The bond energy term is the energy actually required to break a bond, or rather a mole of these bonds. The fragments that you produce depend on what the actual bond is holding together.
There are two terms that are used:
1. Bond enthalpy – this is the average of a bond of a certain ‘type’ For example the C-O bond. This occurs in many diferent molecules and has a slighlty different value on each, as the actul environment of a bond affects the energy needed to break it. Thus the bond enthalpy is the average taken over a whole range of different C-O bonds.
The values that are quoted in data tables are for Bond enthalpy per mole of bonds in kJ mol-1
To understand atomic absorption it is necessary to take a look at some of the concepts involving the nature of light itself.
Light is just a form of electromagnetic radiation. The term electromagnetic radiation refers to the propagation of energy through a medium (or a vacuum) in the form of oscillations of an electrical nature with a corresponding perpendicular magnetic field (motion of electrical fields causes a magnetic field and vice versa)
So electromagnetic radiation is a form of energy that passes from one place to another at the speed of light in the form of waves.
Waves have various characteristics, such as velocity (the speed of light), wavelength (the distance from a point on one wave to the equivalent point on the next wave), frequency and amplitude.
The amplitude is proportional to the intensity of light and not particularly important at the moment. The frequency is the number of wavelengths that pass a specific point per second. There is a very simple relationship between the velocity of a wave, the wavelength and the frequency.
c (the velocity) = lambda (the wavelength) x f (the frequency)
c is a constant (the speed of light)
The energy of a wave is difectly proportional to its frequency and hence, inversely proportional to its wavelength. Both are connected to the energy be the Planck constant ‘h’.
Energy = h x f
Energy = hc/wavelength
In atomic absorption events, the energy of the incoming wave must have the exact energy that is needed for an electron to be promoted to a higher level. It absorbs this energy and gets promoted. Detection systems see this specific wavelength (or frequency) removed from the spectrum.
An absorption spectrum thus appears as a series of dark lines superimposed on a continuous spectral background. This process need not necessarily take place in the visible region of the spectrum (which is tiny compared to all the available electromagnetic spectrum, it occupies the wavelengths from about 400 -700 nm) in which case the detection cannot be visible, it must rely on apparatus designed to detect absorption in specific regions of the electromagnetic spectrum.
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)
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.
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..
2A –> A2 (slow step)
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.
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)Идея за подаръкикони