General Formulae

Posted on 1st Jun 20211st Jun 2021 Posted in General Formulae, Higher Topics, National 5 Topics, Organic Chemistry, UncategorizedLeave a comment

A common question to be asked in the exam, both in National 5 and Higher (and for GCSE and A-Level too) is to provide the general formula for a specific homologous series.

This type of question can come in two forms: specifying the homologous series in question and asking, “what is the general formula for the alcohols?”; or showing you the structure of a molecule and asking you to “write the general formula for the homologous series to which this molecule belongs”, or something similar.

The first type of question is just testing your memory of all the different homologous series that you have encountered – simply recalling the general formula (C_nH_2n+1OH if you were wondering).

In the second type above, you are challenged to recognise what homologous series the molecule belongs to first. A variation of this might be “this hydrocarbon decolourises bromine water – what is its general formula?”. In this case, you need to deduce that decolourising bromine water is a key reaction (and identifying test) for an alkene – from this identification, you then state its general formula – C_nH_2n. Effectively in this type of question, your brain goes through a two-part process – recognition, followed by recall – applying it to the situation of the question.

These types of questions are actually very easy, however many pupils (and teachers dare I say!) forget the general formulae. This is quite understandable since there are a few, and they differ in somewhat counter intuitive ways.

If you are inclined to memorise them, this will provide you an excellent advantage for this type of question, as well as questions asking you to apply them. However, for those of us who struggle to remember them, understanding where the general formula actually comes from and being able to work them out for yourself is vital; and perhaps more beneficial for you in the long run than just simple memory work.

Working it out

Just what is a general formula? I think pupils often miss the point of them – thinking that they are just a simple curiosity associated with a homologous series; and something else to cram into revision before the exam. The general formula of a homologous series describes the relationship between the quantities of each element in the formula of all the molecules in that homologous series. No matter what the size is of the molecule, what structural isomer is being considered, all the molecules will fit the same general formula.

To illustrate this, let’s first consider the alkanes. Pictured below is butane. It has a chemical formula of C₄H₁₀. This means that it contains 4 carbon atoms and 10 hydrogen atoms. Before we look at how isomers of butane also possess this formula, lets first work out the general formula.

In a general formula, it is conventional to set the number of carbon atoms as the variable n. In butane’s case, n = 4. The general formula should allow us to calculate the number of hydrogen atoms also present in the molecule from the given number of carbon atoms. Let’s set this up:

C_nH_something

For butane :

C₄H_something

Lets be a little more scientific than something – let’s say the number of hydrogen atoms is x :

C₄H_x

We know from butane’s formula that x must be 10, so how do we make 10 from 4?

We could add 6 – but does that work?

C_nH_n+6

C₍₄₎H₍₄₊₆₎ = C₄H₁₀

It obviously does for butane but let’s try it with another alkane – methane. We know that methane’s formula is CH₄

C₍₁₎H₍₁₊₆₎ = CH₇ – This is in no way possible!

Simply adding 6 cannot work, so let’s try multiplying. We shall start with doubling:

C_nH_2n

C₍₁₎H_(2×1)= CH₂ – still not right, but 2 is 2 away from 4, so let’s add 2:

C₍₁₎H_{(2×1 +2)} = CH₄ – bingo! It works.

That gives us a potential general formula of C_nH_2n+2, but does it work with other molecules in this homologous series? Let’s return to butane (where n = 4) to check:

C₍₄₎H_{(2×4 + 2)} = C₄H₁₀ – It works!

And with octane? (which should be C₈H₁₈):

C₍₈₎H_{(2×8 + 2)}= C₈H₁₈

This confirms that our method is correct and that we have finally arrived at a working general formula.

Branching out

This works with single chain hydrocarbons, but does it also work for branched hydrocarbon chains?

Let’s take an isomer of butane – 2-methylpropane.

*The structural formula of 2-methylpropane*

Does this molecule fit the same general formula of butane? An interesting question, one which we can answer in two different ways. Firstly, you could simply compare its formula with that of butane – if they match, then the answer must be yes. As we saw before, butane has a chemical formula of C₄H₁₀. To work out the formula of 2-methylpropane, simply count the number of carbon atoms and hydrogen atoms. It should be obvious that the formula of 2-methylpropane is also C₄H₁₀ – since this is the same as butane’s it must therefore have the same general formula as that of butane. An alternative way of approaching this question is to work out its general formula directly and comparing it to that of the alkanes. Since the formula is C₄H₁₀, this is of the form C_nH_2n+2, it therefore does indeed have the same general formula as butane.

Expanding to other functional groups

Something like the alkanes is easy to remember but being able to understand the process to work it out allows you to realise the logic in the concept; making it possible to apply this skill to other homologous series. Let’s look at the alcohols. The general formula was mentioned above – C_nH_2n+1OH. This looks odd in comparison – you will have encountered alkanes, alkenes, and cycloalkanes, each with general formula involving just multiplying or a combination of multiplying and adding two, but this involves adding one. Where does that come from!? This is where most pupils struggle the most – going from the familiar to this slight deviation. Let’s compare two similar sized molecules to see the origins of this.

*The structures of ethane (left) and ethanol (right)*

We do not have to look far to see where this comes from. Look at the above structural formulae. What is the difference? While the difference in the number of hydrogens in the general formulae may be counter intuitive, spotting the difference in the structure should be obvious! One of the hydrogens is replaced in the alcohol with a hydroxyl functional group. Therefore, there will be one less hydrogen in the general formula! (Note here that in total there are the same number of hydrogens present, but one is part of the hydroxyl group – compare this to a molecule with a halide instead of the hydroxyl group if you need to clarify this). This is where the 2n+1 comes from in the general formula – one hydrogen is swapped for the OH group, so we remove it appropriately from the equation.

Let’s experiment with this concept:

Propan-2-ol – does this have the same general formula as ethanol?

The formula is C₃H₇OH. How does 7 relate to 3? It’s 1 more than double it:

3 x 2 = 6, 6+1 = 7

Therefore, the general formula is C_nH_2n+1OH.

Does this fit with something more complex? A challenge now – what is the general formula of the homologous series to which 4‐ethyl‐3,4‐dimethylheptan‐2‐ol, belongs? Hopefully you know how to name branched chain hydrocarbons! If you can, then naming branched chain alcohols is exactly the same (I’ll write an article on naming and drawing structures later – pupils have endless fun with massively complex molecules once they get the hang of the naming process!). The longest chain is a heptane – so 7 carbons, 3 branches, an ethyl on the third to last carbon, then a methyl group on both of the next carbons, followed by the hydroxyl group on the second carbon atom from the other end.

*The structural formula of 4‐ethyl‐3,4‐dimethylheptan‐2‐ol*

Does this have the same general formula as propan-2-ol? Let’s start by working out its formula – count each of the atoms present. The answer is C₁₁H₂₃OH (which incidentally is an isomer of undecanol). In order to answer our original question, lets take a closer look at those numbers. If n = 11 (the number of carbon atoms), how do we get to 23? Always start by doubling for general formula, as you may have noticed!

11 x 2 = 22

So how do we get to 23 from 22? Add 1!

11 x 2 + 1 = 23

So the general formula is indeed C_nH_2n+1OH. The answer is yes this does have the same general formula as propan-2-ol.

Involving the Halogens

Mentioned previously were the halides. Organohalides are not specifically part of the national 5 course, however they are mentioned; and you should be able to recognise them as addition products from alkenes (particularly the organobromides when bromine is added to an alkene).

Consider the structure of 1-chloropentane, which is an example of the chloroalkanes.

What does this remind you of? It bears a striking resemblance to an alcohol, with the hydroxyl group replaced by chlorine.

Therefore, obtaining its general formula must follow the same process. One hydrogen atom has been completely replaced by the chlorine, meaning that there is 1 less hydrogen than in the alkane equivalent molecule, pentane.

What is the general formula of the chloroalkanes? Start as always with working out the chemical formula of this example first. 5 carbon atoms, 11 hydrogens, and 1 chlorine, so the formula is C₅H₁₁Cl. What is the general formula? Well, let’s say that n = 5, how do we get to 11? Doubling it gives us 10 (5 x 2 = 10) which is 1 away from 11, so add 1. 5 x 2 + 1 = 11. This gives us a general formula of C_nH_2n+1Cl for the chloroalkanes. Exactly the same process could be applied to any of the organohalides – fluoroalkanes C_nH_2n+1F; bromoalkanes C_nH_2n+1Br; or iodoalkanes C_nH_2n+1I.

When working out the general formula, consider the similarities. In the 1-chloropentane example, the molecule is almost identical to that of pentane. Since one hydrogen is replaced by chlorine, the number of hydrogens decreases by 1 and is swapped by chlorine. If asked its formula, you could approach this by first giving the formula of pentane, C₅H₁₂, then modifying this to remove a hydrogen, C₅H₁₁, and finally adding on the chlorine atom, C₅H₁₁Cl.

Carbohydrate molecules

Special mention should go the carboxylic acids, aldehydes and ketones as they have some rather special chemistry, meaning that confusion often occurs with the actual number of carbon atoms in their structures.

Starting with the carboxylic acids, what is their functional group? This might seem odd at first, since the carbon atom on the end actually forms a part of this group. We have a hydroxyl group (-OH), but also a carbonyl (C=O). When both of these groups originate on a terminal position on the same carbon atom, that carbon is a part of that functional group, -COOH. My higher pupils affectionately referred to this as the “COOH group”, spoken with the sound a pigeon makes, as a way of remembering it (to each their own). The point being that the carbon is as much a part of this functional group as the oxygen and hydrogen atoms are. This means that the n in the general formula is actually only that of the chain minus that of the carbon attached to the functional group. It can be written two ways:

C_nH_2n+1COOH

C_n-1H_2(n-1)+1COOH

In the first form, n is equal to the number of carbons present in a chain attached to the carbon of the functional group. In the second form, n is the total number of carbons in the whole molecule – functional group and all. Let’s examine why this is. Take ethanoic acid. This has two carbon atoms.

*The structural formula of ethanoic acid*

In the first formula, you have to be careful to take into account the COOH group. If you put the n = 2 into the formula, it will look like this :

C₍₂₎H_{(2 x 2 + 1)}COOH = C₂H₅COOH

This is wrong! It shows that there are 3 carbon atoms in ethanoic acid – we know from the structure (and the eth– prefix) that there should be 2!

Since there is 1 carbon atom already in the COOH group, this means that there is only 1 carbon atom remaining in ethanoic acid to put into the formula. It should look like this:

C₍₁₎H_{(2 x 1 + 1)}COOH

CH₃COOH

If you are happy with remembering to remove one atom from n when using this general formula, then since the sum in the formula is relatively straight forward, you should have no problems with it. On the other hand, the alternative general formula can be useful if you are prone to forgetting such details (its just a little more complex to remember):

Ethanoic acid n = 2

C_n-1H_2(n-1)+1COOH

C_2-1H_2((2)-1)+1COOH

C₁H₂₍₁₎₊₁COOH

CH₃COOH

Aldehydes and Ketones are isomers of one another. Therefore, they will have the same formula in the form C_nH_mO. However, when writing their formulae, it is customary to differentiate them to indicate the presence of their functional group. For aldehydes, we include a -CHO functional group in the formula. This is not to be confused with a hydroxyl group (-OH, or -COH in this form).

*The structural formulae of propanoic acid (left) and propane (right).*

Look closely at the structures of propanoic acid and propanal. They both contain a carbonyl groups (-C=O). Think back to the reactions of aldehydes and ketones. Primary and secondary alcohols can be oxidised to aldehydes and ketones respectively; aldehydes can be oxidised further to carboxylic acids. The process of oxidation involves increasing the oxygen : hydrogen ratio. An aldehyde’s terminal carbon atom possesses bonds to one hydrogen atom and one double bond to an oxygen atom. Oxidising it removes the hydrogen and replaces it with a hydroxyl group – giving one single bond to an oxygen atom and one double bond to another oxygen atom.

This oxidation relationship between the structure of carboxylic acids and aldehydes is reflected in their chemical and general formulae. To differentiate between aldehydes and ketones, as well as indicate their functional groups being similar to carboxylic acids, aldehydes have the general formulae:

C_nH_2n+1CHO

C_n-1H_2(n-1)+1CHO

Once again, there are two possible general formulae to choose from. Remember – you don’t have to remember both, just pick the one you feel the most comfortable using.

So, with the example of propanal, where n = 3 :

C₍₂₎H_{((2 x 2)+ 1)}CHO

C₂H₅CHO

Remembering to save one carbon atom for the -CHO functional group

C_3-1H_{(2 x (3-1))+1}CHO

C₂H_{2 x (2)+1}CHO

C₂H₅CHO

However, it should be noted that the general formula of aldehydes is often simplified to just C_nH_2nO, omitting any information about differentiating between aldehydes and ketones.

The ketones also have a functional group – the same as the aldehydes – the carbonyl group (-C=O). Since it is not located on a terminal carbon atom, the chemical formula of an aldehyde does not reveal its position, it simply contains an oxygen atom within it.

Propanal’s ketone isomer is called propanone:

The general formula of ketones is:

C_nH_2nO

So for propanone, n = 3.

C₍₃₎H_{2 x 3}O

C₃H₆O

Note here that there is no need to subtract one carbon from the value of n since there is no functional group incorporated into the formula.

Application of General Formulae

General formulae can also serve you as a powerful tool for answering questions – particularly those asking you to write the chemical formula.

Let’s apply this to even larger molecules. Pentacontane is a large alkane with 50 carbon atoms, what would its formula be, and how many hydrogen atoms does it have?

We know from its name it is an alkane (the -ane suffix gives this away) and the Greek prefix pentacont- means 50. We simply have to put this number into the general formula for the alkanes:

C_nH_2n+2

C₍₅₀₎H_2(50)+2 = C₅₀H₁₀₂

Therefore, pentacontane it has 50 carbon atoms and 102 hydrogen atoms.

How about hectanoic acid? The hect- prefix means 100 carbon atoms, and it is clearly a carboxylic acid, so it must have a -COOH functional group, so it’s general formula is C_nH_2n+1COOH

Remembering that one of the carbon atoms forms part of the -COOH group, the formula must be:

C₍₉₉₎H_{(2 x 99 + 1)}COOH

C₉₉H₁₉₉COOH

It has 100 carbon atoms (the 99 being in the alkane chain and the remaining on in the COOH group) and 200 hydrogen atoms (once again, 199 being in the alkane chain and one left over on the -OH of the -COOH group).

It may seem like a very minor component of the organic chemistry units but being able to work out the general formula of a homologous series and being adept at using them to work out chemical formulae is a vital skill for all chemists. Learning where general formulae come from and how to use them properly will deepen pupil’s understanding of the structure of organic molecule’s.

If you would like further practice, or extra challenge with general formulae, try the questions in the worksheets linked below.

Rules for using the general formula:

Identify the functional group (is it a carbon chain? Double bond? Carbonyl? COOH? Hydroxyl?)
How many carbon atoms does it have? This gives you the value for n (however if it is an aldehyde or carboxylic acid, remember to either subtract 1 from this, or use the alternative general formulae)
Enter the value for n (the number of carbon atoms) into the chosen general formula and calculate.

If you cannot remember the general formulae or need to work it out :

Identify the functional group (is it a carbon chain? Double bond? Carbonyl? COOH? Hydroxyl?). Use this to assign the end of the formulae – OH, CHO, COOH, or just leave it as it is.
How many carbon and hydrogen atoms does it have? Try to think how these two numbers relate to one another. Almost always you need to double the number of carbons to get the number of hydrogens – from here do you leave it as it is (2n), add one (2n+1), or add two (2n+2)? Consider as well whether one carbon atom is taken up in the functional group as with aldehydes and carboxylic acids. If so, reduce your value of nby one, or replace it in the general formula with n-1.

General Formulae Worksheets

Rate Questions 2 – Interpret

Posted on 8th May 2021 Posted in National 5 Topics, Reaction Rates

Interpreting

Often in the exam, you will be given graphs following some variable over the course of a reaction and will be asked to comment on the change in rate, or to identify where the start or endpoint of the reaction occurs. You may also be asked what effects changing some of the variables considered in the previous article might have on the appearance of the graph – or the opposite – for example, you might be asked to explain the difference between two lines on the graph.

To compare different variables on the rate of reaction, one must make sure that only the variable to be investigated is changed. For example, if you want to compare the effect of concentration on the rate of reaction, two experiments with different concentrations must be set up. However, the temperature of both and the particles size of the reactants and the chemicals used must remain the same in both cases, with only the concentration differing.

Graphs

When monitoring the rate of a reaction, it is common practice to produce a graph illustrating the change in one particular quantity over time. This quantity will result in different looking graphs.

Volume

If a gas is given off during a reaction, the volume of the gas can be measured. Recording it on a graph will result in a curve which increases until it reaches a maximum when there is no more gas being given off. This is when the reaction has finished – no more gas is produced and the graph levels out.

If the rate is sped up by altering the particle size, concentration, temperature or using a catalyst, the graph will be steeper, resulting in the maximum volume reached more quickly.

A change in rate has no effect on the starting volume or end volume mass – only how quickly it takes to get there!

Mass

If a gas is given off during a reaction, then it will escape into the air. This means that we have some chemicals disappearing from the system. This will result in the overall mass on a balance decreasing. The endpoint is where the graph levels out. An increase in the rate will result in a steeper graph, a decrease in rate will result in a graph with a more shallow gradient.

A change in rate has no effect on the starting mass or finishing mass – only how quickly it takes to get there!

Rate Questions 1 – Influence

Posted on 8th May 20218th May 2021 Posted in National 5 Topics, Reaction Rates

A first in this series of blogs I hope to write in the coming weeks, focusses on the types of questions concerned with the reactions rates topic. Now, despite being a PhD student, I come from teaching the National 5 and Higher and Advanced Higher curriculum, so I shall base these articles on the structure of that course, however I intend to be writing this in my words, not confined as I am to a specific curriculum.

The National 5, and indeed GCSE curriculums are designed as a formal introduction to Chemistry. You may well have encountered aspects of the courses before, however now you are expected not only to remember key facts, but to put into practice your learning and prove your worth in the exam.

Reaction Rates

One can categorise most questions in an introductory rates topic into three types : first, explaining the influence of various factors on the overall rate, or indeed suggest how to increase or decrease it, or what effect some suggestion would have on the rate; second, you may observe and interpret a reaction graph and either deduce the overall rate, or explain what is happening after a specific time, or even predict what may happen to the graph should some method be used in an attempt to alter the rate; or third perform a calculation. Of course, this is a vain attempt to simplify the workings of the examiner’s mind, something not to be attempted lightly!

So lets elaborate on these three areas. This post shall concern itself with the chemistry of influence type questions.

Influence

In an introductory course, we consider there to be four ways of influencing the rate of a reaction. We shall explore these here and investigate the why (why they do what they do to the rate), and the how (how you can use them to control the rate of a reaction).

Temperature

Perhaps the first variable a school pupil encounters when speeding up a reaction. Warming up a reaction mixture often speeds things up. A classic is the sodium thiosulphate and hydrochloric acid reaction – done at room temperature, the colour change from clear to milky white can take over a minute (and accompanied by the unmistakable pungent rotten egg’s smell all too common where sulphur is concerned!). Warming up the reactants on a Bunsen flame before hand results in a reaction that takes only seconds!

Why? Think back to the most basic science lessons – solids, liquids and gases. Everything is made of tiny particles (we know them to be atoms and molecules by this point!). In a solid, they are held together by forces or bonds most rigidly; in a liquid, very loosely, allowing them for flow around; and in a gas, not at all – they can float about to their hearts content. As we go from solid to gas, the particles (molecules) gain thermal energy, which is converted into kinetic energy – motion. Very little motion in a solid, but lots of it in a gas. This gradient of motion is very pronounced in a liquid. A cold liquid (ahh this is very relative of course, for liquid helium is mega cold; liquid water – even at 5°C is warm by comparison! Let us define this by saying that a cold liquid is one near its freezing point, and a warm liquid is one nearer its boiling point) has little thermal energy. Its particles can move only slowly, so they trudge around at a snail’s pace (like old age pensioners walking five abreast down the high street with not a care for other pedestrians trying to get to the train station on time…). Warm it up a bit and you give the particles more energy, permitting them to whizz around like manic first years eager to get to the canteen at lunch time. Not only do the particles now have more speed (kinetic energy), but when they inevitably collied with one another, they will do so with greater force. Think of the old age pensioners – if they wonder around slowly and bump into one another, the bump will only be gentle. However if a first year is careering down the corridor, and another one is running towards them, should they crash into one another, there very well could be some physical damage to someone! Hot particles wallop one another with greater force and more often. This high frequency of wallops means that successful reactions are more likely at higher temperatures than at colder temperatures, and therefore the reaction takes place more quickly.

Effect of temperature on particle motion

How can we use this? Well simply put, if you want to slow down a reaction, cool it (in my current work I perform Grignard reagent synthesis which needs buckets of ice to slow down, otherwise it gets so exothermic that explosion is a real risk. Other reactions involving hazardous substances such as n-butyl lithium requires one to use a mixture of petroleum ether and liquid nitrogen to bring the system down to -78°C to make things safe! On the other hand, if you need to speed up the reaction (some can take days or even months otherwise), then increasing the temperature will significantly speed things along.

In your exam, you may be shown an example of two or more reactions and asked to determine which has the fastest rate. If all other variables are the same (concentration, particle size), then whichever has the highest temperature will react the fastest. In other cases, you may be presented with a graph of a reaction. Temperature will not affect the quantity of product, so the final hight of the graph will not change if temperature is altered – what will change is the steepness of it – its gradient – better described in this instance as the time taken for the graph to flatten out and stop changing. The line from the reaction with the highest temperature will be the steepest – reaching its end point quickest – the reaction performed at a lower temperature will be the one which takes the longest to reach the end point.

Particle Size

Ah-ha – time for some maths! I find the best way to explain particle size to my pupils is to run them through a bit of geometry. Nothing too fancy, but a revelation when you get it, and a nice way of bringing in a bit of cross-over between the STEM subjects!

First off, lets summarise it – larger particle sizes, slower reaction – smaller particle sizes, faster reaction.

Why? Well the wording of the explanation, looks a bit counter intuitive since the real reason is not the particle size per say, rather the surface area. Smaller particles have a much larger surface area than larger particles. So a fast reaction – small particles – large surface area. A slow reaction – large particles – small surface area.

The maths and a bit of technical drawing (great for the Graphic Communications pupils!) helps get your head around this.

Imagine a cube of calcium carbonate (marble chip) 10mm wide, 10mm high and 10mm deep

What is its surface area?

Quite simply, the surface area of a cube is the area of each face added together. The area of a face is its length times its height – so here it is 10 mm x 10 mm = 100 mm

A cube has 6 faces so, 100mm x 6 = 600 mm²

So our cube of calcium carbonate has a total surface area of 600 mm²

Now lets take a chemical knife and slash the cube in half, straight down the middle, then slash it again across its waist, right in the middle, and finally through the horizontal plane. We now have 8 mini cubes, each with dimensions half of that of the original cube – 5 mm by 5 mm by 5 mm

Now lets work out the surface area of our smaller particle sizes.

The area of one face is 5 mm x 5 mm = 25 mm,

So far, so smaller than before.

As before, we have 6 faces on a cube, so 25 mm x 6 = 150 mm²

But this is less than before!? But wait! We have 8 cubes this time.

So our total surface area must be 150 mm² x 8 = 1200mm²

So by reducing our surface area by a half, we double our surface area!

This means that with smaller particles, we expose more surfaces from the inside. If we were reacting the calcium carbonate with hydrochloric acid say, this means that we now have more faces exposed on which the acid can attack. Faces that would otherwise have been protected when the particle was larger.

It is unlikely that you will ever be asked to perform this calculation in an exam, by I urge you to try it for yourself with a particle with sizes of your choosing (maybe 20 mm, 25 mm or 5 meters). Better still, get some plasticene, blu-tac or better still, cut up some wood, and measure the dimensions yourself to prove it to you!

How can you use this? If you are given a reaction with magnesium ribbon, you could speed it up by cutting the magnesium into smaller pieces, or better still, grinding it into a powder. Slowing a reaction down can be achieved by the opposite – compacting the substance into a larger particle. Think about sugar cubes – they take longer to dissolve in tea than granulated sugar.

A graph will be altered by particle size thus : if you have smaller particles (larger surface area) the reaction rate will increase, so the time taken to reach the end of the reaction will reduce – i.e. a steeper gradient. Larger particles (smaller surface area) will mean that the reaction rate will decrease – taking longer to reach completion – i.e. a shallower gradient.

Concentration

Perhaps the most obvious and easy to explain effects. A high concentration means that there are more reactant particles present in a set volume of the solvent. A low concentration means that there are fewer reactant particles present in the same volume of solvent.

The more concentrated a solution, the more reactant particles there are, so the chances of collisions is greater, so they take place more frequently, and the reaction rate is faster.

Why? Well imagine if you will a crowded sports hall, so full of people that there is no more than 1 meter between each of them. They are all milling around randomly. Take yet another group of first year pupils and set them running into the hall with the task of reaching the other side of the room in as little time as possible. What happens? Carnage. It is inevitable that most, if not all of the pupils will collide with the people wandering around very quickly (fast rate) as their concentration is so high.

Lets say we repeat this experiment with the same room, the same number of first years, but the number of people wandering aimlessly around the hall is reduced to just 1 person. What happens? Depending upon that sole person’s location in the room, either all the first years make it across the room without incident, or (in the least likely scenario) one pupil collides with them. At the end of the day, minimal carnage (a slow rate). This is because the concentration of people in the room is so low that the chances of a successful collision is minimal.

This analogy is directly transferable to that of chemicals in a reaction – the more you have present, the more likely successful collisions are to occur and the rate is increased. Fewer reactants in the same volume, and the rate of the reaction is reduced.

How does this help me? If you encounter a question asking you how to speed up a reaction, using more concentrated solutions will help. To slow the reaction down, dilute the reactants – make them less concentrated. One such example could be yet again the reaction of hydrochloric acid with marble chips. To slow this reaction down, one could dilute the acid with more water. If you have the choice of HCl with a concentration of 0.1 mol/L, 1 mol/L and 2 mol/L, then the most dilute (0.1 mol/L) will take the longest to react, being the most dilute (least concentrated), whereas the highest concentration, 2 mol/L, will react the quickest.

A nice demonstration of this is to take dilute HCl (0.1 or 0.01 mol/L) and much more concentrated HCl (4 mol/L or stronger) and add a piece of magnesium ribbon to each at the same time. Many mobile phones are made from a magnesium alloy these days, so the reaction which the pupils will see will be quite similar to that they might encounter should the teacher drop a pupil’s phone in a beaker of the stuff (but without the resulting paperwork and fire hazard, and awkward conversation with the head teacher…). The weaker acid might result in a few, tiny bubbles, but the more concentrated acid will dissolve the metal completely in a matter of seconds. I would also suggest repeating this with an item of food – gives a good idea as to how the acid in your stomach works (which incidentally is also HCl).

Catalysts

The final method of influencing the rate of a reaction is to use a catalyst. This differs from the previous three since it involves adding an additional chemical, not related to the reactants.

There are two types of catalysts – homogeneous and heterogeneous. A homogeneous catalyst is in the same state as the reactants (a liquid if the reactants are both liquids), whereas a heterogeneous catalysts is in a different state to the reactants (a solid if both reactants are liquids, or gases).

The way a catalyst works is a bit more complicated, and the actual physics and chemistry of it is too advanced for an introductory course, but basically, they work by reducing the activation energy (the energy required to start the reaction) of the reaction and are not used up in the process.

In a homogeneous catalyst, the catalyst compound might form an entirely new compound with the reactants temporarily, but will be reformed at the end of the reaction.

A heterogeneous catalyst works by creating a surface upon which a reaction can take place – holding reactants in place so that collisions with other reactants occur more readily, then releasing the product upon its formation. At the end of the reaction, the catalyst surface is once again clean with no reactants attached to it.

If you were to weight out 10g of a catalyst, perform numerous reactions on it, at the end of the day if you were to reweigh it, you would still have exactly 10g of catalyst, regardless of how much reactant you used. Unfortunately, this would not be the case should you lose any through spillage or poor filtering, or general bad lab technique.

Catalysts can however suffer from poisoning. When a catalyst is poisoned, a chemical sticks to its surface, or reacts with it to bond so strongly, that it is impossible to be removed in the course of general use. This means that the catalyst is less effective at speeding up a reaction. When this happens, the catalyst is removed and has to undergo regeneration (further reactions to remove poisons) before it can be used again.

One example of a catalyst is the platinum / rhodium catalyst used in the catalytic converter in a car’s exhaust. This involves a honeycomb lattice of the metals upon which the products of incomplete combustion – carbon monoxide for example – are oxidised to produce the safer carbon dioxide. Impurities in the fuel can lead to poisoning of the catalyst. Leaded petrol is particularly hazardous to the health of a catalyst, as the lead binds irreversibly to the catalyst, poisoning it and preventing it from working effectively.

Chemistry F2

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