12 April
3 January 2019
The fertiliser industry has many different chemical processes involved in synthesising fertilisers to boost crop yields, however there are two that stand out as being the most significant in human history.
In the early 1900s, Fritz Haber and Carl Bosch developed a technique for reacting hydrogen gas with nitrogen gas to form the chemical ammonia. Ammonia is a vital precursor for developing nitrate based fertilisers.
Methane or steam is used as a source for hydrogen and the nitrogen is acquired from the air.
The two gases are passed over a hot iron catalyst (at 400℃) and at 200atm pressure.
The reaction is in dynamic equilibrium - meaning that altering the reaction conditions can cause the reaction to shift, reforming the products.
N2(g) + 3H2(g) ⇌ 2NH3(g)
This temperature is used because, while the reaction rate would be greater at higher temperatures, ammonia decomposes back into its reactants at higher temperature. The reaction favours the products at higher pressure, however it is more expensive to maintain a high pressure, so a compromise is used. The reaction uses a catalyst to help speed up the reaction and allow it to be efficient at the lower than desirable temperature and pressure.
Under these conditions the percentage yield of the reaction is less than 20%. This is because much of the ammonia formed decomposes back into its reactants. To make it more efficient, the nitrogen and hydrogen produced by the decomposition of ammonia are reused as reactants again.
Nitrate fertilisers are salts of reactions of nitric acid and a base. Nitric acid can be made from the ammonia produced in the Haber Bosch Process in a reaction known as the Ostwald Process.
The Ostwald Process involves taking oxygen from air and ammonia and reacting them over a platinum gauze catalyst at 800℃. This reaction is exothermic so the catalyst does not need to be heated. This produces nitrogen oxide (NO). This is further reacted with more oxygen to produce the brown gas nitrogen dioxide (NO2). Additional oxygen is added and the gas reacted with water to produce nitric acid.
Nitric acid has a variety of uses from a precursor to forming other chemicals, to making fertilisers and explosives.
An Analytical Technique Using the Mole
23 December 2018
A titration is designed to work out the concentration of one chemical if one reacts an accurately measured volume of it with an accurately measured volume with known concentration of another solution.
To get an accurate volume of a solution with an unknown concentration, one uses a pipette. This is a glass tube with a single graduation mark which indicates a specific volume when filled to this point. Pipettes come in a variety of volumes - 1 mL, 5 mL, 10 mL, 20 mL, etc. Using a pipette, a specific volume is pipetted into a conical flask. This need not matter what the volume is, so long as it is known and of a sufficient quantity that an end point to a reaction be readily observable. The flask is often placed on a white tile so as to make any colour changes clearly visible.
To this solution, an indicator is added. This indicator must change colour distinctly when the reaction is complete. However, if the products of the reaction are clearly visible, i.e. if a distinct colour change occurs, then there may be no need for an indicator to be used.
A burette is used to gradually add a chemical of known concentration to the unknown concentration solution. This apparatus has volume graduation marks down its side to measure the volume of a liquid which has been released from it into a conical flask placed underneath it.
The starting volume of liquid in the burette is recorded and then the valve is opened and the reactant pours into the solution of unknown concentration. This is continued until there is a colour change in the indicator. At this point the valve is closed and the flow of reactant ceased. In the first run of the titration, a rough titre is recorded - i.e. the volume which passed out of the burette. This is read from the bottom of the meniscus of the burette’s solution and subtracted from the original volume of solution in the burette.
This is considered the rough titre and is never used in any subsequent calculations as it is inaccurate. Its purpose is to show the user roughly where the end point of the reaction is and where to stop the burette to get a more accurate reading.
The second run will continue until just before the value of the rough titre (possibly aiming for 3cm3 less than it). After this point, the solution in the burette is added dropwise (one drop at a time) until the user observes a clear change in colour. This point is noted as the titre.
The experiment is repeated until at least 2 concordant results (values similar to one another) are obtained.
These concordant results are used to calculate and average volume of the solution of known concentration required to react fully with the solution of unknown concentration.
The next step involves a calculation to ascertain the concentration of the solution with the hitherto unknown concentration.
For this, the number of moles of the solution in the burette is calculated.
This is perhaps best elucidated through a worked example.
Let’s say that a scientist desired to know the concentration of a solution of an alkali (perhaps sodium hydroxide), however the label on its container has worn off and its concentration is unknown.
The scientist sets up a titration where hydrochloric acid with a concentration of 0.1molL-1 is titrated into a conical flask containing 20mL of the solution of NaOH.
The results of the experiment are recorded in the following table :
The rough titre can be ignored as it differs from the main two runs significantly. Therefore the average titre can be calculated from the 1st run and the 2nd run thus :
To continue, we must first establish the molar ratio of both the acid and the alkali. To do this we require the balanced equation of the reaction :
HCl + NaOH → NaCl + H2O
From this it is clear that there is a 1 : 1 molar ratio. Therefore whatever the number of moles of the acid is, then there will be the same number of moles of alkali.
Therefore, in the burette, we added 0.00256 mol of HCl to the alkali. Since the balanced equation shows a 1 : 1 molar ratio in the reaction, there was 0.00256 mol of NaOH in the conical flask. Since the 0.00256 mol of NaOH took up 20cm3 we have a concentration of 1.28molL-1.
This example covers a simple 1 : 1 molar ratio. What happens when we don’t have such a simple reaction?
Let’s look at a titration with sulphuric acid and potassium hydroxide.
H2SO4 + 2KOH → K2SO4 + 2H2O
Here we have 1 mole of H2SO4 for every 2 moles of KOH.
This gives us a molar ratio of 1 : 2.
Suppose we performed a titration experiment and the average titre for H2SO4 was 24.8cm3 and its concentration was 0.1molL-1. What would the concentration of the KOH be if 10cm3 was used in the conical flask?
In this example, there were 0.00248 mol of HCl in the burette. This reacted with double this amount of KOH, meaning that the molar ratio was 1 : 2, meaning that we had 0.00496 mol of KOH. As this was in 10 cm3 in the conical flask, the concentration was 0.496 molL-1.