Why is potassium used in matches

The match head contains an oxidising agent, commonly potassium chlorate, and glue to bind it to further abrasive materials and other additive compounds. When the match is struck, a small amount of the red phosphorus on the striking surface is converted into white phosphorus, which then ignites.

The heat from this ignites the potassium chlorate, and the match head bursts into flame. This is because they contain phosphorus in the match head, in the form of phosphorus sesquisulfide.

Other than this difference, however, they still function in much the same way. You can see the chemical reaction that occurs when a match is struck happening in super-slow motion in this amazing video by American cinematographer Alan Teitel or with chemistry commentary from ACS Reactions here.

You probably know about chemical reactions. Heat can help kick off some chemical reactions or make them happen faster. To our eyes it just looks like a red powder.

But if you zoomed right in to see how all its atoms are arranged, it would look like a bunch of triangles and other shapes stuck together into a long chain. When you rub the match on the box, you get friction, which means you get heat. This heat causes a small amount of the red phosphorus chain to be broken apart. It reacts immediately with a gas in the air called oxygen. This will create a lot more heat.

So the story so far: the friction breaks the red phosphorous chain, which allows the white phosphorous to react with oxygen and the match starts to get hot. Friction and white phosphorus have provided the starting heat, and now the match needs fuel and oxygen to continue to burn. The side of the box contains red phosphorus, binder and powdered glass.

The heat generated by friction when the match is struck causes a minute amount of red phosphorus to be converted to white phosphorus, which ignites spontaneously in air. This sets off the decomposition of potassium chlorate to give oxygen and potassium chloride.

The sulfur catches fire and ignites the wood. Applying these values and the mole ratios to the equation above, the mass of phosphorus sesquisulfide needed is seen to Hence the two chemicals must be combined in a ratio of In the final part, part 5 of part a , you are given the standard molar enthalpy changes of formation for each of the reactants and products involved in the reaction on the match head, and asked to calculate the standard enthalpy change for this reaction. The standard enthalpy change of formation is the enthalpy change when one mole of a compound is formed from its constituent elements under standard conditions.

Thus the standard molar enthalpy changes of formation can be represented by an arrow going from the elements to the compound whose enthalpy change of formation you are given. A value can be added to that arrow to represent the enthalpy change for the formation of the number of moles of compound in question. For the reactants, you need to go from the compound to the elements from which it is formed. This is the reverse of the enthalpy of formation and hence you these values must be subtracted.

For the products, you need to move from the elements to the compounds and hence these values are added. With calculations such as this it is very easy to make a silly mistake when placing the numbers into the calculator.

It is good practice therefore to write out the answer for each individual calculation and then complete the final calculation in a separate step. You notice here that this is a very large negative enthalpy change showing us that the reaction is highly exothermic as you would expect. The question now moves on to look at the phosphorus sulfides in more detail.

You are told that phosphorus sulfides can be made by heating white phosphorus with sulfur. When this reaction is carried out at low temperature, a range of products from P 4 S 3 to P 4 S 10 are made.

In part b of the question you are asked to predict how many peaks would be seen in the 31 P NMR spectrum of three of these phosphorus sulfides. To answer this question, we need to look at each of the three sulfides in turn, you need to look for any symmetry in the molecules which would create phosphorus atoms in equivalent environments.

In P 4 S 4 you can see that there is a plane of symmetry through the middle of the molecule. Therefore the two phosphorus atoms highlighted are in equivalent chemical environments and so you would expect to see 3 peaks in the 31 P NMR spectrum. Finally in P 4 S 6 the two P atoms highlighted are in identical chemical environments owing to the plane of symmetry which cuts through the middle of the P-P bond. Therefore you would expect to see 3 peaks in the 31 P NMR spectrum as a result of the 3 different P environments highlighted.

Finally you are told that P 4 S 4 can exist in two different isomeric forms, only one of which is shown. Part c of the question asks you to suggest a structure for this second isomer. When drawing the structure it is important to ensure that each element has the correct number of bonds. Phosphorus is in group 5 and so needs to form three covalent bonds to fill its outer shell. Sulfur is in group 6 and so needs to form only two covalent bonds.

Since the 31 P NMR spectrum of the second isomer shows only a single peak, the molecule must have a high degree of symmetry. From maths perhaps or studies on the shapes of molecules you may realise that the most symmetric way to arrange four atoms is at the four corners of a tetrahedron.