A Christmas special!

I know… I know, it isn’t Halloween. I just couldn’t help myself.

Filed under Uncategorized

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:

E_{n} = -2.18 x 10-18 (Z^{2}/n^{2})

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

Filed under Chem 2, Physical Chemistry

**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 20^{th} 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

Filed under Chem 2, Physical Chemistry

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:

NO_{2(g)} + CO_{(g)} → NO_{(g)} + CO_{2(g)}

From looking at the overall equation, it would appear that two NO_{2} 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:

NO_{2} + NO_{2} → NO_{3} + NO

NO_{3} + CO → NO_{2} + CO_{2}

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 NO_{2} molecules collide to give the intermediate NO_{3} and NO. The intermediate, NO_{3} is then immediately consumed in the second collision to give the product CO_{2} and NO_{2}.

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* 5^{th} edition.

2NO_{2} + F_{2} → 2NO_{2}F

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[NO_{2}][F_{2}]

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 = k_{1}[NO_{2}][F_{2}]

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:

2O_{3} → 3O_{2}

Rate = k[O_{3}]^{2}[O_{2}]^{-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 = k_{2}[O_{3}][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

k_{1}[O_{3}] – k_{1}^{’}[O_{2}][O] – k_{2}[O_{3}][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 k_{1}^{’ }>> k_{2}, so the denominator simplifies:

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

Filed under Physical Chemistry

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 E_{a}:

Filed under Chem 2, Physical Chemistry

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, 5^{th} Edition) is shown below:

Iodide ions are oxidised in acidic solution to form triiodide ion (I_{3}^{–}), by hydrogen peroxide.

H_{2}O_{2(aq)} + 3I^{–}_{(aq)} + 2H^{+}_{(aq)} → I_{3}^{–}_{(aq)} + 2H_{2}O_{(l)}

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

We give the rate law the form:

Rate = k[H_{2}O_{2}]^{m}[I^{–}]^{n}[H^{+}]^{p}

To calculate the order with respect to H_{2}O_{2}, we look at experiments 1 and 2 where the concentration of H_{2}O_{2} 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 = 2^{m} therefore m = 1.

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

Rate = k[H_{2}O_{2}][I^{–}]

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

Rate = 1.2×10^{-2}[H_{2}O_{2}][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…

Filed under Chem 2, Physical Chemistry