Atomic Structure

1.8 recall that atoms consist of a central nucleus, composed of protons and neutrons, surrounded by electrons, orbiting in shells

The centre of an atom is the nucleus, which is composed of protons and neutrons. Their masses are roughly equal and since the mass of the electron is pretty much negligible, most of the mass of an atom is in the nucleus. 
The electrons are found in a series of energy levels which you call shells at IGCSE. Each ‘shell’ can only hold a certain number of electrons, these shells can be thought of as getting progressively further from the nucleus. Electrons will always go into the lowest possible energy level, provided there is space. The first shell can only hold 2 electrons, then the shells after that can hold a maximum of 8. 
In a diagram, the electrons are shown on circles around the nucleus. Beware that these circles are just imaginary lines to help you understand that the electrons orbit around the nucleus, at IGCSE level you just need to accept that. 
For those moving on to higher levels of Chem, the ‘truth’ is different, it’s actually not really possible to plot the path of an electron. For more information look at the following link, to be honest I don’t really understand it myself, it’s so confusing. 

Later in the course you will come across  dot-and-cross diagrams. Dots or crosses are used to represent electrons, in the above, Carbon has 4 outer shell electrons. They are drawn far apart even though you could draw them close to each other like in the first shell, this is because electrons would repel each other as they have the same negative charge. (Remember like charges repel). So only if you have more than 4 OSE, then do you draw them in pairs. Remember, draw the 4 OSE like in the diagram of carbon above, then pair up any OSE left. The following diagram might help you understand:
This way of drawing electrons is clear and makes it easy to count too. When you learn about ions, dot-and-cross diagrams are useful and they help you see how the electrons are transferred. Like in the above diagram, the Chlorine atom gains one electron (the cross) from the sodium atom to become a chloride ion (Cl), whilst the sodium atom becomes a sodium ion (Na+). It is positive because it lost one electron, so it has one more proton than electron now.🙂

1.9 recall the relative mass and relative charge of a proton, neutron and electron

Relative mass
Relative charge
1/1840 (negligible)

Extra (you don’t need to know this): 
Protons and neutrons don’t actually have exactly the same mass – neither of them has a mass of exactly 1 on the carbon-12 scale (the scale on which the relative masses of atoms are measured). On the carbon-12 scale, a proton has a mass of 1.0073, and a neutron a mass of 1.0087. 

1.10 understand the terms atomic number, mass number, isotopes and relative atomic mass (Ar)
Atomic number is the number of protons there are in the nucleus, it is sometimes called the proton number, though atomic number should be more accurate because atoms are electrically neutral, the number of protons and electrons are equal. (Protons have a charge of +1 whilst electrons are -1, so they cancel each other out.) So the atomic number tells you the number of protons and the number of electrons. 
Mass number is the number of protons and neutrons. It is sometimes known as the nucleon number, because protons/neutrons are nucleons. So if a question asks you for the number of neutrons in an atom, mass number – atomic number = no. of neutrons
Isotopes: these are atoms which have the same atomic number but different mass numbers, i.e. same number of protons but different number of neutrons.
The number of neutrons in an atom can vary a little. For instance, there are three kinds of carbon atoms 12C, 13C and 14C. Their number of neutrons varies but they have the same number of protons, because each element’s atomic number is unique. If it has a different number of protons, it wouldn’t be the same element anymore. So these atoms are isotopes of carbon. Bear in mind that the fact that they have varying numbers of neutrons makes no difference whatsoever to the chemical reactions of the carbon. Though their physical properties may vary. 

Mass number

Relative atomic mass (web definition): the ratio of the average mass per atom of the naturally occurring form of an element to one-twelfth the mass of an atom of carbon-12. Symbol Ar Abbreviation r.a.m. I find it easier to think of it as:
The number of times the mass of one atom of an element is heavier than 1/12 of the mass of a carbon-12 atom. Anyways, this number is always at the top of an element’s symbol, it’s always either mass number or RAM at the top, atomic number at the bottom – as shown below:

mass number is always on the top, atomic number at the bottom, don’t mix them up. at least you know that the mass number is always greater than atomic number

1.11 calculate the relative atomic mass of an element from the relative abundances of its isotopes
You multiply the relative abundance of each isotope by its mass number, add these together, and divide by 100. It’s easier to understand through an example, in this case I’ll use chlorine, since it’s pretty common.


Chlorine consists of 75% Chlorine-35 and 25% Chlorine-37. You can think of the data as 100 atoms, 75 having a mass of 35 and 25 with a mass of 37. So the calculation is:
[(75 x 35) + (25 x 37)] / 100 = 35.5
So the RAM of chlorine, or Ar(Cl) is 35. 
(There are tiny percentages of other chlorine isotopes but the two shown above are the most common, and so the rest are ignored at IGCSE level.)

The RAM of an element will be closer to the mass number of the more abundant isotope. For example, the RAM of chlorine is 35.5, which is closer to chlorine-35, because it is the more abundant isotope. Obviously 75% > 25%!.

1.12 understand that the Periodic Table is an arrangement of elements in order of atomic number
The number of protons in the element’s atom increases across the Periodic Table as you’ve probably noticed in yours.

1.13 deduce the electronic configurations of the first twenty elements from their positions in the Periodic Table
To work out the electronic arrangement of an atom:

  • Look up the atomic number in the Periodic Table – making sure that you choose the right number if two numbers are given. The atomic number will always be the smaller one and tends to be below the symbol.
  • This tells you the number of protons, and hence the number of electrons.
  • Arrange the electrons in levels, always filling up an inner level before you go to an outer one. 

e.g. to find the electronic arrangement in oxygen

  • the Periodic Table gives you the atomic number of 8.
  • Therefore there are 8 protons and 8 electrons.
  • The arrangement of the electrons will be 2,6. (First shell only holds 2 electrons, then there’s 6 left which occupy the second shell.)

1.14 deduce the number of outer electrons in a main group element from its position in the Periodic Table
If you look at the patterns in the table:

  • The number of electrons in the outer level is the same as the group number. (Except with helium which has only 2 electrons. The noble gases are also usually called group 0 – not group 8.) This pattern extends throughout the Periodic Table for the main groups (i.e. not including the transition elements). 
  • So if you know that barium is in group 2, it has 2 outer shell electrons (btw, outer shell electrons which I will abbreviate to OSE are also known as valence electrons); iodine is in group 7, so it has 7 OSE, lead is in group 4, so surprise surprise, it has 4 OSE. 
  • Noble gases have full outer shells. Thus they are unreactive. More on them in this post:

More on Electrolysis

Note: This post is mainly for Single Science although it could be good background information for Double Award anyway.🙂

SS 1.53 describe simple experiments for the electrolysis, using inert electrodes, of aqueous solutions of sodium chloride, copper (II) sulfate and dilute sulfuric acid and predict the products

So, electrolysis can be used to decompose molten compounds as described in an earlier post on Electrolysis: 
However, for Single Science, you also need to know about electrolysis of compounds in aqueous solutions. Predicting the reactions and working out the products for aqueous solutions are less straightforward than for molten compounds. 
An aqueous solution of a compound is a mixture of two electrolytes, it’s a compound dissolved in water really (so water is the solvent). For example, an aqueous solution of copper (II) sulphate contains two electrolytes: copper (II) sulphate and water. It therefore contains copper (II) sulphate ions (Cu2+) and sulphate ions (SO42-), and also small amounts of hydrogen ions (H+) and hydroxide ions (OH) from the dissociation of water molecules.
H2O (l) à H+(aq) + OH(aq)
These ions compete with the ions from copper (II) sulphate for discharge at the electrodes. 
In general, when an aqueous solution of an ionic compound is electrolysed, a metal or hydrogen is produced at the cathode. At the anode, a non-metal, for example oxygen or a halogen, is given off. 

Let’s see if this is the case in the electrolysis of dilute sodium chloride solution. 
Electrolysis of Dilute Sodium Chloride Solution
Note: There is a difference in the products between the electrolysis of dilute sodium chloride solution and concentrated sodium chloride solution. I will elaborate later.
An aqueous solution of sodium chloride contains four different types of ions. They are
  • Ions from sodium chloride – Na+ (aq) and Cl (aq)
  • Ions from water – H+(aq) and OH (aq)

When dilute sodium chloride solution is electrolysed using inert electrodes, the Naand Hions are attracted to the cathode. The Cl– and OH– ions are attracted to the anode. 
At the cathode: 
The Hand Na+ ions are attracted to the platinum cathode. Hions gains electrons from the cathode to form hydrogen gas. (The hydrogen ions accept electrons more readily than the sodium ions. As a result, Hions are discharged as hydrogen gas, which bubbles off. I will explain why Hions are preferentially discharged later.)
2H+(aq) + 2e à H2(g)
Na+ ions remain in solution. 
At the anode:
OHand Cl– are attracted to the platinum anode. OH– ions give up electrons to the anode to form water and oxygen gas. 
4OH(aq) à 2H2O(l) + O2(g) + 4e
Cl– ions remain in solution. 

The overall reaction is: 
  2H2O(l)   à 2H2(g) + O2(g)

Since water is being removed (by decomposition into hydrogen and oxygen), the concentration of sodium chloride solution increases gradually. The overall reaction shows that the electrolysis of dilute sodium chloride solution is equivalent to the electrolysis of water.
Another important thing to note is that twice as much hydrogen is produced as oxygen. This is because for every 4 electrons that flows around the circuit, you would get one molecule of oxygen. But four electrons would produce 2 molecules of hydrogen. Hence in a diagram, you would see the volume of hydrogen produced is twice that of oxygen. Refer to the equations above and note the number of electrons involved to help you understand. 
This diagram is just to illustrate how twice as much hydrogen gas is produced. 
Electrolysis of Concentrated Sodium Chloride Solution
The only difference is that at the anode, Cl–  ions are more numerous than OH– ions. Consequently, Cl– ions are discharged as chlorine gas, which bubbles off. 
2Cl (aq) à Cl2(g) + 2e
The OH– ions remain in solution.   
One volume of hydrogen gas is given off at the cathode and one volume of chlorine gas is produced at the anode. The resulting solution becomes alkaline because there are more OH– than Hions left in the solution.    

Compare the electrolysis of molten sodium chloride, dilute sodium chloride solution and concentrated sodium chloride solution:

Molten sodium chloride:
  • Cathode: Na+ions discharged
  • Anode: Cldischarged
Dilute NaCl solution: 
  • Cathode: H+ions discharged
  • Anode: OHdischarged

Concentrated NaCl solution:
  • Cathode: H+ discharged
  • Anode: Cldischarged
So you can see that Naand Cl– ions are not always discharged even though in all 3 of the above, the electrolytes contained these ions. For example in the electrolysis of dilute NaCl solution, Hare discharged in preference to Naions. OH ions are discharged in preference to Cl– ions. Before I talk about the electrolysis of copper(II) sulfate and dilute sulfuric acid, I will discuss why one type of cation (or anion) in the electrolyte is more readily discharged than another type. If you already know this, just scroll right down.🙂
Note: Most of the following is taken from a G.C.E. ‘O’ Level textbook, but I find it useful. :) 
Reactivity Series and Selective Discharge of Ions
In electrolysis, when more than one type of cation or anion is present in a solution, only one cation and one anion are preferentially discharged. This is known as the selective discharge of ions. 
How do you predict which ions are discharged in the electrolysis of a compound in aqueous solution?
If inert electrodes are used during electrolysis, the ions discharged and hence the products formed depend on three factors:
  1. The position of the metal (producing the cation) in the reactivity series. 
  2. The relative ease of discharge of an anion. 
  3. The concentration of the anion in the electrolyte. 
The ease of discharge of cations and anions during electrolysis is shown below.

NB: Ease of discharge increases as you go down the table
Potassium ion, K+
Chloride ion, Cl
Sodium ion, Na+
Bromide ion, Br
Calcium ion, Ca2+
Iodide ion, I
Magnesium ion, Mg2+
Hydroxide ion, OH
Zinc ion, Zn2+
Note: sulphate ions (SO42-) and nitrate ions (NO3) will not be discharged during electrolysis.
Iron ion, Fe2+
Lead ion, Pb2+
Hydrogen ion, H+
Copper ion, Cu2+
Silver ion, Ag+

Selective discharge of cations during electrolysis
The cations of an element lower in the reactivity series are discharged at the cathode in preference to cations above it in the solution. This is because cations of a less reactive element accept electrons more readily. For example, if a solution containing Naand Hions is electrolysed,  Hions are discharged in preference to Naions. The more reactive the metal, the more stable its compound. They have lost a lot of energy and have lost electrons to form stable cations, so cations lower down the reactivity series are more readily discharged.

Selective discharge of anions during electrolysis
Sulphate (SO42-) and nitrate (NO3) ions remain in the solution and are not discharged during electrolysis. If a solution containing SO42-NO3– and hydroxide (OH) ions is electrolysed, the OH– ions will be discharged in preference to SO42- and NO3– ions. The OH– ions give up electrons most readily during electrolysis to form water and oxygen.

4OH (aq) à 2H2O (l) + O2 (g) + 4e  
Effect of concentration on selective discharge of anions
An increase in the concentration of an anion tends to promote its discharge. For example, in the electrolysis of concentrated sodium chloride solution, two types of ions are attracted to the anode: Cl– and OH– ions. According to their relative ease of discharge, OH– ions should be discharged preferentially. However, in concentrated sodium chloride solution,  Cl– ions are far more numerous than OH– ions and so are discharged at the anode instead. 

2Cl (aq) à Cl2 (g) + 2e
What are the general rules for predicting selective discharge?
The following rules can be applied when predicting the products of electrolysis of any aqueous solution (using inert electrodes):
Rule 1
Identify the cations and anions in the electrolysis. Remember that an aqueous solution also contains H+ and OH ions from the dissociation of water molecules.
Rule 2
At the anode, the product of electrolysis is always oxygen unless the electrolyte contains a high concentration of the anions, Cl, Br­- or I ions.
Rule 3
At the cathode, reactive metals such as sodium and potassium are never produced during electrolysis of the aqueous solution. If the cations come from a metal above hydrogen in the reactivity series, then hydrogen will be liberated (liberate=release). If the cations come from a metal below hydrogen, then the metal itself will be deposited.
Rule 4
Identify the cations and anions that remain in the solution after electrolysis. They form the product remaining in solution. Summarise the reactions.
For example, in the electrolysis of dilute sodium chloride solution, Na+ and Cl ions remain in solution after H+ and OH ions have been discharged. Hence the solution of sodium chloride becomes more concentrated after electrolysis.

Electrolysis of Copper (II) sulphate solution
Copper (II) sulphate solution can be electrolysed using inert platinum electrodes. (Sometimes inert carbon electrodes in the form of graphite are used.)

During electrolysis, the cathode is coated with a layer of reddish-brown solid copper. The blue colour of the solution fades gradually as more copper is deposited. The resulting electrolyte also becomes increasingly acidic.

An aqueous solution of copper (II) sulphate contains four types of ions:

  • Ions from copper (II) sulphate: Cu2+ and SO42-
  • Ions from water: Hand OH

At the anode:
OH– ions and SO42- ions are attracted to the anode. OH– ions give up electrons more readily than SO42- ions. Consequently, OH– ions are preferentially discharged to give oxygen gas.

4OH (aq) à 2H2O (l) + O2 (g) + 4e

The SO4ions remain in solution. 
At the cathode:
Hions and Cu2+ ions are attracted to the cathode. Copper is lower than hydrogen in the reactivity series. Cu2+ ions accept electrons more readily than Hions. As a result, Cu2+ ions are preferentially discharged as copper metal (atoms). 
Cu2+ (aq) + 2e à Cu (s)
The Hions remain in solution. 
When aqueous copper (II) sulphate is electrolysed using platinum electrodes, copper metal is deposited at the cathode and oxygen gas is given off at the anode. The overall reaction is:
2CuSO4 (aq) + 2H2O (l) à 2Cu (s) + O2 (g) + 2H2SO4(aq)

Electrolysis of dilute sulfuric acid solution
Inert carbon or platinum electrodes are used.
At the cathode:
In this case, the only positive ions arrive at the cathode are the hydrogen ions from the acid and the water. (Adding acid to water forces it to split up/hydrolyse.) These are discharged to give hydrogen gas.

2H+ (aq) + 2e à H2(g)
At the anode:
At the anode, SO42- ions  and OH– ions (from the water) accumulate. OH– ions are discharged to give O2 gas. 
4OH (aq) à 2H2O (l) + O2 (g) + 4e
The amount of hydrogen produced is twice that of oxygen. Just like in the electrolysis of dilute sodium chloride solution.  For every 4 e that flows around the circuit, you would get one molecule of O2 . But four electrons would produce 2 molecules of H2

Ionic Compounds

f) Ionic compounds

1.28 describe the formation of ions by the gain or loss of electrons
Ions are atoms or molecules with an electric charge due to the gain or loss of electrons. If electrons are lost, the ion has a positive charge. Metals tend to do this, so they form cations (positive ions), so normally elements from group 1-3 will form cations.
If electrons are gained, the ion has a negative charge. Non-metals tend to do this, and they form anions (A-Negative-ION – ANION). So elements from group 5-7 will form anions. Group 0/8 are the noble gases and are inert + unreactive, so they do not form ions.
1.29 understand oxidation as the loss of electrons and reduction as the gain of electrons
OILRIG –  Oxidation Is Lost, Reduction Is Gain
1.30 recall the charges of common ions in this specification
Positive ions/Cations
Negative ions/Anions
Name of ion
Name of ion
Copper (I)
Copper (II)
Iron (II)
Lead (II)
Nickel (II)
Iron (III)

1.31 deduce the charge of an ion from the electronic configuration of the atom from which the ion is formed
So if the electronic configuration is 2.8.1, you can see that the atom has one outer shell electron only. And so it only needs to lost that to have a full outer shell. So its ion would have a positive 1 (1+) charge, as the electronic configuration would be 2.8. Basically, if there are less outer shell electrons to lose to have a full outer shell, then the charge will be positive. (Here, it is easier to lose than gain, because the ion would have to gain SEVEN electrons to have a full outer shell!)
Another example, if the electronic configuration is 2.8.7, then the atom only needs to gain 1 outer shell electron to have a full outer shell. So the ion formed would have the electronic configuration of 2.8.8, and so the charge would be negative 1 (you’re gaining one electron, which has a negative charge (1-) ). Thus if it is easier to gain electrons, (it is here, rather than losing 7 outer shell electrons), then the charge will be negative.

Here, the Sodium (Na) has lost one electron. It doesn’t have equal numbers of protons and electrons anymore; it has one less electron than protons (or you can think of it as one more proton than electrons), so it has a 1+ charge. 

The chlorine, on the other hand, has gained one electron. So it has one more electron than proton, thus it has a negative 1 charge (1-). The formula for sodium chloride is NaCl. 

The magnesium atom loses 2 electrons to an oxygen atom, and they both have full outer shells now. It is common that ions form noble gas structures like this to become more stable and unreactive like the group 0/8 elements. 

The magnesium oxide is held together by very strong attractions between the ions. The ionic bonding is stronger here than in sodium chloride as this time you have 2+ ions attracting 2- ions. The greater the charge, the greater the attraction.

The formula for magnesium oxide is MgO.

1.32 explain, using dot and cross diagrams, the formation of ionic compounds by electron transfer, limited to combinations of elements from Groups 1, 2, 3 and 5, 6, 7

Hydrogen and water

Note: I’m adding Labels at the end of posts so that when you click on them, it will show ALL the posts in that section related to it. This will make this blog easier to use, as I don’t post in a specific order, rather I answer or post according to what people needed. Hope this helps and please feedback. :) 

Section 2: Chemistry of the Elements; part e) Hydrogen and water

2.26 describe the reactions of dilute hydrochloric and dilute sulfuric acids with magnesium, aluminium, zinc and iron

Metals above hydrogen in the reactivity series will react with acids to form a salt (e.g. magnesium sulfate or zinc chloride) and hydrogen. The metals are ‘displacing’ hydrogen. The higher the metal in the series, the more violent the reaction. (This is why if you put copper in acids, you won’t see a reaction, as it is below hydrogen in the reactivity series. However, it does react with concentrated nitric acid but we’re not concerned with that now.)
Metal + dilute hydrochloric acid à metal chloride + hydrogen
Metal + dilute sulfuric acid à metal sulfate + hydrogen

Magnesium reacts vigorously with cold dilute acids, and the mixture becomes very warm as heat is produced. There is rapid fizzing (effervescence) and a colourless gas is evolved, which pops with a lighted splint (the test for hydrogen). The magnesium gradually disappears and a colourless solution of magnesium sulfate or chloride is formed. 
–The reactions between magnesium and hydrochloric acid or sulfuric acid are similar because it is reacting with the hydrogen ions. All acids in solutions have hydrogen ions. Although hydrochloric acid has chloride ions, and sulfuric acid has sulfate ions, these are spectator ions. They do not participate in the reaction and are unchanged by it. 
You can rewrite the equations as ionic equations. In the case of hydrochloric acid:
Mg (s) + 2H+ (aq) + 2Cl (aq) à Mg2+(aq) + 2Cl (aq) + H2 (g)
You can see that the chloride ions weren’t changed by the reaction. It is a spectator ion, so we leave it out of the ionic equation. Leaving out the spectator ions produces the ionic equation:
Mg(s) + 2H+(aq) à Mg2+ (aq) + H2 (g)
Repeating this with sulfuric acid:
Mg (s) + 2H+ (aq) + SO42- (aq) à Mg2+ (aq) + SO42- (aq) + H2(g)
Again, leaving out the spectator ion which is the sulfate ion in this case.
Mg(s) + 2H+(aq) à Mg2+ (aq) + H2 (g)

So the reactions look the same because they are the same. All acids in solution contain hydrogen ions. That means that magnesium will react with any simple dilute acid in the same way. 

Aluminium is slow to start reacting, but after warming it reacts very vigorously. There is a very thin, but very strong, layer of aluminium oxide on the surface of the aluminium, which stops the acid from getting to it. On heating, the acid removes this layer, and the aluminium can show its true reactivity. With dilute hydrochloric acid:
2Al (s) + 6HCl (aq) à 2AlCl3 (aq) + 3H2 (g)

Zinc and Iron
Zinc and iron react slowly in the cold, but more rapidly on heating. Their reactions are less vigorous than that of aluminium, and iron less than zinc of course, as it is below zinc in the reactivity series. Zinc forms zinc chloride or sulfate and hydrogen. The iron forms iron (II) sulfate or iron (II) chloride and hydrogen. For example:

Zn (s) + H2SO4(aq) à ZnSO4 (aq) + H2 (g)
Fe (s) + 2HCl (aq) à FeCl2 (aq) + H2 (g)

2.27 describe the combustion of hydrogen

Hydrogen reacts violently with oxygen in the presence of a flame to give water. It could explode if there was a lot of hydrogen. But a lighted splint placed at the mouth of a test tube of hydrogen will just give a squeaky pop as the hydrogen reacts with oxygen in the air. The lighted splint and a squeaky pop heard is the test for hydrogen. 

2.28 describe the use of anhydrous copper (II) sulfate in the chemical test for water

Anhydrous copper (II) sulfate is white, anhydrous being without water, it is dry (an–without, hydrous–related to water). Whereas hydrated copper (II) sulfate crystals are bright blue, the water is what gives it the colour, and is part of the structure. To show that the water is part of the structure, there is a ‘.’ [dot] in the formula: 
^You see the dot in the middle? That shows the water is part of the copper sulfate crystal structure. 

So that is a chemical test for water, just add it to anhydrous copper (II) sulfate and watch it turn blue!

Adding water to anhydrous copper sulphate

2.29 describe a physical test to show whether water is pure

Heat the water and use a thermometer to check if it boils at exactly 100°C. Pure water boils at exactly 100°C. Or you can cool it until it freezes, it should freeze at exactly 0°C. My teacher said it’s safer to state both, as pure water is the only substance that has these specific boiling and freezing points, whereas another substance might boil/freeze at either temperature. 
Hope this helped!

Contact Process-Manufacture of sulfuric acid

d) The industrial manufacture of chemicals

This part is for SINGLE SCIENCE.

5.25 recall the raw materials used in the manufacture of sulfuric acid
The raw materials are:

  • sulphur
  • air (oxygen)
  • water

5.26 describe the manufacture of sulfuric acid by the contact process, including the
essential conditions:
i a temperature of about 450 °C
ii a pressure of about 2 atmospheres
iii a vanadium(V) oxide catalyst

Stage 1: making sulfur dioxide

You can either burn sulfur in air:
S(s) + O2(g) à SO2(g)
or heat sulfide ores strongly in air:
4FeS2(s) + 11O2(g) à Fe2O3(s) + 8SO2(g)
(FeSis pyrite or iron pyrite)
Iron pyrite crystals

Stage 2: Making sulfur trioxide
Now the sulfur dioxide is converted into sulfur trioxide using an excess of air from the previous processes. 
2SO2(g) + O2(g) 2SO3(g)     H= -196 kJ/mol 
An excess of oxygen is used in this reaction, because it is important that as much sulfur dioxide as possible is converted into sulfur trioxide. Having sulfur dioxide left over at the end of the reaction is wasteful, and could cause possibly dangerous pollution. (Remember sulfur dioxide can dissolve in water and form acid rain, this can kill plants and animals-by altering pH of water fish live in for example. It will corrode limestone which is basically calcium carbonate. It can also leach nutrients from the soil.)
As the forwards reaction is exothermic, there would be a higher percentage conversion of sulfur dioxide into sulfur trioxide at a low temperature. (Remember your equilibrium stuff, go to my equilibrium post if you want to revise that first.)
However, at a low temperature, the rate of reaction would be very slow. 450°C is a compromise. Even so, there is already about a 99.5% conversion. 
There are 3 gas molecules on the left-hand side of the equation, but only 2 on the right. Reactions in which number of gas molecules decrease are favoured by high pressures. (Remember Le Chatelier’s principle where you’re trying to remove the change, if you increase pressure, moving the equilibrium to the side with less gas molecules would decrease pressure.) In this case though, the conversion is so good at low pressures already that it isn’t economically worthwhile to use higher ones. So a pressure of 2 atmospheres is sufficient.
The catalyst, vanadium (V) oxide, has no effect on the percentage conversion, but helps to speed up the reaction. Without the catalyst, the reaction would be extremely slow. 
Remember, catalysts remain chemically unchanged at the end of the reaction. They help to speed up the rate of reaction, by providing an alternative pathway with a lower activation energy. Activation energy is the minimum amount of energy needed for a reaction to take place. So if the activation energy is lowered, more particles will have the required activation energy so a greater number of the collisions will be effective. Effective collisions are ones where reactions actually take place. Sometimes particles collide without reacting because they don’t have the minimum activation energy required.

Sulfur dioxide is converted into sulfur trioxide

Stage 3: Making the sulfuric acid

In principle, you can react sulfur trioxide with water to make sulfuric acid. 
SO3(g) + H2O(l) à H2SO4(aq)

In practice, this produces an uncontrollable fog of concentrated sulfuric acid. Instead, the sulfur trioxide is absorbed in concentrated sulfuric acid to give fuming sulfuric acid (also called oleum). 
H2SO4(l) +SO3(g) à H2S2O7 (l)
This is converted into twice as much concentrated sulfuric acid by careful addition of water. 
H2S2O7(l) + H2O(l) à 2H2SO4(l)
I’m not sure which equation you guys have learnt, so I’ve included both the principle and the real life one.🙂

5.27 recall the use of sulfuric acid in the manufacture of detergents, fertilisers and paints

Sulfuric acid has a wide range of uses throughout the chemical industry. The highest single use is in making fertilisers (including ammonium sulfate and ‘superphosphate’-essentially a mixture of calcium phosphate and calcium sulfate).
It is also used in the manufacture of detergents and paints. If you look at the list of ingredients on any industrial or domestic detergents (including shampoos and liquid ‘hand-soaps) and find the words ‘sulfate’ or ‘sulfonate’, then sulfuric acid was used in the manufacturing process. Even those simply labelled as containing ‘anionic surfactants’ almost certainly contain these sorts of ingredients, even if they don’t name them.
In paint manufacture, sulfuric acid is used in extracting the white pigment titanium oxide, TiOfrom titanium ores.

Group 1 elements-lithium, sodium and potassium

Update: I’ve included the trends, physical and chemical properties of Group 1 elements as requested by someone. Some information taken from Edexcel Chem textbooks. :) 

Section 2: Chemistry of the elements
part b) Group 1 elements-lithium, sodium and potassium

2.6 describe the reactions of these elements with water and understand that the reactions provide a basis for their recognition as a family of elements
2.7 recall the relative reactivities of the elements in Group 1

Alkali metal
Hydroxide solution produced
Gas produced
Rate of gas produced
Lithium hydroxide
Fairly vigorous
Sodium hydroxide
Potassium hydroxide
Very vigorous
Rubidium hydroxide
Caesium hydroxide
Extremely explosive

As you can see, the reactions get more violent as you go down the group, telling you that the metals are more reactive down the group. 

Describe what happens when sodium is added to water:
When sodium is added to water, it reacts very quickly and vigorously. The reaction is exothermic and the heat produced melts the sodium. The molten sodium darts around the water surface and a yellow flame is seen. You may see a bit of fizzing/bubbling (effervescence) as hydrogen is evolved. 
Remember MM-FF. Melts, moves, floats, fizzes 
Li, Na and K are all less dense than water, hence they float on water. 

Write word equations to show the reactions of lithium, sodium and potassium with water:
Lithium + water àlithium hydroxide + hydrogen
Sodium + water à sodium hydroxide + hydrogen
Potassium + water àpotassium hydroxide + hydrogen

So you see they all form hydroxides, and since all group 1 metal compounds are soluble it dissolves to form an alkali. (All alkalis are just soluble bases, but not all bases are alkalis, for example metal oxides are bases but not all of them are soluble to form alkalis.. e.g. Copper (II) oxide.)
If they ask you about what colour the solution will turn if universal indicator is added, it will turn blue or purple. This is because the solution is alkaline, and if they ask you to state what ion causes this, it’s the OH ion (hydroxide ion). They are called alkali metals for a reason!

2.8 explain the relative reactivities of the elements in Group 1 in terms of distance between the outer electrons and the nucleus. (single science)

As you go down the group the metals become more and more reactive. This is because their atoms get bigger, so the outer shell electrons are further away from the nucleus. So the electrostatic force between the nucleus and the outer shell electron (OSE) is weaker, hence it is easier to lose the OSE (alkali metals have only 1 OSE). The atoms want to lose the OSE to form full outer shells so they are more stable and unreactive after that. 

Also, you can think of how as the atoms get bigger, there are more electron shells in between the OSE and the nucleus. You can think of them as ‘shields’, so the force of the nucleus on the OSE is weaker the bigger the atom. 

The following is extra information someone has requested. 

Physical Properties

Melting Point (°C)
Boiling Point (°C)
Density (g/cm3)
Lithium – Li 
Sodium – Na
Potassium – K
Rubidium – Rb
Caesium – Cs
  • You will notice that the melting and boiling points of the elements are very low for metals, and decrease as you go down the group.
  • Their densities tend to increase, but potassium has a lower density than sodium, so the densities don’t increase that tidily. Lithium, sodium and potassium are all less dense than water, hence will float on it. 
  • The metals are very soft and can be easily cut with a knife. They get softer as you go down the Group. They are shiny and silver when freshly cut, but tarnish within seconds on exposure to air. 
Storage and handling
All these metals are extremely reactive, and get more reactive as you go down the Group. They all react quickly with air to form oxides, and react between rapidly and violently with water to form strongly alkaline solutions of the metal hydroxides.

To prevent them from reacting with oxygen/water vapour in the air, lithium, sodium and potassium are stored under oil. Rubidium and caesium are so reactive that they have to be stored in sealed glass tubes to stop any possibility of oxygen getting to them. 

Great care must be taken not to touch any of these metals with bare fingers. There could be enough sweat on your skin to give a reaction producing lots of heat and a very corrosive metal hydroxide. 

Compounds of the alkali metals
All Group 1 metal ions are colourless. That means that their compounds will be colourless or white unless they are combined with a coloured negative ion. For instance,  Potassium dichromate (VI) is orange, because the dichromate (VI) ion is orange. Group 1 compounds are typical ionic solids and are mostly soluble in water.

Summary of the main features of the Group 1 elements:
Group 1 elements:
  • are metals
  • are soft with melting points and densities very low for metals
  • have to be stored out of contact with air or water
  • react rapidly with air to form coatings of the metal oxide
  • react with water to produce an alkaline solution of the metal hydroxide and hydrogen gas
  • increase in reactivity as you go down the Group
  • form compounds in which the metal has a 1+ ion
  • have mainly white compounds which dissolve to produce colourless solutions

Haber Process

Section 5: part d) The industrial manufacture of chemicals

-for double award you only need to know the Haber Process, Single Award people, you also need to know the Contact Process for the manufacture of Sulphuric Acid!

5.21 recall that nitrogen from air, and hydrogen from natural gas or the cracking of hydrocarbons, are used in the manufacture of ammonia
So if they ask what are the raw materials used in the Haber process, you know it. It’s nitrogen from the air and hydrogen either from natural gas, which is methane (CH4); or from the cracking of hydrocarbons. (Cracking is in the Crude Oil post.) 

5.22 describe the manufacture of ammonia by the Haber process, including the essential conditions:
i. a temperature of about 450°C
ii. a pressure of about 200 atmospheres
iii. an iron catalyst

Remember these conditions!! And remember that the reaction is reversible. Also, the forwards reactions is exothermic
N2 + 3H2 à 2NH3

So decreasing the temperature would actually increase the yield, however, it is still done at a fairly high temperature to speed up the reaction. It makes the rate of reaction faster so the manufacturers get their ammonia quicker, as they say, time is money. The reaction would be too slow otherwise at low temperatures. It would be useless to have a low temperature and achieve a high yield of ammonia if it’s going to take ages. You need the gases to reach equilibrium within the very short time that they will be in contact with the catalyst in the reactor.So 450°C is a compromise, and still produces a reasonably high proportion of ammonia. 

The catalyst does NOT affect the amount of products made. The yield of ammonia stays the same, you just get it faster because it speeds up the reaction by lowering the activation energy needed for the reaction. :) 

Increasing the pressure would favour the forwards reaction which is what is wanted, to get more ammonia. This is because if you look at the balanced equation, there are 4 moles of gas on the reactants side (left) but 2 moles of gas (ammonia) on the right hand side (products). So according to Le Chatelier’s principle where you try to remove the change, if you increase pressure, the equlibrium would move to the right hand side to decrease pressure. And the products have less pressure because there are only 2 moles there.

Also, just as an extra, just thought that this will be useful to know and is very logical:🙂

Rate considerations
Increasing the pressure brings the molecules closer together. In this particular instance, it will increase their chances of hitting and sticking to the surface of the catalyst where they can react. The higher the pressure the better in terms of the rate of a gas reaction.

Economic considerations
Very high pressures are very expensive to produce on two counts.
You have to build extremely strong pipes and containment vessels to withstand the very high pressure. That increases your capital costs when the plant is built.
High pressures cost a lot to produce and maintain. That means that the running costs of your plant are very high.

The compromise
200 atmospheres is a compromise pressure chosen on economic grounds. If the pressure used is too high, the cost of generating it exceeds the price you can get for the extra ammonia produced.

5.23 understand how the cooling of the reaction mixture liquefies the ammonia produced and allows the unused hydrogen and nitrogen to be recirculated

Separating the ammonia
When the gases leave the reactor they are hot and at a very high pressure. Ammonia is easily liquefied under pressure as long as it isn’t too hot, and so the temperature of the mixture is lowered enough for the ammonia to turn to a liquid. The nitrogen and hydrogen remain as gases even under these high pressures, and can be recycled.

At each pass of the gases through the reactor, only about 15% of the nitrogen and hydrogen converts to ammonia. (This figure also varies from plant to plant.) By continual recycling of the unreacted nitrogen and hydrogen, the overall conversion is about 98%.

5.24 recall the use of ammonia in the manufacture of nitric acid and fertilisers

So ammonia is used to make nitric acid and fertilisers, as you know from bio, plants need nitrates to grow.

Just in case you want to know, here are some properties of ammonia:
  • alkaline gas (turns damp red litmus paper blue, which is the test for ammonia!)
  • extremely soluble in water–it forms a weak alkali–>ammonia solution
  • less dense than air
  • colourless gas with pungent odour

Metallic Crystals

1.45 describe a metal as a giant structure of positive ions surrounded by a sea of delocalised electrons

That’s what a metal is, a lattice of positive ions in a sea of delocalised electrons. Basically delocalised electrons is as the name suggests, it’s not attached to any particular atom and is able to move freely. 

1.46 explain the malleability and electrical conductivity of a metal in terms of its structure and bonding

Metals can conduct electricity because the delocalised electrons are free to move and carry the charge. The key words here are that the electrons are free to move/mobile and can move through the structure. 
Metals are malleable and ductile because they are a lattice structure and arranged in layers, which are able to slide off each other easily when a force is applied. [The layers are the sheets of positive ions.] Thus making it easy to bend and shape–>malleable. Easily pulled out into wires–>ductile.  
Metals can be made harder by alloying them with other metals. An alloy is a mixture of metals-for example, brass is a mixture of copper and zinc. In an alloy, the different metals have slightly different sized atoms. This breaks up the regular arrangement and makes it more difficult for the layers to slide. 

a pure metal, with neat layers that would easily slide over each other
the atoms have different sizes, so it disrupts the regular packing and makes it much more difficult for particles to slide over each other when a force is applied. this tends to make alloys harder than the individual metals that make them up. 

A common example of an alloy is stainless steel, which is iron mixed with chromium and nickel. The chromium and nickel form strong oxide layers which protect the iron. This is why stainless steel is so resistant to corrosion. Obvious uses include kitchen sinks, saucepans, knives and forks and gardening tools. But there are also major uses for it in the brewing, dairy and chemical industries where corrosion-resistant vessels are essential.

I’m sure you all know the difference between elements, compounds and mixture by now. So I’d just like to point out that alloys are considered mixtures rather than a compounds because of the totally variable proportions of the metals. Whereas in water (a compound), every single water molecule has two hydrogen atoms combined with one oxygen atom. In the alloys, the metals can be mixed in any proportions…

Tests for ions and gases

g) Tests for ions and gases
2.38 describe simple tests for the cations:
  1. Li+, Na+, K+, Ca2+using flame tests
  2. NH4+ using sodium hydroxide solution and identifying the ammonia evolved
  3. Cu2+, Fe2+ and Fe3+using sodium hydroxide solution

1. Flame tests:

  • Li+ (Lithium)à Red flame
  • Na+ (Sodium)à Orange/Yellow flame
  • K+ (Potassium)à Lilac
  • Ca2+ (Calcium)à Brick red flame

2. NH4+(Ammonium ion)à Add sodium hydroxide solution (aqueous NaOH) and do NH3 (ammonia) test on the fumes evolved. You use damp red litmus paper, and the NH3 evolved will turn it blue.

3. When you add sodium hydroxide solution to the following:
  • Cu2+ à Light blue precipitate formed
  • Fe2+ à Green precipitate formed
  • Fe3+ à Orange/brown precipitate formed

2.39 describe simple tests for the anions:
  1. Cl, Br and I, using dilute nitric acid and silver nitrate solution
  2. SO42-, using dilute hydrochloric acid and barium chloride solution
  3. CO32-, using dilute hydrochloric acid and identifying the carbon dioxide evolved

1. To test for the halide ions. When you add dilute nitric acid (aqueous HNO3) followed by silver nitrate solution (aqueous AgNO3) to the following:
  • Cl–  à white precipitate formed – insoluble AgCl
  • Brà cream precipitate formed – insoluble AgBr
  • I–     à yellow precipitate formed – insoluble AgI

2. To test for sulphate ions (SO42-).  When you add dilute hydrochloric acid followed by barium chloride solution (aqueous BaCl2): a white precipitate of insoluble BaSO4 is formed

3. To test for carbonate ions (CO32-). You can:
  • Add an acid (e.g. hydrochloric acid) and test any gas evolved with limewater. You should observe effervescence (fizzing), and the gas will turn limewater milky white as it is CO2 that is given off. Remember that acid + metal carbonate à salt + water + carbon dioxide
  • Heat it strongly, and bubble gas into limewater. Here, thermal decomposition is occurring and CO2 is given off, hence the limewater will turn milky white too. The most common example is with calcium carbonate, calcium carbonate à calcium oxide + carbon dioxide + water

2.40 describe simple tests for the gases:
  1. hydrogen
  2. oxygen
  3. carbon dioxide
  4. ammonia
  5. chlorine.

  • Hydrogen: apply a litsplint and you will hear a squeaky pop sound
  • Oxygen: apply a glowing splint and the splint relights (there’s also the option of applying a lit splint, and the flame just gets brighter but relighting the glowing splint one is much better)
  • Carbon dioxide: bubble it into limewater and it goes milky white
  • Ammonia: use a damp red litmus paper and it turns blue
  • You can also hold it near fumes of concentrated HCl and you  will observe cloudy fumes evolving, which is NH4Cl (ammonium chloride) and this gas has a characteristic pungent smell
  • Chlorine: use damp blue litmus paper and it goes red (then bleaches it white)

To remember which ones uses blue litmus paper and which ones use red litmus paper, remember the periodic table. The metal elements on the left form alkaline substances so they must turn red litmus paper blue – like the universal indicator turns blue in alkaline solutions. And vice versa with the non-metal elements on the right side of the periodic table, they form acidic substances so use blue litmus paper, and observe that it turns red. With the gases, the litmus paper must be damp so that the gases dissolve then act on it. For instance with the chlorine gas.

As an extra, I think you all should know the test for pure water too. You must heat it to boiling point and it boils at exactly  100°C, but you should also cool it till freezing point and it should freeze at 0°C. The reason for adding the freezing bit too is to make it fool-proof, as pure water is the only substance that has these exact boiling and freezing points whilst other substances may boil at 100°C too etc.

For testing the mere presence of water, you just need to add a few drops to anhydrous copper sulphate crystals, which are white, and they turn blue to become hydrated copper sulphate (formula: CuSO4.H2O the dot ‘.’ shows that the water is part of the structure, it is what makes it turn blue–called the water of crystallisation)

Water also turns blue cobalt chloride (CoCl2) pink. 

Crude Oil

Crude Oil

5.6 recall that crude oil is a mixture of hydrocarbons
Hydrocarbons are compounds that contain hydrogen and carbon atoms only.

5.7 describe how the industrial process of fractional distillation separates crude oil into fractions

Fractions are groups of hydrocarbons with similar boiling points. They are separated by fractional distillation.

Fractional distillation is carried out to separate crude oil into its fractions. How is it carried out? (3-4m)

Crude oil is heated, vaporised, and pumped into a fractionating column, which is coolest at the top and hottest at the bottom. The larger hydrocarbons sink whilst the smaller hydrocarbons rise up the column. As they near the area where the temperature is close to their boiling point, they condense, and the fractions are collected. The refinery gases rise straight up and are collected as gases. The longer the chain of hydrocarbons, the higher its boiling point, thus the further down the fractionating column it will be.

5.8 recall the names and uses of the main fractions obtained from crude oil: refinery gases, gasoline, kerosene, diesel, fuel oil and bitumen

Note: Use the name in bold, gasoline is also known as petrol and kerosene is also known as paraffin but use gasoline and kerosene.

Name of fraction (in order of increasing molecule size)
Main uses
Refinery gases
Fuels for lighters and cooking
Gasoline  (petrol)
Vehicles e.g. cars
Kerosene (paraffin)
Jet fuel, paraffin heaters
Fuel oil
Fuel for ships, factories, central heating
Roads and roofs

5.9 describe the trend in boiling point and viscosity of the main fractions

As the size of the hydrocarbon molecules increases, the boiling point and viscosity increases. This is because with increasing chain length (more carbon atoms and therefore more hydrogen atoms), van der Waal’s forces of attraction between the molecules increases.

5.10 recall that incomplete combustion of fuels may produce carbon monoxide and explain that carbon monoxide is poisonous because it reduces the capacity of the blood to carry oxygen

Incomplete combustion of hydrocarbons produces carbon monoxide, water and soot.
Carbon monoxide is a poisonous gas. It combines with haemoglobin in the red blood cells and prevents it from carrying oxygen, hence reducing the capacity of the blood to carry oxygen.

5.11 recall that, in car engines, the temperature reached is high enough to allow nitrogen and oxygen from air to react, forming nitrogen oxides

Just recallthis. And if you want to know, nitrogen oxides are major pollutants, I mean just look at the exhaust fumes coming out of the vehicles on the road…

5.12 recall that fractional distillation of crude oil produces more long-chain hydrocarbons that can be used directly and fewer short-chain hydrocarbons than required

When they say the long-chain hydrocarbons can’t be used directly it’s basically telling you that they’re not very useful. With increasing chain length, the hydrocarbons become less flammable, more viscous and therefore less useful. So there isn’t much use for these heavy fractions, but there’s more of it, whilst there’s a higher demand for short-chain hydrocarbons.
Thus chemists have found a way to convert these large, less useful, heavy fractions into smaller, more useful ones…:

5.13 describe how long chain alkanes are converted to alkenes and shorter-chain alkanes by catalytic cracking, using silica or alumina as the catalyst and a temperature in the range of 600-700°

Crude oil is a mixture of hydrocarbons. It can be separated by fractional distillation. In the petrochemical industry some larger hydrocarbon molecules are broken down into smaller hydrocarbon molecules by cracking.

Describe how cracking is carried out: (2m)
A long chain hydrocarbon is put into a catalytic cracker where there is an absence of air, high temperatures of 600-700°C and a catalyst of aluminium oxide to speed up the process. The result is a mixture of smaller saturated and unsaturated hydrocarbons/alkanes+alkenes.
Explain why it is necessary to do this: (3m)
Smaller hydrocarbon molecules are less viscous and more flammable. Thus they are more useful and in higher demand. (Petrol, a short chain alkane is in higher demand than say, diesel.) Cracking also produces alkenes as a by-product which can be used to make plastics. This brings about a bigger profit. (I believe hydrogen is also obtained for the Haber process, to manufacture ammonia.)