## How does the iodine clock reaction work?

The iodine clock reaction times how long it takes for a fixed amount of thiosulphate ions to be used up, i.e. the time taken for the iodide ions to reach a fixed number of moles produced in the reaction between potassium iodide and an oxidising agent (usually hydrogen peroxide, or sodium peroxodisulphate).

The system is as follows:

The oxidising agent reacts with the iodide ions (usually introduced in the form of potassium iodide).

H2O2 + 2I + 2H+ 2H2O + I2

The iodine produced is then absorbed by reaction with a fixed amount of thiosulphate ions:

2S2O32- + I2 S4O62- + 2I

As soon as the thiosulphate ions are used up the free iodine then reacts with some starch indicator that is added right at the beginning. The reaction produces an almost instantaneous blue/black colour. Thus it is possible to time fairly accurately the time taken for a fixed amount of iodine to be produced (from the thiosulphate moles present initially)

If the oxidising agent concentration is kept constant and all other factors are constant except the iodine concentration then the rate equation simplifies to:

Rate = k[iodide]x

where x is the order of the reaction with respect to [iodide], and k is a kind of super constant combining all of the other constant factors, the rate constant and the other components of the reaction.

So, as rate is porportional to 1/time taken then

1/t = k[iodide]x

if you take logs throughout

log(1/t) = xlog[iodide] + logk

This now has the form y=mx+c (a straight line graph)

So a plot of log(1/t) on the y-axis against log[iodide] on the x-axis gives a straight line of gradient x and intercept k (although k is pretty useless in this case)

## Which factors determine the percentage yield in a laboratory preparation?

Experimental preparations of compounds usually follow the standard methodology, i.e.
1. reaction
2. work-up
3. purification

The theoretical yield can be simply calculated froma knowledge of the stoichiometry of the reaction(s) involved in the preparation.  The final percentage yield is calculated from actual yield/theoretical yield x 100%

For example
In the preparation of nitrobenzene by nitration of benzene the nitrating agent (concentrated sulphuric / nitric acid mixture) is added in slight excess to prevent over nitration of the benzene ring. The yield is dependent only on the number moles of benzene used:
C6H6 + NO2+ C6H5NO2 + H+

It may easily be seen that 1 mole of benzene s expected to produce 1 mole of nitrobenzene. However, this is rarely the case.

Any discrepancy bertween theoretical yield and practical results is due to the following causes:

The reaction may not produce a stoichiometric amount of product. This could be due to equilibrium, slow rate, decomposition of reactant, or product, or evaporation. The reactants may also not be 100% pure. A common problem in synthesis is the issue of side-reactions producing an undesired product.

During a work-up which may involve solvent extraction or crystallisation some of the product is invariably discarded.

During the purification stage some of the product may be lost in distillation, or by decomposition, or even evaporation. Some may remain inside the reaction, or purification vessels, filtration systems, chromatographic columns etc etc.

All of the above will lead to a reduced yield.

Higher than expected yield can only be due to the presence of impurities caused by inadequate purification. (assuming that no mathematical errors have been committed).

It just remains to ascertain how such impurities have arrived there.

It is usually because the product still contains solvent, but clearly it could still be mixed with other products or even reactants. It could absorb components from the air such as water vapour or carbon dioxide. The vessel used to collect the product could even be contaminated.

## What is the molecular formula of sulfur?

The formula of the elements is best left to the simplest possible form, except in the case of the gases and the halogens, that are shown as diatomic molecules by tradition. The remaining elements are simply expressed as a single atoms, although it must be emphasised that elements, or rather the atoms of elements, do not ‘go around’ as single entities, unless they are noble gases.

All other atoms are unstable on their own under normal conditions. Metal atoms exist in giant structures, giant covalent elements exist as just that, giant and covalent, simple covalent molecules are arrangements of atoms in molecules etc. etc. Nothing exists on its own.

So, if we want to represent carbon we don’t write down C(very big number) we just understand that it is a single macromolecule and express it as C, pure and simple.
The same goes for all giant structures, be they metals, or macromolecules. In the case of phosphorus, sulfur and certain other simple molecular elements we are faced with a choice. Do we refer to them by the molecular formula, or do we understand that they exist as finite simple molecules, but continue to express them as if they were free entities? Most texts and chemists choose the latter.

It is understood (within the chemists fraternity) that sulfur is usually found as S8 molecules and that phosphorus is P4 pyramids, but that there is no real advantage to be had expressing them as such. Indeed sulfur changes its form several times while being heated through to the vapour state.

S8 crowns give way to S8 strings that entangle into S8 cross linked structures that turn to S8 vapour.

When sulfur is reacting with (for example) oxygen which is the ‘correct’ way to represent it? What form does it actually have at the moment of reaction? We don’t know (and probably don’t care).
The sensible answer to the first question is that there is no ‘right’ way to express the chemical structure. For this reason the majority of texts (and chemists) simply refer to sulfur as ‘S’, using the same logic as used for giant and similar structures.

## How does copper react with nitric acid?

Copper is an unreactive metal and doesn’t react in normal circumstances with dilute acids. However it does react with nitric acid. Why is this?

Nitric acid is an oxidising agent and the reaction is not the usual acid + metal reaction. The products are oxides of nitrogen instead of hydrogen. The actual nitrogen oxide formed depends on the concentration and temperature of the acid.

There are actually two equations for the reaction of copper with nitric acid. It depends on whether the nitric acid is concentrated or not. If it is concentrated and in excess then the ratio is 1:4 copper to nitric acid. If it is dilute then the ratio is 3:8.

Cu + 4HNO3 –> Cu(NO3)2 + 2NO2 + 2H2O

3Cu + 8HNO3 –> 3Cu(NO3)2 + 2NO + 4H2O

Nitric acid when concentrated is a strong oxidising agent so it makes sense that a higher oxidation state of nitrogen (IV) oxide is formed when the nitric acid is concentrated.

## Which factors contribute to the boiling point of ethanoic acid?

Boiling points are a function of the intermolecular forces.

The ethanoic acid dimer has an effective RMM = 120. This means that considerable Van der Waals force would be expected between particles. Compare with a totally non-polar molecule with a similar relative mass such as decane. The boiling point of decane is 174ºC.

Were ethanoic acid simple monomers (RMM = 60) then comparison with pentane (bp=36ºC) suggests a much higher degree of intermolecular force.

It seems that the boiling point of ethanoic acid (118ºC) is somewhere between that expected from a simple monomer (with hydrogen bonds) and a dimer.

Clearly the situation is not simple and there will be contributions from hydrogen bonding, Van der Waals forces, monomers and dimers. It is difficult to state categorically which is the most important.