The quantum casino: tutorials |
Perhaps the most fundamental question in chemistry is ‘why do particular chemical reactions occur?’ Put more specifically, why does one reaction occur while the reverse one does not? For example, if we place a piece of magnesium in dilute hydrochloric acid, hydrogen gas is evolved and magnesium chloride solution is produced. However, no amount of bubbling hydrogen into magnesium chloride solution will get back the magnesium and the hydrochloric acid.
More formally
Mg(s) + 2HCl(aq) H_{2}(g) + MgCl_{2}(aq)
occurs, but
H_{2}(g) + MgCl_{2}(aq) Mg(s) + 2HCl(aq)
does not.
Mg(s) + 2HCl(aq) H_{2}(g) + MgCl_{2}(aq) | H_{2}(g) + MgCl_{2}(aq) Mg(s) + 2HCl(aq) |
There seems to be some direction associated with this (and other) reactions – we can say that this reaction is irreversible. What governs the ‘direction’ of this, and other chemical reactions, and can we predict it or even change it?
The absorption tube fitted to the flask contains calcium oxide to absorb carbon dioxide |
Some reactions obviously are reversible and we can push them in one direction or the other by changing the conditions such as temperature. At room temperature (around 298 K), calcium oxide will react with carbon dioxide (in the air, for example) to form calcium carbonate. Calcium oxide is used in absorption tubes to protect other chemicals from CO_{2} in the air.
CaO(s) + CO_{2}(g) CaCO_{3}(s)
However, above about 1200 K the opposite reaction occurs, calcium carbonate decomposes to form calcium oxide and gives off carbon dioxide. This is what happens in lime kilns where limestone (calcium carbonate) is heated to form lime (calcium oxide).
CaCO_{3}(s) CaO(s) + CO_{2}(g)
So it might make more sense to write
CaCO_{3}(s) CaO(s) + CO_{2}(g)
Notice that this reversal of the reaction has nothing whatsoever to do with the fact that increasing temperature makes reactions go faster. Here temperature has changed the direction of the reaction, not just its speed.
Limestone was once heated in kilns like this to turn it into lime |
The answer is chance!
A little thought will lead us to a surprising answer to our original question – ‘why do particular chemical reactions occur?’. It is that the outcome of chemical reactions is governed by chance alone. This has to be the case because chemicals cannot ‘know’ what the outcome of a reaction should be – calcium, carbon and oxygen atoms cannot ‘know’ that at a certain temperature they are ‘supposed’ to arrange themselves as CaCO_{3} , while at some other temperature they ‘should’ be arranged as CaO and CO_{2}. Atoms, molecules and ions arrange themselves in the way that is most probable by chance alone.
Initially this may seem unlikely – the outcomes of chemical reactions are totally predictable, how can they be driven by chance alone? Magnesium always reacts with hydrochloric acid to produce hydrogen gas and magnesium chloride in entirely predictable quantities - it does not seem to be down to chance in any way.
However, statistics tell us that events governed by chance are predictable if there are enough of them. We cannot predict the outcome of a single throw of a die but if we throw it very many times, we can predict with great confidence that there will be an equal number of occurrences of each number. Very large numbers is the key and, of course, in chemical systems we are always dealing with vast numbers – typically in a school experiment we might be dealing with quantities containing up to a mole of atoms or molecules. A mole is 6 x 10^{23} particles – an unimaginably large number. Industrially, reactions use many times more than this.
Entropy and disorder |
We can use a computer simulation to investigate chance behaviour. Initially we will use a simple system which is not even a chemical reaction. We’ll look at how objects arrange themselves if they flip at random between two boxes. The simulation lets us look at what happens with different numbers of objects and different starting positions. This is analogous to the situation of two gas jars separated by a cover slip. One contains a gas, and the other a vacuum. What happens when the partition is removed and why?
Before doing this simulation, teachers might like to do a more concrete demonstration of the inevitability of mixing by shaking similar numbers of marbles of two different colours (say a dozen of each) in a large measuring cylinder (preferably a plastic one to avoid the risk of breakage). Start with all the marbles of one colour at the top. Does it ever occur that, after shaking, all the marbles of one colour end up at the top? |
This shows that if the simulation is run for long enough, the number of objects in each box (as shown by the histograms) becomes equal (with small statistical deviations). This happens every time the simulation is run and happens by chance alone. Thus we have an entirely predictable outcome by chance alone. In other words chance leads to predictability. The more molecules in the simulation, the less significant are the statistical variations. With the vast number of particles in any real chemical situation, these variations become totally insignificant.
Note. It is important to realise that in the case of real molecules in a gas gar, the molecules are not ‘sucked’ into the vacuum in the empty jar. They move at random, and if they happen to be travelling in the direction of the empty jar, they will arrive there. Once there, collision with the sides of the jar or other particles may send them back again.
The outcome of the simulation is, of course, consistent with what happens in real life. If we remove the cover slip between a gas jar full of gas and an empty one, we know with absolute certainty that a gas will spread evenly between two jars. The important point to realise is that this happens by chance alone and not because the molecules of gas ‘know’ where they should be. It is also important to realise that although the numbers of molecules in each jar remain constant, they are not the same molecules – there is continued movement of molecules back and forth but the rate of movement from left to right is exactly the same as that from right to left. We call this situation a dynamic equilibrium.
Arranging objects |
A further simulation allows you to look at the probability of different arrangements of objects between two boxes with different numbers of particles. For example for 100 particles, an arrangement of 1 particle in the left hand box and 99 on the right is very unlikely while arrangements of 50 : 50 and others close to it such as 49 ; 51 are very likely. This is because there are many more ways of these distributions happening. The reason that particles spread evenly between the two boxes is that there are more arrangements of particles between the boxes that correspond to mixing than ones that correspond to separation. We say that the all–in–one–box arrangements are more ordered (less random) than the evenly–spaced–out ones.
For just two particles, A and B, the arrangements in Table 1 are possible. This leads to three possible observable ‘states’ of the system – we cannot distinguish between the two 1/1 arrangements because the particles are identical.
Arrangements | State | Number of arrangements that lead to this state | |
Box 1 | Box 2 | ||
A and B | none | 2/0 | 1 |
A only | B only | 1/1 | 2 |
B only | A only | 1/1 | |
none | B and A | 0/2 | 1 |
Table 1 The possible arrangements of two particles between two boxes
A 1/1 mixing is obtainable in two ways – there are two arrangements that lead to this state. It is twice as likely as either of the ‘unmixed’ states which each have only one arrangement. There is a total of four arrangements.
For three particles, A, B and C, the arrangements in Table 2 are possible, leading to four observable states.
Arrangements | State | Number of arrangements that lead to this state | |
Box 1 | Box 2 | ||
A, B, C | none | 3/0 | 1 |
A, B | C only | 2/1 | 3 |
B, C | A only | 2/1 | |
C, A | B only | 2/1 | |
A only | B, C | 1/2 | 3 |
B only | C, A | 1/2 | |
C only | A, B | 1/2 | |
none | A, B, C | 0/3 | 1 |
Table 2 The possible arrangements of four particles between two boxes
Students might be asked to do a similar table for four particles in two boxes – there is a total of 16 (2^{4}) arrangements and five observable states. In general there are x^{y} arrangements where x = number of boxes and y = number of particles.
With six particles, there is still only one way of having all the particles in box A but 20 ways of arranging three in box A and 3 in box B.
Notice how states with more mixing are more probable than those with less because there are more arrangements that lead to them. We say that these mixed arrangements are more disordered or random than the ‘unmixed’ ones.
The direction of chemical reactions |