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)
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:
C
6H
6 + NO
2+ 
C
6H
5NO
2 + 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.
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, sulphur 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 sulphur 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 sulphur changes its form several times while being heated through to the vapour state.
sulphur crown
S8 crowns give way to S8 strings that entangle into S8 cross linked structures that turn to S8 vapour.
When sulphur 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 strucure. For this reason the majority of texts (and chemists) simply refer to sulphur as ‘S’, using the same logic as used for giant and similar stuctures.
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.
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.
Hydrogen peroxide has the ability to gain or lose electrons, as its oxygen atoms are in the -1 oxidation state. By gaining electrons they can go to the -2 oxidation state, and by losing electrons they can go to the zero oxidation state (the element)
When someting acts as an oxidising agent is gains electrons (removing them from the oxidised species). This can be shown by the relevant half-equations:
H2O2 + 2e –> 2OH-
or in the presence of acid:
H2O2 + 2H+ + 2e –> 2H2O
Well just finished the upgrade to what I hope is a more secure ChemBlog after being attacked by the spam insertion php robots. It only took me about 8 hours to sort it out… grrr
Now damn well stay out!
CC
Once again it’s time for the merry-go-round, all change, super new look shiny syllabus. Out with the old and in with the new. But what do you think of the changes?
Where has spectroscopy and instrumental analysis gone from the main syllabus? It’s been exiled to the options. This seems strange to me as it’s probably the most used (and useful) area of chemistry in research and industry.
What about radioactivity? It’s been banished altogether. There’s no interest in nuclear chemistry nowadays in the world is there?
Perhaps it’s all physics… I don’t think so!
CC
Original post: Friday, May 18th, 2007
The study of kinetics is to find out the things that affect how fast reactions occur and then ultimately to explain the actual mechanics of the process (what actually happens and in which order)
On the basic level, simply observation shows us that changing the concentration of reactants affects the rate of formation of products (or disappearance of reactants) - the question is, “what exactly is the relationship between reactant concentration and rate?”
Mathematically speaking there are only three logical possibilities:
1. There is no relationship
2. There is a linear relationship
3. There is a non-linear relationship
All three possibilities can be covered by one generic equation, called the rate equation:
For a one component (A) system…
Rate = k [A]^x
for a two component (A + C) system
Rate = k[A]^x[C]^y
At the very beginning of a reaction the initial rate can be found (or a close approximation) by experiment. We also know the concentration of the reactants at this time (we choose them), so if two experiments are carried out on a one component system we have two unknowns (k and x) and two equations - simultaneous equations that can then be solved for k and x
For two component systems (A+C) things get a little more complicated - but only a little. Two experiments must now be carried out, keeping the concentration of one of the reactants constant (say, C). The second equation can then be written as:
Rate = constant [A]^x
where the constant is (k[C]^y)
and once again you have two simultaneous equations with two unknowns. The constant doesn’t concern us at this time so you just solve for x.
Now you carry out two reactions keeping [A] constant when the rate equation can be simplified to:
Rate = constant[C]^y
now solve for y
Now you can choose any of the experimental results to substitute for x and y to find k (the rate constant)
Why do we go through all of this?
The answer lies in the fact that the orders of reaction x and y give us hints about likely mechanisms as they show the molecularity (number of particles ) involved in the rate determining step (slowest step) of the mechanism (and possibly any prior equilibria)
For example, if the rate equation comes out to be
Rate = k[A]^2[C]^0
This tells us that the rate determining step (slowest) does not involve C in any way and probably involves a step with two particles of A..
possibly:
2A –> A2 (slow step)
Conclusion:
Kinetics is a wholly experimental branch of chemistry that seeks answers to what actually is happening during the course of a reaction. It does so by ’solving’ the rate equation - a general equation that can be adapted for 1, 2 or 3 component systems.
The orders (x and y) give clues as to the mechanism of the reaction.
Once the orders of reaction with respect to the individual components of the reaction are found, further investigations at different temperatures can be carried out to find the activation energy of the reaction using the Arrhenius equation.
Advantages? All the calculations in exam questions involve VERY simple maths and once the concepts are grasped it’s easy marks.
Friday, May 18th, 2007
Think about it this way…
the reactants (left hand side) go on to make the products (right hand side)
Imagine that this reaction proceeds VIA the elements in their standard states, (according to Hess’s law we can go any way we want providing that we get to the desired products)
First of all, the reactants have to return to their elements. This process is the opposite of the enthalpy of formation, i.e. the negative or reverse of delta Hf = -delta Hf
The products are then formed from the elements - this is simply the enthalpy of formation, i.e. + deltaHf
So, if you sum these two processes to go from reactants to products you have:
Reaction enthalpy = - delta Hf(reactants) + deltaHf(products)
rearrange this to get:
Reaction enthalpy = deltaHf(products) - delta Hf(reactants)
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