FAO: Zhao Xuan. Re: 2010 oct/nov Paper 53 Q1 a) ii)

Updated Wed 20th may 11:34am

Dear Zhao  Xuan.
Re: 2010 Oct/Nov Paper 53 Q1 a) ii)

2010 Oct-Nov Paper 53 Q1 a) ii)

You asked “Why is it a straight line through the origin and not a curve since the catalyst is limiting?”

It has to pass through the origin because [H2O2] is on the X axis; With no [H2O2) then rate has to be zero. But why a straight line and not a curve???

I’m sure you spotted the slightly unusual rate(y-axis) against conc(x-axis) form of the graph, which isn’t in the syllabus!

(Usually they ask for the more common conc(y-axis) vs. time(x-axis) ), however the three ‘one stop more advanced’ kinetics graphs you will get a A-level (which some other exam boards do cover) is are given below (and we did these in class).

Rate(y-axis) vs. conc(x-axis) graph
This type of graph and from what the mark scheme says means that the examiners are looking for a first order type of kinetics. The clue is here: “The rate of decomposition depends on the number of hydrogen peroxide molecules present in a given volume of solution.”

There must be some kind of proportional response, although they don’t use the specific word “proportionality” they are giving a description of some kind of proportionality.

Clearly it cannot be zero order. There is NO change in rate with [ ] for zero order reactions.

I offer this: Until CIE state in the the learning content of the syllabus something to the effect of

“… at this point, students should now adopt explicit or even implicit references to ‘proportionality’ to include exponential proportionality”

then otherwise it is linear proportionality that should be assumed. It is reasonable to take the simplest proportionality, and that is linear proportionality. Look up the helpful philosophical point called ‘Occams razor’

ALSO

But despite all that, if you did offer a correct curve then it could still gain credit, simply because it could be correct and the question doesn’t convey enough info to purely settle on one answer. I really hope an examiner would realise this and have given a proper curve credit. I really hope they would not blindly follow the mark scheme.

The “limiting” aspect of the catalyst (the max reactions it can catalyse per second – described by how many ‘active sites’ / ‘pores’ it has, was addressed by allowing students to draw the plateau which will happen once all the pores are full. this is why TOO HIGH a PRESSURE for heterogeneous catalysis has a ‘cut off limit’ – the gases condense in the pores which effectively blocks them and prevents catalysis. I have a feeling the catalyzed reaction would have a curve going the other way, but I’m finding it difficult to say which way it curves with total certainly as the ‘non-catalysed reaction’ would also have to be considered.

I admit, If I was doing this in an exam, I would probably not have picked up on that only a straight line is accepted.

But, truth be told, I think this is a silly question in the manner that it’s asked. I don’t see much direct chemistry knowledge being probed here but whether the student can pick up on the examiners “encoding” / “disguising” of the question.

At the moment, that’s the best I can do to try and answer your question. I hope it helps.

FINAL ADVICE: Try to answer such unclear questions by drawing upon the simple info that you have seen in the course. If you have an option of difficult or simple info, rule of thumb is to ‘keep it simple’, like bullet points do.

Very best wishes for the exam tomorrow, later today.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s