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.