GGCCSSEE RReevviissiioonnnn 110011 GGuuiiddee ttoo CChhhheemmiissttrryy AAQQAA SSppeecciiffiiccaattiioooonn AA UUnniitt CChheemmiissttrryy CCCC33 CCHHYY33HH Daniel Holloway Contents 1 Acids & Bases 3 2 Energy Calculations 6 3 Water & Solubility 11 4 The Development of the Periodic Table 16 5 Chemical Analysis 22 End of Unit Questions 28 Copyright © 2009 Daniel Holloway Significant contribution Nelson Thornes AQA Science [GCSE Chemistry] 2 Proton Donors & Acceptors When an acid dissolves in water, it forms H+ions. This is a hydrogen atom which has lost an electron –in other words, it is a proton. These produced protons become surrounded by water molecules to keep them in solution –we call it hydrated. Hydrated hydrogen ions are shown with H+(aq). An alkaliis a basewhich dissolves in water, and produces OH-ions (hydroxide ions). Because acids act as a source of protons, we call them proton donors. The hydroxide ions from an alkali combine with protons to form water: OH-(aq)+ H+(aq)→ H O (l) 2 And because alkalis behave like this, we call them proton acceptors. Strength of Acids & Alkalis The strength of an acid depends on the extent to which it ionises in water. A strong acid or alkali is one which is 100% ionised in water. Hydrochloric acid, sulphuric acid and nitric acid are all strong acids. Sodium hydroxide and potassium hydroxide are both strong alkalis. A weak acid or alkali is only partly ionised in water. Ethanoic acid, citric acid and carbonic acid are all weak acids; and ammonia solution is a weak alkali. We can detect strong and weak acids using their pH. This scale is a measure of the concentration of hydrogen ions in a solution. 3 A strong acid, e.g. hydrochloric, will be completely ionised, so the concentration of hydrogen ions is 1 mol/dm³. However, a weak acid, such as citric acid is only partly ionised, so the concentration of hydrogen ions will be much lower than 1 mol/dm³ Titrations Adding an acidic solution to an alkaline solution will produce a neutralisation reaction. They react together and neutralise each other, producing a salt in the process. When a neutralisation reaction takes place, the quantities of each solution used must be correct, because if a very strong acid and a very strongalkali were mixed, if there was more acid solution, the whole alkali solution would be neutralised, but not all of the acid solution would be –so the mixture would become slightly acidic overall. We can measure precise volumes of acids and alkalis neededto react with each other using titrations. In the neutralisation reaction, the point at which the acid and the alkali have completely reacted is called the end point. We can show the end point using a chemical indicator. Indicators change colour over different pH ranges. We have to choose suitable indicators when carrying out titrations with different combinations of acids and alkalis:  strong acid + strong alkali –use any indicator  weak acid + strong alkali –use phenolphthalein  strong acid + weak alkali –use methyl orange These are the steps to carry out a titration to calculate how much acid is needing to react with an alkaline solution: 1 Measure an known volume of the alkali solution into a conical flask using a pipette 2 Add an indicator solution to the alkali in the flask 3 Now put the acidic solution into a burette. This long tube has measurements down the side, and a tap on one end and can accurately measure the amount entering the flask. So record the reading on the burette (i.e. starting volume) 4 Open the tap to release the acid solution. The solution from the burette is released one drop at a time, alongside swirling of the flask to ensure the solutions are mixed 5 Keep repeating Step 4 until the indicator changes colour to let you know the acid and the alkali have completely mixed 6 Record the amount of acid you entered by reading the measurement on the burette Be sure to repeat the entire process two or three times at least to ensure accuracy. 4 Calculations InvolvingTitrations When talking about concentration, we tend to describe it as the amount of the solute (in terms of moles) dissolved in the solution(in one cubic decimetre), so the units are mol/dm³ so if we know the amount of a substance dissolved in a known amount of solution we can calculate the concentration. For example, imagine we were making a sodium hydroxide solution in water by dissolving exactly 40g of sodium hydroxide to make 1dm³ of solution:  We know that the mass of one mole of NaOH is the sum of the atomic masses of sodium, oxygen and hydrogen: 23 + 16 + 1 = 40g  Because 40g is in the solution, we know that there is exactly one mole of NaOH in the solution  And we know that the solution is 1dm³, so the concentration is 1 mol/dm³ The worked examples below are more complicated calculations involving titrations: 5 Energy & Reactions An exothermic reactionreleases energy. We use exothermic reactions in burning fuels as a source of energy. However, some reactions give off more energy than others, so we can calculate how much energy is released in a given reaction. There is apparatus available to do this called a calorimeter. In a school lab, a simple calorimetermight be used –but a more accurate instrument is available, called a bomb calorimeter. A bomb calorimeter works by measuring the temperature of the water inside it –because the energy produced in an exothermic reaction increases the temperature of its surroundings, in this case the water. The change in energy is calculated using the temperature change and amount of water (see later on for calculations using energy). A simple calorimeter however involves very basic apparatus. We don’t use this to measure energy change necessarily because it isn’t very accurate –but we can use it to compare energy changes from different fuels. When a reaction takes place, bonds are broken and new chemical bonds are made:  breaking bondsis an endothermic process, because energy has to be taken in from the surroundings to break the bonds (remember energy is needed to break bonds)  making bondsis an exothermic process, because energy is released in the formation of new chemical bonds Because a reaction makesand breaksbonds, reactions are sort of both exo-and endothermic. For this reason, it is the balance between exo-and endothermic reactions which decides the overall reaction type; for example if more energy is released in the making of new bonds than is taken in to break the bonds, it is overall exothermic –because the exothermic > endothermic. Energy Level Diagrams We can draw energy level diagramsto show energy changes in a reaction. These diagrams show the relative amounts of energy stored in the products and reactants of a reaction, measured in kJ/mol. 6 This is the energy level diagram for an exothermic reaction. The products are at a lower energy level than the reactants, so energy has been released as the reactants form the products. In this release of energy, temperature of the surroundings increases. In such an exothermic reaction, we say that the change in energy is negative –which we writeas ΔH -ve (see below). This is so because energy is released –so there is less energy in the products than the reactants. So this is the energy level diagram for an endothermicreaction. With an endothermic reaction, more energy is needed to break to bonds of the reactants than is released in forming products. Here, temperature ofthe surroundings decreases. Because the change in energy this time is positive, we say ΔH +ve (see below). The Greek letter “delta” (written as Δ) is often used in the sciences and maths to represent change. In chemical energies, we use ΔHto abbreviate energy change. So +ΔH means energy increases, -ΔH means energy decreases. The amount of energy needed to start a reaction is called the activation energyof the reaction. Adding a catalystwill significantly reduce this amount of energy (see Rates of Reaction, C2). This in turn increases the proportion of reacting particles which will have enough energy to react. This has many advantages, especially industrially, as it means reactions are more efficient –and the catalysts are economical also. Calculating Energy Changes Looking back at the calorimeters above, when chemicals react and give off/take in energy, we can use calculations to work out exactly how much energy has changed. There is one vital piece of information we need to know to do this… 7 4.2 joules of energy raises 1g of water by 1°C Hence the units involved in this energy change will be kJ/g/°C (kilojoules per gram per degree). A simple calorimeter is used to measure energy change in a reaction A+ B→ C. So let’s calculate an example of such a reaction: Question: 60cm³ of a solution containing 0.1 moles of Ais mixed with 40cm³ of a solution containing 0.1 moles of B. Prior to mixing, their temperature was 19,6°C. After mixing, the maximum temperature reached was 26.1°C. 1 First, calculate the temperature change: 26.1°C –19.6°C = 6.5°C 2 Since 60cm³ of Aadded to 40cm³ of Bmakes 100cm³ overall, we are looking at 100g (assuming the density of the solution is the same as water density). And we know that 4.2J raises 1g by 1°C 3 So energy change = 100g x 6.5°C x 4.2J/g/°C = 2,730J = 2.73kJ 4 BUT –don’t forget the solutions are only 0.1 molar –so we have to multiply our value by 10 to find out a 1.0M solution 2.73kJ x 10 = 27.3kJ 5 So the final energy change was -27.3kJ [We know that the temperature increased, so the reaction was exothermic -where energy gets released. That is how we know the energy change will be negative] Bond Energies The energy required to break apart a bond between two particular atoms is known as bond energy. Bond energies are measured in kJ/mol and we can use them to work out ΔH in energy calculations. Some of the most common bond energies are displayed below: 8 To calculate energy change we need to know: a) the amount of energy needed to break the bonds between the atoms; and b) the amount of energy released in the formation of new chemical bonds. For example, the bond energy for an H-H bond is 436kJ/mol. This means that the bond energy for forming a new H-H bond is -436kJ/mol. These two energy level diagrams show H-H bonds being made and broken. The left diagram shows an already bonded H-H bond being broken. This has a bondenergy of +436kJ/mol, so we write ΔH = +436kJ/mol on the diagram next to the change in height arrow. The right side is a diagram representing two separate hydrogen atoms bonding. Obviously, this is bond making –which releases energy –so the energy change is -436kJ/mol, written the same way as before, except with a minus sign. To clarify, the left diagram is endothermic, the right is exothermic. Making and breaking the same bond alwaysinvolves the same amount of energy, just different + and –signs. More Complicated Energy Level Diagram Calculations Although bond making and breaking is always the same energy levels back and forward – different chemical reactions mix and match the type of bonds being made and broken. This is why energy levels can begin to look a bit more complicated. A good example of a chemical reaction where different bonds are involved is the Haber process. This is the making of ammoniafrom nitrogen and hydrogen. Question: Ammonia is made from nitrogen and hydrogen in the Haber process. The balanced chemical equation for this reaction is: N + 3H → 2NH 2 2 3 Calculate the overall energy change for this reaction. 9 Solution: 1 Notice we need to break 1 mole of nitrogen and 3 moles of hydrogen. Nitrogen molecules are held together by an extremely strong triple bond (N≡N) with a bond energy of 945kJ/mol. Hydrogen molecules are held together by a single bond with a bond energy of 436kJ/mol. 2 Since there are three moles of hydrogen, the energy needed to break one mole nitrogen and three hydrogen moles is: 945kJ + (3 x 436)kJ = +2253kJ 3 We can therefore draw on the first line taking us to our 2 nitrogen atoms and 6 hydrogen atoms, as shown 4 When these atoms form ammonia, NH , 6 N-H bonds are made as 2 moles of NH are 3 3 formed (N-H has a bond energy of 391kJ/mol) 6 x 391kJ = 2346kJ 5 Because we are making bonds, energy is released, so it becomes -2346kJ 6 We can draw on a second arrow taking us to our two moles of NH , 3 7 We can then calculate the difference (+2253) –2346= -93kJ 8 So the ΔH = -93kJ 10