Category Archives: Physical Chemistry

Quantum theory 2: The Bohr model of the atom

At the beginning of the 20th century, the atom was explained using Rutherford’s model of a small, central nucleus with orbiting electrons like planets around a star. You still see this type of atom used in symbols today:

This model doesn’t work – orbiting electrons would lose energy and eventually (in about 10-10s) fall into the nucleus. Niels Bohr came up with the solution to this problem by considering the absorption and emission spectra of hydrogen. When you shine light through hydrogen gas, you don’t see a continuous spectrum, but a line spectrum (as shown below).

Bohr was able to explain that by postulating two things.
1) An electron can only exist in certain energy levels within the atom and 2) the electron can undergo a transition from one energy level to another. It is these transitions that we see on the line spectrum.

The energy of each energy level can be calculated using the following:


where Z is the atomic number and n is the energy level. This equation can then be used to find the energy of the transitions.

The transitions are grouped together in “series”. Each series corresponds to all the transitions to/from a certain energy level.

The Bohr model accurately describes the energy levels of single electron atoms, but fails for anything larger. For that, we need to look at exactly how the electrons behave when they are near the nucleus.


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Quantum theory 1: The dual nature of light

The wave nature of light:

One way of looking at light is as an electromagnetic wave, consisting of perpendicular, oscillating electric and magnetic fields.




We characterise it by is wavelength, (and therefore its frequency). The wavelength, (λ) is the distance between two waves, and the frequency (ν) is the number of waves that pass through a point per unit time.

The wavelength and frequency are related to one another by c, the speed of light:

c = λν

The range of electromagnetic radiation is called the electromagnetic spectrum, (shown below) and ranges from wavelengths smaller than atoms (gamma rays) to kilometres (radio waves). Visible light makes up a very small section, ranging from about 400 nm (violet) to 700 nm (red).

There is significant evidence for the wave nature of light – we see diffraction effects, it can be polarised and it can interfere with itself.

The particle nature of light:

Isaac Newton was a big proponent of the particle, (or, in his words “corpuscular”) theory of light, but this view was largely abandoned in favour of the wave model until the beginning of the 20th century. Einstein won his Nobel prize for his paper on the photoelectric effect which showed the energy in light was “quantised”, based on the work of Max Planck.

Planck showed that there was a close relationship between the light emitted by a solid and its temperature. A material will glow “white hot” at 1200°C, but red at 750°C. He found that, when using Boltzmann’s statistical version of the second law of thermodynamics, the energy emitted by a solid could only be E = nhν, where n is an integer value – ie, the energy is quantised. The value h is Planck’s constant and equal to 6.63 x 10-34 J.s.

Einstein took this idea and showed that it applied to the photoelectric effect. This was an experiment where light is shone on the clean surface of a metal and the electrons released are measured. Electrons are ejected from an atom when enough energy is absorbed; however this only occurs when light above a certain frequency is used. This threshold frequency is characteristic of each metal.

To explain this, the electron must be ejected using the energy from a single photon, and that photon must have greater energy than the ionisation energy of the atom. If the photon doesn’t have enough energy, no matter how many photons hit the metal, the electron won’t be ejected. So, light must be quantised and behave as a particle!

How do we reconcile these two behaviours? Not in any really satisfying way. We say that light behaves as both a wave and a particle, (known as the wave-particle duality).






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Kinetics 3: Reaction mechanisms


 Elementary reactions:

Take the example:



NO2 + NO2  →  NO3 + NO

NO3 + CO  →  NO2 + CO2



The following example is modified from Ebbing’s General Chemistry 5th edition.

 2NO2 + F2  →  2NO2F

 The above reaction is believed to occur via two elementary steps:



Rate = k[NO2][F2]




Rate = k1[NO2][F2]






For example, let’s look at the destruction of ozone:

2O3 → 3O2

Rate = k[O3]2[O2]-1

 The following mechanism is proposed:



Rate = k2[O3][O]






 Rate production = rate consumption

Rate production – rate consumption = 0


rearrange for [O]:



and we assume that

We get the experimental rate law!


Some links:


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Kinetics 2: Collision theory

It is well established that increasing the temperature, increases the rate of a reaction. That dependence can be explained by looking at what is happening on a molecular level during a reaction. You might want to go back and read up on the kinetic theory of gases before we go on.

According to collision theory, for a reaction to occur, reactant molecule must collide with sufficient energy (activation energy) and the correct orientation. Activation energy is the potential barrier that the molecule has to overcome for bonds to break and reform.

The frequency of “correct” collisions is related to the temperature, as molecules will have higher kinetic energy at higher temperatures as well as the activation energy and an orientation factor. Combining these gives us the Arrhenius equation:


 We can make this equation more useful by taking the log:


We can now measure the rate at two temperatures and use this to find Ea:


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Kinetics 1: Rates of reaction

Chemical kinetics is the study of reaction rates, which includes how fast a reaction occurs, how the conditions change the rate and what molecular interactions are involved in a reaction step.
We define the reaction rate as the amount of reactant used, or amount of product used per unit time, and always has the units M/s. As rates are dependent on concentration, rate laws are typically written like so:

For that imaginary reaction, we have defined the rate as the decrease in concentration of A (which is why it is negative) over change in time. The rate is proportional to the concentrations raised to some power, (which must be determined experimentally, it is not the stoichiometric value!) and k, the rate (proportionality) constant.


The exponent values are referred to as the order of the reaction with respect to whichever species it is associated with, (in the previous example, the rate is nth order with respect to the concentration of B). The overall order of the reaction is the sum of the exponents.


Determining the rate law (initial rates method):


To determine the rate law, you need to know the order of reaction with respect to each reactant, and find the rate law. Typically, the experiment measures the initial rate at different concentrations, so the order can be calculated using simultaneous equations. Once the orders have been determined, it is easy to find k. An example (taken from General Chemistry, Ebbing, 5th Edition) is shown below:


Iodide ions are oxidised in acidic solution to form triiodide ion (I3), by hydrogen peroxide.

 H2O2(aq) + 3I(aq) + 2H+(aq) → I3(aq) + 2H2O(l)

 The reaction was run at a variety of concentrations and the rate recorded:

We give the rate law the form:


Rate = k[H2O2]m[I]n[H+]p


To calculate the order with respect to H2O2, we look at experiments 1 and 2 where the concentration of H2O2 doubles, but the other two stay the same. We divide one by the other:

We substitute the values in and see that the I, H+ and k values cancel leaving:


2 = 2m therefore m = 1.

This procedure is repeated for [I] and [H+], giving the rate law

 Rate = k[H2O2][I]

 The values for one of the experiments can be substituted in to find k, giving us:

 Rate = 1.2×10-2[H2O2][I]


Integrated rate laws:

The initial rate law is all well and good, but it doesn’t tell you what the concentration of a species will be at a given time during the reaction. To find this, we integrate the rate law and get the following:

For a first order rate law, (eg. Rate = k[A]a) we get:

ln[A]t = -kt + ln[A]0

and a half life, (the time taken for half of your reactant to react) of:

For a second order reaction, we get the following:


Collision theory and reaction mechanisms to come…

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Acid-base equilibria



 HA ⇄ H+ + A



The same thing applies to bases;

B + H2O ⇄ HB+ + OH

Leading to the Kb expression:





H2CO3 ⇄ H+ + HCO3





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Chemical equilibrium

Not all chemical reactions go to completion, most go backwards and forwards around a point somewhere in the middle, eventually coming to a stop. Of course, the reaction doesn’t cease at this point, the reaction continues to hover around the equilibrium point, which is why it is known as dynamic equilibrium.

Technically, equilibrium is defined as the point at which the rates of the forward and reverse reactions are equal.

We express the point of equilibrium using the equilibrium expression:

for the reaction aA + bB ⇄ cC + dD. K is the equilibrium constant, the larger it is, the more the reaction favours the products. Obtaining the equilibrium constant is simply a matter of substituting the equilibrium concentrations into the equilibrium constant.

Alternatively, you may be given K and the initial concentrations and asked to find the equilibrium concentration. There is an example here,

because I can’t get the example to format here. The procedure to answer these problems is as follow:

  1. Set up a table of concentrations showing the initial, change and final concentrations for all products and reactants.
  2. Substitute the “equilibrium concentrations” (including the x) into your equilibrium expression.
  3. Solve!

And for those of you who have forgotten year 10 maths – the quadratic equation and a reminder of how to use it is here. For those of you who do remember, you can see a cat on a vacuum cleaner.


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