though?

What is wrong with the statement: If only the surface area of reactants is changed, there will be no difference in the area under the line.

(If I am wrong here, id aprecaite to know what i have wrong)

Rob.

[Borek]

Its cool, do you mind if i argue (debate? ) it though?

No problem, fire away. It may even happen that you will convince me I am wrong

If only the surface area of reactants is changed, there will be no difference in the area under the line.

What's reasoning behind this statement?

I am attaching picture we are referring to just to make discussion easier for others.

[robzy]

Well, that graph tells you how many particles have a specific energy, right?

Lets call the lower limit of the graph (on the x-axis) 0 energy, and the upper limit 8 energy. (We could then approximate that 50% of the particles present have an energy between 3 and 5, but that isnt relevant to the question, its still a fact)

This would mean that the area under the graph corrosponds to how many particles exist (albeit in arbitrary units).

Notice how the dotted line has more area under it than the solid line? That would mean more particles would exist in that one. (There would be appr. 30% more particles having energy 5 for instance)

Increasing surface area would not increase the number of react-able particles at any time. For that reason, if we drew a new line which showed the affect of increasing surface area, it would have to have the same area under it as the original (solid) line.

Rob.

[Borek]

Increasing surface area would not increase the number of react-able particles at any time.

Why not? They have some surface density, twice the surface, twice the number of particles exposed to the reaction environment.

[robzy]

Increasing surface area would not increase the number of react-able particles at any time.

Why not? They have some surface density, twice the surface, twice the number of particles exposed to the reaction environment.

Argh, sorry. I meant: Increasing surface area would not increase the number of particles to be reacted.

Origional post edited. And my logic (in my head at least) still stands.

However - I could still be missing something

Rob.

[Borek]

Argh, sorry. I meant: Increasing surface area would not increase the number of particles to be reacted.

It will. Try to explain why not.

[robzy]

Okay, take a block of NaCl. There are a million Na particles and a million Cl particles.

When you put it in water it will take quite a while to disolve because very little of the NaCl is actually exposed to the water (Lets say a minute). There will be a million "reactions" in the end though.

If you take the same NaCl and crush it up the surface area will be greated and thus it will dissolve quicker (It might only take 15 seconds). After all is said and done though, there will still be a million "reactions".

Whether it is crushed up or not there is the same number of particles that end up being reacted.

The number of particles reacted per unit time changes, but the number reacted total does not.

Rob.

[Borek]

it will dissolve quicker

That's where your reaction speed is surface dependent.

Million reactions you are referring to is what happens when the reactant is already dissolved.

Think what will happen if your reaction is between solid Zn and some strong acid. Imagine you have used 1 g Zn in one piece. Now imagine you have used it in the form of powder.

[robzy]

The area under the graph will be representative of the total number of particles involved in a reaction though. It's in the notes our teacher gave us:

http://members.optusnet.com.au/~robzy/rates2.jpg

Rob.

[Borek]

Not exactly. If one of the reactants is in the solid phase point 7 of your notes starts to play most important role. Number of collisions will be proportional to the solid surface. Note that atoms on the surface are more or less equivalent - differences in their energies are much smaller then in the case of atoms (ions) in liquid (gas) phase. Note also, that even if there are large enough differences between atoms on the surface (so that only some of them have energy high enough to react) their number will be still proportional to the surface of solid - say only 10% of surface particles have energy high enough - if you double the surface you also double number of these particles.