Archive for the ‘Inorganic’ Category
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)
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.
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.
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
SO2 seems to be a little confusing as there are several different descriptions circulating. Some books claim that it has an expanded octet, although there is no real evidence to support this. It can be simply described without octet expansion in that there are only 8 electrons around the central sulphur atom.
On a simplistic Lewis structure basis this can be achieved with one lone pair, one dative covalent bond and one double bond. According to bond measurements, however, the two S-O bonds are in resonance and have the equivalence of one and a half bonds each (three electrons per bond). Both of the S-O bonds are of equal length and energy as predicted by this model. The bond angle is that expected from three regions of electron density giving a slightly distorted trigonal planar electronic arrangement and an O-S-O bond angle of 119º.
May 21st, 2007
Hydrogen peroxide decomposes according to the equation: 2H2O2 –> 2H2O + O2
If you start with 100 volume H2O2 then 1 dm3 of solution will release 100 dm3 oxygen (at RTP)
This means that 1 litre of H2O2 solution releases 100/24 moles of oxygen and as 2 moles of H2O2 decompose to release 1 mole of oxygen then there must be 2 x 100/24 moles of H2O2 in the solution
This is equal to 8.33 mol/dm3
May 21st, 2007
All liquids contain particles in which the energy distribution is governed by the laws of statistics. This energy distribution may be plotted as a curve, called the Maxwell Boltzmann graph that shows some particles with very little energy, some with large amounts of energy and the bulk of the particles with energy somewhere in between. Those particles with large amounts of energy can escape from the body of the liquid producing a gas formed of liquid particles above the surface. This is called the vapour and the pressure that it exerts is called the vapour pressure.
The boiling point of a liquid is the temperature at which the vapour pressure equals the outside (atmospheric pressure). At this point bubbles of vapour form in the body of the liquid at this temperature – we call this “boiling”
The bubbles of vapour forming in the liquid can literally push the water apart as they have the same pressure as the atmosphere.
If the external pressure decreases then the vapour pressure will be able to equal it at a lower temperature. i.e. the liquid boils at a lower temperature.
May 26th, 2007
There are several misconceptions about electrolysis, such as the idea that electricity flows across the electrolysis cell.
The conditions required for electrolysis are an electrolyte with ions that are free to move. They may be in solution or in the form of a molten salt. There must then be an electrical potential applied by means of electrodes across the electrolyte. This electrical potential creates a large electrostatic force field that attracts the charged ions.
The actual charges on the ions are what attracts them to the electrodes (that are charged electrically) the positive ions migrate to the negative electrode (the cathode) and the negative ions migrate to the positive electrode (the anode).
When a negative ion arrives at the positive electrode it has its extra electron stripped off by the electrode. This electron continues on to the battery just like normal electricity (all electrons are, of course, indistinguishable)
Cl- –> Cl + 1e
When a positive ion arrives at the negative electrode it finds billions of electrons just waiting to fill up its positive ‘hole’. One of these extra electrons jumps onto the positive ion cancelling it out.
Na+ + 1e –> Na
The overall effect of these two processes, or electrode reactions, is that one electron has left the cathode and one electron has arrived at the anode. From the point of view of the battery and external circuit one electron has left the negative side and has arrived at the positive side. i.e. a current has flowed around the circuit.
Nothing actually flows across the electrolysis cell, the electrons are absorbed by one electrode and produced at the other by two different reactions. For these reactions to be able to occur the ions in the cell must be able to move, i.e. in solution or molten.
The net result for the electrolyte is that the sum of the two electrode reactions has happened in the cell.
2Cl- –> Cl2 + 2e
2Na+ + 2e –> 2Na
2Cl- + 2Na+ –> 2Na + Cl2
NaCl(l) —-> 2Na(l) + Cl2(g)
The chlorine reaction was doubled up to show that chlorine is released as Cl2 molecules. The sodium is double up to equalise the charge. Under the conditions of electrolysis of molten sodium chloride the sodium is released as a liquid.
May 31st, 2007
The term oxidation state is a convenient way of describing the effective condition of an atom in a compound in terms of the charge that it would have if it were ionic. It is not intended to suggest that atoms have a specific charge, but it is a useful way of describing the state or condition that an atom of an element is in within a compound; useful for the purposes of redox chemistry.
Whether an element is assigned a specific oxidation state, for example, +1 or -1 oxidation state depends on what it is bonded to. THIS IS IMPORTANT
If an atom of a specific element is bonded to an atom that is MORE electronegative, then it is assigned a positive oxidation state, and vice versa.
In reality this assignment is for descriptive terms only – it doesn’t really imply anything about the atoms or the bonds apart from the fact that one atom will tend to attract electrons more than the other.
Take the following nitrogen example:
In ammonia, NH3, nitrogen is more electronegative than hydrogen and so is assigned a negative oxidation state of 3- (assuming that each hydrogen is +1)
When bonded to oxygen however, as is N2O, the nitrogen is assigned a positive oxidation state of +1 (as oxygen is -2 (usually except for in peroxides) when attached to something less electronegative)
The extreme case that highlights this is F2O.
What is the oxidation state of oxygen here? well since oxygen is LESS electronegative that fluorine, and fluorine always has an oxidation state of -1, it must be +2.
In summary, the most important factor to establish is which is the more electronegative of the two attached atoms. This atom takes the negative oxidation state, which can then be calculate from the valencies of the atoms in question.
May 31st, 2007
Colour is caused by selective absorption of certain wavelengths and reflection of the rest. The most familiar example would be that of phosphorus, which has three allotropes each of which have a different colour; white, red, and black, and several variations within these three basic forms. The structure of the red form is not certain, although one form of red phosphorus has been elucidated there is no clear evidence that the other red forms are the same.
All of the allotropes have molecular structures with corresponding molecular orbitals. It is the electrons within these orbitals that absorb light that promote electrons to higher molecular orbitals giving the characteristic colour by subtraction from white light.