M. Bodenstain and S.C. Lind studied the reaction of hydrogen and bromine gas in 1906 by heating the mixture.
They reported the rate of formation of HBr as;
| d[HBr] | k'[H2][Br2]� | ||
| = | |||
| dt | k" + [HBr] [Br2]‾� |
Where k' and k" are constants.
It took chemists thirteen years later to come up with the mechanism. It was Christiansen, Herzsfeld and Polanyi who proposed that;
| (1) | Br2 | → | 2 Br | k1 | |
| (2) | Br + H2 | → | HBr + H | k2 | |
| (3) | H + Br2 | → | HBr + Br | k3 | |
| (4) | H + HBr | → | H2 + Br | k4 | |
| (5) | 2Br | → | Br2 | k5 |
Whereas the chemical reaction equation represents the sum total of what is happening in the reaction, this sequence of elementary steps represents the actual reactions; where the species collide with each other.
Tutorial 1
- Why are all the elementary steps involved exclusively only one or two reagents. Why not three?
- What you understand about an elementary step involving only one molecule? Answer
Since the elementary steps represents the actual reactions then the rate of each step can be represented by
For example for the second elementary step we can write;
| d[Br] / dt | = | - k2[Br][H2] | Or | d[H] / dt | = | k2[Br][H2] |
Looking at the mechanism, the production of HBr is from elementary steps 2 and step 3, but consumed in step 4. So the rate of formation of HBr should be represented by;
| d[HBr] / dt | = | k2[Br][H2] + k3[H][Br2] - k4[H][HBr] |
The intermediate species in this mechanism would be Br and H. At all time their concentrations should be zero - the steady state approximation. Sort of "now I see you now I don't". So the next item of business will be to substitute all the intermediate species in this equation.
Accounting for sum total of the production and consumption of Br and H in all the elementary steps will give us;
| d[Br] / dt | = | 2k1[Br2] - 2k2[Br][H2] + k3[H][Br2] + k4[H][HBr] - 2k5[Br][Br] = ; 0 |
| d[H] / dt | = | k2[Br][H2] - k3[H][Br2] - k4[H][HBr] = 0 |
Tutorial 2
Why is the rate of formation of Br in step (1) has a coefficient of 2 in front of the rate constant? Answer
| (k1 / k5)� [H2][Br2]� | ||||||||
| [Br] | = | [ (k1 / k5) [Br2] ]� | [H] | = | k2 | |||
| k3[Br2] + k4[HBr] |
| d[HBr] | 2 k3 k2 k4‾� k1� k5‾� [H2][Br2]� | ||
| So | = | ||
| dt | k3 k4‾� + [HBr] [Br2]‾� |
THE IMPORTANCE of UNDERSTANDING THE MECHANISM OF A REACTION
Unless we understand the mechanism of a chemical reaction we cannot claim to understand the chemistry of the reaction. Let us see what we can gain from the mechanism of the reaction between hydrogen and bromine.
It is very clear that with the thermal dissociation of a small amount of bromine molecule, step (1), all the hydrogen present can be reacted to hydrogen bromide. This is because; for each atom of bromine formed in step (1) it will react with one molecule of hydrogen and eventually reproduce an atom of bromine in step (3). This can go on and on if not for step (5). Since reaction depends on the frequency of collision the probability of Br colliding with hydrogen molecule is very much higher then Br purely on concentration consideration. Intermediate species is "not suppose even to exist" remember.
The cycle of a Br atom reacting to finally regenerate itself resulted in the reaction being known as a chain reaction. A chain reaction consists of;
- an initiation step - step (1);
- a propagation step - steps (2) and (3); and
- a termination step - step (5)
Step (4) does no actual harm to the formation of HBr as it gives a Br atom which goes on to produce HBr. Just a delay action known in chain reaction as an inhibition step.
It must always be understood that not all the steps are of equal importance although they must all follow the sequence. For example the inhibition and termination reactions become important only when the amount of hydrogen and bromine molecules have been greatly reduced. By this time the frequency of collision according to steps (4) and (5) significantly increase. In other words the inhibition and termination reactions are only significant at the end of the reaction. For the commercial production of HBr just remember to continuously add bromine and hydrogen molecules into the process and you can forget about steps (4) and (5).
Bromine atom is very reactive because of an unpaired electron in its valence orbital. Active atoms (or any chemical species) with an unpaired electron in its valence orbital are known as a free radical.
So this particular reaction is referred to as a free radical chain reaction.
The heat needed for the reaction is for the dissociation of the bromine molecules into bromine atoms. The rest of the elementary steps involve reactions with very reactive free radicals and the energy barriers are very low.
Elementary steps Activation Energy Br2 → 2Br 188.3 kJ mol‾� Br + H2 → HBr + H 75.31 kJ mol‾� H + Br2 → HBR + H 4.184 kJ mol‾� The energies were determined by separate experiments.
If we can only get bromine atom we do not need heat for the reaction. Bromine atom can be obtained by UV radiation.
Br2 + hν → 2Br So if we mixed hydrogen and bromine and direct a UV radiation at the mixture for a short period of time we should be able to make the reaction to work. This was indeed the case.
I hope you see the importance of studying the mechanism of reactions in chemistry. It is through such understandings that we can make the reaction go faster, proceed with less energy, increase the yield of the desired products, reduce by-reactions, and help in designing catalysts.
