AQA - Unit 5 - Energetics, Redox and Inorganic Chemistry

JUST noticed this!

In the mark scheme when you have to write the reaction 2I- + S2O8 (2-) ---> I2 + 2SO4 (2-) or the reactions between vanadium(V)oxide and SO2/SO3...

the mark scheme says instead of the arrow...it has a weird half circle with a line through it :s WHY?! do we have to write this weird symbol in the exam or is the arrow fine?!

JUST noticed this!

In the mark scheme when you have to write the reaction 2I- + S2O8 (2-) ---> I2 + 2SO4 (2-) or the reactions between vanadium(V)oxide and SO2/SO3...

the mark scheme says instead of the arrow...it has a weird half circle with a line through it :s WHY?! do we have to write this weird symbol in the exam or is the arrow fine?!

Hey, I'm just going over the Ligand Substitution Chapter, and in the Chelation section the book is asking about how many entities are on either side of the equation. What are they

It's referring to particles. It's like saying NH3 and Cl- are species. Particles are a type of entity.

I prefer to use the term moles rather than particles though since questions on it tend to use an equation.

Hope that's cleared things up

lols :P

Thank you so much that makes so much more sense now, do you use them to show the increase in entropy (chelate effect)?

Yup. Though the Chelate effect is referring to the increased stability of complexes which is caused by the increase in entropy, not referring to the increase in entropy itself.

Example time!

[Cu(H2O)6]2+ + EDTA4- ---> [Cu(EDTA)]2- + 6H2O

There is an increase in entropy as there is an increase in the number of moles (2 moles on the left to 7 moles on the right) so [Cu(EDTA)]2- is more thermodynamically stable (you're not going to be able to easily reverse the reaction). This happens with all reactions where multidentate ligands substitute unidentate ligands.

It's like comparing the amount of disorder of bouncy balls in a box.

10 bouncy balls bouncing in a box is going to be way more chaotic than a box containing one bouncy ball bouncing around.

You might also want to know if you don't already that the enthalpy change is near zero because the same number and type (coordinate) of bonds are broken as are formed. Hence why entropy change has such an effect. E.g. with the above reaction six coordinate bonds with the six H2O ligands are broken, but six coordinate bonds are formed with EDTA4-

Thanks so much!!!! I understand that much better now and also I'm liking the bouncy ball analogy
Btw you seem well on your way to acing the exam, hope by friday that I can too

in the lattice MgO between the Mg2+ and O2- are there electrostatic force of attraction or ionic bonds.

you're very welcome. I'm sure you'll be more than fine

Sadly I have a habit of going straight to answering the questions in exams rather than reading the question so I often end up giving the wrong answer Need to think of a way of forcing myself to read the questions properly.

I only recently realised the exam is in the morning rather than in the afternoon, shows how on the ball I am

I'm pretty sure both of them are out of 120. 96/120 was 80% on the last paper.

For the last three papers an A* (i.e. 90% ums points) has consistently been around 82-84 marks out of 100.

good ways to remember the colours of the metal compounds in solution?