An equilibrium reaction is one in which the reactants, say A + B, collide successfully and react together to form the products, C + D. The 'reverse' also occurs simultaneously, in which C + D react to form A + B once again. When this system reaches a point in which the concentrations of A, B, C & D are constant, the reaction is said to be in Dynamic Equilibrium. At this point, the reaction appears to have stopped, but both the forward and backward reactions are still occurring at equal rates. However, it is very rare for this equilibrium position to be at the half-way point.

The above equilibrium reaction can be represented by the equation:

A + B ⇌ C + D

Le Chatelier's principle can be used to predict the changes that will occur if the conditions are changed in a chemical equilibrium reaction. His principle states that:

"If a system at equilibrium is subject to a change in either pressure, temperature or concentration, then the system will move to oppose that change."

Using this principle is relatively straightforward; whatever you change in the reaction conditions, the system will move in the opposite direction to try and restore the equilibrium condition.

For example, let us consider the reaction in which Nitrogen Dioxide forms Dinitrogen Tetroxide

2 NO₂(g) ⇌ N₂O₄(g) ΔG = 45.53 kJ/mol

Temperature: It can be seen that the reaction here is endothermic (takes in heat from the surroundings). If the reaction conditions are changed, such that the temperature is increased, the equilibrium will attempt to oppose the change by decreasing the temperature and thereby shifting in the endothermic direction, to the RIGHT. Conversely, if the temperature is decreased, then the equilibrium will oppose the change by moving in the exothermic direction to raise the temperature once again, and hence the system will move to the LEFT.

Think about the rates of the forward and reverse reactions. The endothermic reaction (in this case, the forward reaction) is more sensitive to changes in temperature. Increasing temperature will increase the rates of both forward and reverse reactions, but the rate of the endothermic forward reaction will increase by a bigger factor.

Pressure: To determine the way an equilibrium reaction will shift when the overall pressure is changed, one must look at the number of moles of each gas species present in the reaction. In the above reaction concerning Nitrogen Dioxide it can be seen from the equation that the side with the greatest number of moles of gas is the left. Therefore, increasing the pressure on this reaction will cause the equilibrium to move right, as this is the side with fewer moles of gas, which thereby reduces the pressure once again.

Again, you can think about the rates of the forward and reverse reactions. In this case, the forward reaction has two gas molecules (2 NO₂(g) ) reacting but the reverse reaction has only only gas molecule reacting (N₂O₄(g) ). Increasing pressure will increase the rates of both forward and reverse reactions, but the rate of the forward reaction will increase by a bigger factor.

Concentration: If the concentration of one of the chemicals in an equilibrium reaction is changed, then the system will oppose the change by moving to either increase or decrease the concentration, depending on what change was made. Consider the following reaction:

H₂ + I₂ ⇌ 2 HI

If you were to add lots of Hydrogen Iodide to the above equilibrium, the system will move left to remove the excess HI concentration added. Likewise, if a large excess of Iodine was added, then the system would move to the right to reduce the concentration.

And again, you can think about the rates of the forward and reverse reactions. Adding HI will increase the rate of the reverse reaction. Adding I₂ will increase the rate of the forward reaction.

Catalysts;

Some reactions require the use of a catalyst in order to reach equilibrium in a feasible, efficient time. A catalyst does not affect the position of an equilibrium; it speeds up both the forward and backward reactions equally and therefore only increases the rate at which equilibrium is reached.

Catalysts work by providing an alternate reaction pathway of lower activation energy for both the reactants and products in an equilibrium reaction, and the overall reaction energy remains unchanged.

Ammonia

For many reactions, a compromise in reaction conditions must be met in order to achieve the best yield of product. Although, for example, a reaction may yield the greatest amount of product at extremely high temperatures, it is not feasible to generate such high temperatures in industry.

Consider the Haber Process, in which Hydrogen & Nitrogen react to form Ammonia:

3 H₂ + N₂ ⇌ 2 NH₃ ΔG = -92.4 kJ/mol

The forward reaction is exothermic, and therefore the reaction is favoured by low temperatures. Even though this is the case, having temperatures that are too low will make the reaction too slow to be commercially useful. It is therefore necessary for a compromise temperature of approx 450 °C to be made, resulting in a mere 10-20 % yield of Ammonia.

The Haber Process is also favoured by high pressures, since there are more moles of gas on the left (reactant) side of the equilibrium equation. Generating pressures that are extremely high is a difficult and costly process, since the energy required to generate such pressures can be high, as can the cost of the vessel used to withstand such high pressures. The compromise pressure here is 250 atm.

Ethanol

This is the alcohol in alcoholic drinks and therefore is unsurprising that it has been made for thousands of years by fermentation of sugars such as glucose while using the enzymes in yeast as a catalyst.

C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂

Ethanol is also used in drugs, cosmetics, detergents and inks. At present the main source of ethanol is ethene, which is made from crude oil obtained by fractional distillation then cracking.

Ethanol is made by the reversible reaction of hydration (adding of water) to ethene and is speeded up by the catalyst of phosphoric acid absorbed on silica.

H₂C=CH₂ + H₂O ⇌ CH₃CH₂OH ΔH = -46 kJ mol⁻¹

The products and reactants are all gaseous at the temperature used. Applying Le Chatelier's principle to this equilibrium predicts that the maximum yield will be produced with:

• a high pressure forcing the equilibrium to move to the right with fewer gas molecules
• a low temperature forcing the equilibrium to the right to give out heat
• excess steam to force equilibrium to the right again to reduce the steam concentration.

However the practical problems include;

• the low temperature will reduce the reaction rate and thus how quickly the equilibrium is reached
• the high pressure tends to cause ethene to polymerise (to polyethene)
• the high pressure increases costs of building the plant and energy cost to run it
• too much steam will simply dilute the catalyst.

In practice conditions of about 570 K and 6500 kPa pressure are used obtaining a yield of ethanol at around 5 %. The unreacted ethene is separated from the reaction mixture and recycled over the catalyst again until a 95 % conversion is obtained.

An acid is a substance which in an aqueous solution will release H⁺ ions. Acids are also known as proton donors as they release a "proton" in the form of a H⁺ ion. Common laboratory acids include hydrochloric acid (HCl), sulfuric acid (H₂SO₄), nitric acid (HNO₃) and ethanoic acid (CH₃COOH).

A base is a proton receiver as it readily accepts H⁺ ions ("protons") from an acid. Examples:

• OH^- + H⁺ → H₂O
• O^2- + 2 H⁺ → H₂O
• NH₃ + H⁺ → NH₄⁺

Common bases are metal oxides, metal hydroxides, and ammonia. An alkali is a soluble base, and will release OH^- ions in an aqueous solution. Common alkalis include sodium hydroxide (NaOH), potassium hydroxide (KOH), and aqueous ammonia (NH₃(aq)).

A salt is produced during neutralisation: the reaction of an acid with a base, alkali or carbonate. The H⁺ ion of an acid is replaced by a metal ion or NH₄⁺.

Here are some common reactions:

• Acid + Base → Salt + Water
• Acid + Alkali → Salt + Water
• Acid + Metal → Salt + Hydrogen
• Acid + Carbonate → Salt + Carbon Dioxide + Water

Solid salts consist of a lattice of positive and negative ions, sometimes water molecules are incorporated into the lattice. The water in a lattice is called the water of crystallisation. When a salt contains water the salt is hydrated, if a salt does not contain water of crystallisation the salt is anhydrous.

A mole of a particular hydrated salt usually has the same number of moles of water of crystallisation, its formula shows how many water molecules are present for every molecule of salt e.g. copper sulfate, CuSO₄, has 5 moles of water for every mole of salt, therefore its hydrated formula is CuSO₄ ⋅ 5 H₂O, (a dot separates the salt's formula from the water of crystallisation).

Examples:

• CuSO₄ ⋅ 5 H₂O
• Na₂CO₃ ⋅ 10 H₂O
• MgSO₄ ⋅ 7 H₂O

Many hydrated salts lose their water of crystallisation when heated.

Water of crystallisation is confusing when working out concentrations of salt solutions. If we want to make 250 cm³ of a 0.100 mol dm^-3 solution of sodium carbonate, then how much sodium carbonate do we need?

The easy bit is (or should be) n = c x V = 0.250 dm³ x 0.100 mol dm^-3 = 0.0250 mol.

What is the mass of 0.0250 mol of sodium carbonate? If we are using the decahydrate (Na₂CO₃ ⋅ 10 H₂O) then its M_r is 286.2 and the mass is m = n x M_r = 7.16 g

If we are using the anhydrous salt (Na₂CO₃) then its M_r is 106.0 and the mass is m = n x M_r = 2.65 g

So 7.16 g of Na₂CO₃ ⋅ 10 H₂O dissolved in 250 cm³ is an Na₂CO₃ solution (notice the water is no longer mentioned) with a concentration of 0.100 mol dm^-3.