## Attack of the php insertion robots

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

## 2009 IB Chemistry syllabus review

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!

CCикони

## Why do we study kinetics of reactions?

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.

## How can enthalpy of formation values be used to calculate reaction enthalpy?

Friday, May 18th, 2007

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)Идея за подаръкикони

## What is the structure of sulphur (IV) oxide?

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.

## How is the volume measurement of hydrogen peroxide related to its molarity?

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

## How can I test for an alcohol?

May 21st, 2007
To detect the presence of the OH group regardless there are several simple tests that can be done, however they all have the disadvantage of interference by water.
A small piece of sodium metal can be placed in the alcohol and a steady stream of hydrogen bubbles gives a positive indication. Remember that the presence of water will also cause hydrogen to be evolved.
The suspected alcohol may be mixed with some PCl5 and the evolution of misty HCl gas is a positive indication of an OH group. Once again water interferes.
—————————————————————————
Primary, secondary and tertiary alcohols
There are three different types of alcohol depending on what is attached to the carbon holding the OH group.
· Primary – one alkyl group attached to the carbon holding the -OH
· Secondary – two alkyl groups attached to the carbon holding the -OH
· Tertiary alcohols – three alkyl groups attached to the carbon holding the -OH
To differentiate between the three it is possible to use oxidation with sodium dichromate in dilute sulphuric acid and heat.
The 1º alcohol oxidises to an aldehyde (alkanal) and then to a carboxylic acid (alkanoic acid)
The 2º alcohol oxidises to a ketone (alkanone) and stops there.
The 3º alcohol cannot be oxidised under these conditions.
Another (infrequently used) test is Lucas’ test for 1º, 2º, 3º alcohols.
It involves shaking the alcohol with ZnCl2 and dilute HCl. The 3º alcohol goes cloudy almost at once, the 2º alcohol goes cloudy after a few minutes whereas the 1º alcohol needs concentrated HCl to go cloudy. The cloudiness is due to the formation of the haloalkane that is immiscible with the aqueous solution of the zinc chloride and forms tiny droplets of an organic phase within the aqueous phase.

## What actually happens when a liquid boils?

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.
ikoni

## How does electrolysis work?

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
or
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.

## What does oxidation state actually mean?

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.