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:

En = -2.18 x 10-18 (Z2/n2)

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.

Links:

http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html
http://csep10.phys.utk.edu/astr162/lect/light/bohr.html
http://regentsprep.org/Regents/physics/phys05/catomodel/bohr.htm

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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).

 

Links:

http://hyperphysics.phy-astr.gsu.edu/hbase/mod1.html

http://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality

 

 

 

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Kinetics 3: Reaction mechanisms

The balanced chemical equations used to describe a reaction are a description of the overall process, but usually doesn’t give an accurate picture of what actually happens. There may be several steps represented by the overall equation, which are important to understand the process. Kinetics can help determine what these smaller steps are.

 Elementary reactions:

Take the example:

NO2(g) + CO(g)  →  NO(g) + CO2(g)

 From looking at the overall equation, it would appear that two NO2 molecules collide with a CO molecule and BAM! you get a reaction. In reality, a successful tri-molecular collision is fairly unlikely and this reaction would sit far to the left. However, it is believed that this reaction actually occurs in two smaller steps, known as elementary reactions:

NO2 + NO2  →  NO3 + NO

NO3 + CO  →  NO2 + CO2

 An elementary reaction is a single step, involving one molecular interaction that results in a reaction. The sum of all the elementary steps gives the overall equation. In this case, two NO2 molecules collide to give the intermediate NO3 and NO. The intermediate, NO3 is then immediately consumed in the second collision to give the product CO2 and NO2.

We can use the elementary reactions to derive the overall rate law, or use the overall rate law to confirm a mechanism. For an elementary reaction, (but NOT for an overall reaction), we can assume that the rate is proportional to concentration of the reactants, raised to their stoichiometric values.

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:

Adding the two equations together gives you the overall reaction, with F as an intermediate, so that satisfies that requirement. However, the mechanism must also be in agreement with the experimental rate law. In this case, the rate law was found to be:

 

Rate = k[NO2][F2]

 

To show that the mechanism is in agreement, we look at the rate determining step, which is the slowest step. In this case, the first step is shown to be slower, so that is the reaction that will determine the overall rate, (you can only move as fast as the slowest step – for a practical demonstration of this try to hurry down Bourke St during the Christmas shopping season). We assume that the overall rate is the same as that of the rate determining step and can write:

 

Rate = k1[NO2][F2]

 

This is the same as the experimentally determined step, and so we can say that this mechanism is consistent.

 

Of course, it isn’t always that easy. Sometimes, the first step is the fast step and we need to invoke the steady state approximation to find the rate law.

 

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

2O3 → 3O2

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

 The following mechanism is proposed:

We can see that the two add together to give the overall reaction, and that the rate determining step is the second step.

 

Rate = k2[O3][O]

 

 

 

The problem with this is that there is an intermediate in the rate law, and we can’t have that. So we apply the steady state approximation – assume that the concentration of the intermediate remains constant.

This means that we are creating it at the same rate as we are using it:

 Rate production = rate consumption

Rate production – rate consumption = 0

k1[O3] – k1[O2][O] – k2[O3][O] = 0

rearrange for [O]:

We can then plug this into the rate equation to get a rate law without the intermediate:

Because the second step is much slower than the first, we can assume that k1>> k2, so the denominator simplifies:

and we assume that

We get the experimental rate law!

 

Some links:

http://www.science.uwaterloo.ca/~cchieh/cact/c123/steadyst.html

http://www.science.uwaterloo.ca/~cchieh/cact/c123/steadys2.html

http://depts.washington.edu/chemcrs/bulkdisk/chem162A_sum04/handout_Lecture_06.pdf

 

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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:

Simplify:

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:

and:

Collision theory and reaction mechanisms to come…

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Acid-base equilibria

Acids and bases are easy to deal with when they are strong, as you can assume that any reaction goes to completion. However, weak acids and bases don’t always ionise completely, so you need to look at the equilibria to fully understand them.

Acids dissociate in water to give a proton, (or hydronium ion) and the conjugate base ion.

 HA ⇄ H+ + A

 We can put this into an equilibrium expression, which gives us Ka, the acid ionisation constant:

Ka is an indication of the strength of the acid, the stronger the acid, the larger the Ka.

The same thing applies to bases;

B + H2O ⇄ HB+ + OH

Leading to the Kb expression:

Ka and Kb are the inverse of one another for a given equilibrium.

 Buffers:

Most of the stuff for acid-base equilibria is the same as for any other chemical equilibrium problem, however buffers do require a bit more discussion.

A buffer is a solution that can resist changes in pH when small amounts of acid or base are added. They are very important in a broader sense as they regulate the pH of your body and the oceans. Buffers contain a conjugate pair of a weak acid or base, generally in equal parts, for example a carbonate buffer would look like this:

H2CO3 ⇄ H+ + HCO3

where you have added H2CO3 and NaHCO3 to water. The buffer can shift its equilibrium between the acid and conjugate base when small amounts of acid or base are added. The pH will be regulated, because you have such large amounts of H2CO3 and HCO3 to start with, any small change will not affect the overall equilibrium to a large degree.

The Henderson-Hasselbalch equation is used to calculate the pH of a given buffer, and is just a rearrangement of the Ka expression:

 

 Links:

http://www.shodor.org/unchem/basic/ab/

http://www.nauticus.org/chemistry/chemacidbase.html

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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.

Links:

http://library.thinkquest.org/10429/low/equil/equil.htm

http://www.shodor.org/unchem/advanced/equ/index.html

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