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MANUAL CHEMICAL REACTIONS Apply knowledge of chemical reactions in a processing environment Unit Standard 244241 NQF Level 3 Credits: 6 Compiled by: Kgomotso Matobole Rev.1 – September 08 Moderated by: Learner Name: Learner Number: Rev.1 – Sept 08 Sparrow Consulting © September 08 Table of Contents UNIT 1: ELEMENTARY CHEMICAL CALCULATIONS .................................................... 5 1.1 Instructions ...................................................................................................... 5 1.2 Introduction ...................................................................................................... 7 1.2.1 Atomic number and mass number ................................................................... 7 1.2.2 Isotopes ........................................................................................................... 9 1.2.3 Atomic mass and relative atomic mass ............................................................ 9 1.2.4 Relative formula mass ................................................................................... 10 1.2.5 Mole .............................................................................................................. 11 1.2.6 Avogadro's number ........................................................................................ 12 1.2.7 Molar volumes of gases ................................................................................. 13 1.2.8 Chemical equation based calculations ........................................................... 14 UNIT 2: ACIDS AND BASES .......................................................................................... 16 2.1 Instructions .................................................................................................... 16 2.2 Introduction .................................................................................................... 18 2.2.1 Properties of acids and bases ........................................................................ 18 2.2.2 Definitions for acids and bases ...................................................................... 18 2.2.3 Strong and weak acids .................................................................................. 20 2.2.4 Water ionisation and pH ................................................................................ 22 2.2.5 Acid reactions ................................................................................................ 24 2.2.6 Acid-base titration calculations ...................................................................... 26 2.3 Experiment .................................................................................................... 29 UNIT 3: OXIDATION-REDUCTION (REDOX) REACTIONS............................................ 31 3.1 Instructions .................................................................................................... 31 3.2 Introduction .................................................................................................... 33 3.3 Oxidation number .......................................................................................... 33 3.3.1 Rules for assigning oxidation numbers .......................................................... 34 3.4 Balancing redox reactions.............................................................................. 35 3.4.1 Oxidation number method.............................................................................. 35 3.5 Standard electrode potential .......................................................................... 39 3.5.1 Galvanic cell .................................................................................................. 39 3.6 Standard potentials ........................................................................................ 41 3.7 Experiment .................................................................................................... 44 UNIT 4: CHEMICAL REACTION RATES ........................................................................ 46 4.1 Manual Instructions .................................................................................................... 46 2 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 4.1.1 Chemical reaction rates ................................................................................. 48 4.1.2 Factors affecting the rate of reaction .............................................................. 49 4.2 Experiment .................................................................................................... 51 UNIT 5: CHEMICAL EQUILIBRIUM ................................................................................ 53 5.1 Instructions .................................................................................................... 53 5.1.1 Equilibrium ..................................................................................................... 55 5.1.2 Equilibrium Constant...................................................................................... 56 5.1.3 Factors affecting chemical equilibrium ........................................................... 59 5.2 Experiment .................................................................................................... 61 Figures and tables Figure 1: Periodic table of elements with atomic and mass numbers ................................ 8 Figure 2: Carbon isotopes ................................................................................................. 9 Figure 3: Acids turn litmus paper red, and bases turn litmus paper blue .........................19 Figure 4: Titration setup ...................................................................................................26 Figure 5: example of regular batteries ..............................................................................39 Figure 6: Simple galvanic cell .........................................................................................40 Figure 7: Rate of reaction curve .......................................................................................48 Figure 8: Equilibrium of a trioxide producing reaction .......................................................56 Table 1: List of elements with their atomic mass and number .........................................10 Table 2: Acids and bases with their conjugates...............................................................20 Table 3: Common strong acids .......................................................................................21 Table 4: The ionisation constants of common acids ........................................................22 Table 5: Hydronium and hydroxide ions concentrations in various solutions ....................22 Table 6: pH values of common acids and bases .............................................................23 Table 7: Reduction potentials at 25ºC .............................................................................43 Manual 3 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Module overview Introduction This module provides an introduction to elementary chemistry as it is applied in the process industry. In Unit 1 the learner is prepared to perform a range of elementary chemical calculations. This includes calculating relative atomic masses, relative formula masses, masses for a mole of different substances and the mass and percentages of substances. In Unit 2 the learner is introduced to the properties of acids and bases and provided with a general understanding of reactions with acids and bases. In Unit 3 the principles and working of oxidation-reduction reactions are discussed. Unit 4 concerns chemical reaction rates and its effect on the industry. In Unit 5 learners are introduced to the principles of chemical equilibrium, factors that affect chemical equilibrium and the effect it has on industry. Learning outcomes By the end of this module, you should be able to: Perform elementary chemical calculations. Demonstrate an understanding of acids and bases. Demonstrate an understanding of oxidation-reduction (redox) reactions and their industrial applications. Demonstrate an understanding of chemical reaction rates. Demonstrate understanding of chemical equilibrium. US specific outcomes The following specific outcomes are covered in this module: SO1: Perform elementary chemical calculations. SO2: Demonstrate an understanding of acids and bases. SO3: Demonstrate an understanding of oxidation-reductions (redox) reactions and their industrial applications. SO4: Demonstrate an understanding of chemical reaction rates. SO5: Demonstrate understanding of chemical equilibrium. Manual 4 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 UNIT 1: ELEMENTARY CHEMICAL CALCULATIONS 1.1 Instructions Ref. No SO1 AC1-7 Resources Learning materials CCFO2, 3, 6, 7 Learning Methodology Read through Unit 1 of the learning materials. Make notes of things you do not understand and/ or need more information on. Workbook Assessment N/ a N/ a Ex. 1 Ass. 1 Act. 1 SO1 AC1-7 Classroom Learners attend a lecture on the following: CCFO 2-7 Facilitator atomic number, mass number, atomic mass, mole and Avogadro constant relative atomic masses relative formula masses isotopes calculation of the masses for a mole of different substances calculation of the volumes for a mole of different gases calculation of the mass and percentages of substances. Manual 5 Act. 2 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Ref. No Resources Learning Methodology SO1 AC5-7 Classroom In pairs, discuss and calculate the: CCFO1-7 Facilitator masses for a mole of different substances Calculators volumes for a mole of different gases mass and percentages of substances. SO1 AC5-7 CCFO1-7 Learning materials and workbook PoE Facilitator/ SME Manual Revise the work that you have done up to this point. Make sure that you have completed the CCFO checklist and obtained the required evidence for your PoE. If there is anything that you do not understand, ask your facilitator. 6 Act. 3 Workbook Assessment Ex. 2 Ass. 1 CCFOs CCFOs Act. 4 US 244241 Rev.1 – Sep 08 Sparrow Consulting © September 08 1.2 Introduction Chemistry has its own language and it is therefore important to familiarise yourself with the terminology before you perform elementary calculations. 1.2.1 Atomic number and mass number The atomic number of an atom is the number of protons in the nucleus of an atom. If the atom is neutral, there are an equal number of electrons and protons in an atom. It is important to note that the number of electrons is closely related to the element's properties. Every atom has a unique atomic number. For example, the only element with an atomic number of 6 is carbon. The atomic number also indicates the position of the element on the periodic table. The symbol Z is used to represent an atomic number. The mass number is the sum of the protons and neutrons in the nucleus. These neutrons are the neutral particles also contained in the nucleus and are indicated by the symbol N. Mass number is indicated by the symbol A. Therefore mass number can be calculated as follows: A=Z+N. It is a norm to represent the nuclear composition of an element (E) as ZA E. For instance, the symbol 8 16O means that the oxygen atom has a mass number of 16 and atomic number of 8. Figure 1 below shows the elements with their atomic and mass numbers: Manual 7 US 244241 Rev.1 – Sep 08 Sparrow Consulting © September 08 Figure 1: Periodic table of elements with atomic and mass numbers Manual 8 US 244241 Rev.1 – Sep 08 Sparrow Consulting © September 08 1.2.2 Isotopes Isotopes are atoms that have the same atomic number, but a different atomic mass. They have the same number of protons and electrons, but a different number of neutrons. For example, most carbon has six protons and six neutrons, which gives it an atomic mass of 12. Some carbon atoms have an atomic mass of 14, implying that it has six protons and eight neutrons. This is illustrated in Figure 2, where the red dots indicate the protons and the grey ones the neutrons. All isotopes of the same element have the same chemical properties. Figure 2: Carbon isotopes 1.2.3 Atomic mass and relative atomic mass By definition the atomic mass of the carbon isotope 12C is exactly 12 atomic mass units (amu). It follows that 1 atomic mass unit = 1/ 12 x mass of one 12C atom. Since one atom has a mass of 1.9926 x 10-26 kg, it follows that 1 amu = 1.9926 x 10-26/ 12 = 1.6605 x 10-27 kg. This value is known as the atomic mass constant, u. Therefore atomic mass is the total mass of protons, neutrons and electrons in a single atom (when the atom is motionless). The relative atomic mass of an element is the ratio of the absolute mass of an atom of that element to that of the atomic mass unit. The atomic mass of the carbon isotope 12C is exactly 12 atomic mass units. It is therefore a relative number that indicates how many times heavier an atom is than a carbon atom. This is not a whole number because the atoms in nature have isotopes of different masses. It is represented by the symbol Ar. Note that relative atomic mass does not have units. Manual 9 US 244241 Rev.1 – Sep 08 Sparrow Consulting © September 08 The table below lists a few elements with their atomic mass and number: Name Symbol Atomic number Relative atomic mass Hydrogen H 1 1.008 Helium He 2 4.003 Lithium Li 3 6.939 Beryllium Be 4 9.012 Boron B 5 10.811 Carbon C 6 12.011 Nitrogen N 7 14.007 Oxygen O 8 15.999 Fluorine F 9 18.998 Table 1: List of elements with their atomic mass and number 1.2.4 Relative formula mass The smallest units of chemical compound are called formula units, regardless of whether they are molecules or ions. The total mass of all the atoms in a molecule is called the formula mass. Relative formula mass can be calculated by adding up all the relative atomic masses of the elements that make up the compound. It is represented by the symbol Mr and it is measured in g/ mol. Example The relative formula of marble (CaCO3) can be calculated as follows: Mr. (CaCO3) Manual = [40+12+(3x16)] = 100g/ mol 10 US 244241 Rev.1 – Sep 08 Sparrow Consulting © September 08 1.2.5 Mole Scientists discovered that by simply determining the mass of the substance, it was possible to count particles or atoms. A mole (mol) is the amount of a pure substance that contains the same amount of chemical units as there are atoms in exactly 12 grams of carbon, namely 12. In order to avoid confusion, it is necessary to specify the elementary units, which may be a molecule, an atom, an electron, an ion, etc. For example, a mole of hydrogen can mean either a mole hydrogen atoms (H) or hydrogen molecules (H2). The symbol n is used to represent the amount of the substance. It is critical to note that the amount of the substance is not the same thing as the mass of the substance, because the amount (n) of a substance is proportional to the mass (m) of the substance. This proportionality can also be written as an equation: n= m/ Mr Note the units of mole: g/ g.mol-1= g.mol/ g = mol Example 1 How many moles of atoms are there in (a) 50g of aluminium and (b) 0.55g of chlorine? (a) n= m/ Mr = 50/ 27 = 1.85 mol (b) n= m/ Mr = 0.550/ 35.5 = 0.22 mol 2 How many moles of hydrogen molecules are there in 10g of water? In each water molecule there are two hydrogen atoms, therefore each mole of water contains 2 moles of hydrogen. Mr (H2O) =(1x2) + 16 = 18g/ mol n(H2O) =10/ 18 Manual 11 US 244241 Rev.1 – Sep 08 Sparrow Consulting © September 08 = 0.56 mol In this case 0.56 mole of water contains 2 x 0.56 mole of H atoms, which works out to be 1.12 mol. 3 What is the relative atomic mass of an element of 0.1 mol and a mass of 1.6g? Also name the element. M=m/ n = 1.6/ 0.1 =16g/ mol From the periodic table it can be seen that oxygen has a relative atomic mass of 16g/ mol. 1.2.6 Avogadro's number The mass of a carbon atom was determined to be 1.9926 x 10-23g. Therefore, the number of carbon atoms in 12g 6 12C is = 12g/ 1.9926 x 10-23g = 6.02x1023 This number is known as the Avogadro's number or Avogadro's constant (named after the Italian scientist, Amadeo Avogadro). This number equals 1 mol. Avogadro's number is represented by NA or L and is measured in units of mol-1 because it is the number of elementary particles in one mol of a substance. Example 1 How many molecules are there in 50g of carbon dioxide? Take as Ar (C) =12 and Ar (O) =16 from the periodic table Mr (CO2) = (12 +2x16) = 44g/ mol n (CO2) = 50g/ 44g.mol-1 = 1.14 mol This carbon dioxide contains 1.14 mol x6.02x1023 molecules/ mol = 6.86 x1023 molecules Manual 12 US 244241 Rev.1 – Sep 08 Sparrow Consulting © September 08 2 How many atoms are there in 0.25 mol of copper? = 0.25 mol x 6.02x1023 atoms/ mol Number of atoms = 1.51x1023 atoms 3 Calculate the mass of 2 x1024 aluminium atoms n (Al) = 2 x1024 atoms/ 6.02x1023 atoms/ mol = 3.32 mol Ar (Al) = 27g/ mol m (Al) = 3.32 mol x 27g/ mol =89.7g 1.2.7 Molar volumes of gases The molar volume of a substance is the volume occupied by one mole of that substance. It can easily be calculated from the known density of the substance, since: Density = Mass/ volume Gases are of particular interest since their density is greatly dependent on temperature and pressure. Using the results obtained from experiments performed with various gases, it was determined that 1 mol of a gas occupies 22.414 dm3 (cubic decimetre) at standard temperature and pressure (STP). Standard temperature occurs at 0º C and standard atmospheric pressure is101.3kPa. To put it in another way, one mole of any gas at STP will occupy a volume of 22.4 dm3. Example 1 How many litres will 0.350 moles of HCl occupy at STP? Use the short method to solve for x: 22.414 dm3/1 mol =x/ 0.35 mol x=7.84 dm3 This gas occupies 7.84 litres. 2 Manual Calculate the volume occupied by 5g of hydrogen gas at STP. 13 US 244241 Rev.1 – Sep 08 Sparrow Consulting © September 08 Mr (H2) = 1x2 = 2g /mol n (H2) = 5/ 2 = 2.5 mol Short method 22.414 dm³/ 1 mol = x/ 2.5mol x=56.04 dm³ 1.2.8 Chemical equation based calculations In the industry, chemical equations are of great importance, because it allows you to determine whether a certain chemical manufacturing process will produce the desired product economically. Chemical equations express the net change in composition associated with a chemical reaction by showing the number of moles of reactants and products. 1.2.8.1 Mass calculations Example Sodium chlorate decomposes into sodium chloride and oxygen according the following chemical equation: 2NaClO3 => 2NaCl + 3O2 Calculate the mass of oxygen when 50g of sodium chlorate decomposes completely into sodium chloride. 1 Calculate the formula masses for the substances concerned. Mr (NaClOз) Mr (O2) 2 = 103.5g/ mol = 32g/ mol Multiply the balancing number by the formula mass. 2NaClOз => 2NaCl + 3O2 Manual 2x103.5 3x32 =207 =96 14 US 244241 Rev.1 – Sep 08 Sparrow Consulting © September 08 3 Use the mass unit in the problem statement and calculate the mass of oxygen by means of a simple proportion: 207g of NaClOз produces 96g of oxygen Therefore 50g of NaClOз yields (50g/ 207g) x 96g = 23.1g of O2 1.2.8.2 Percentage calculations Example 1 Calculate the percentage of hydrogen in sulphuric acid. The formula for sulphuric acid is H2SO4 2 Mr (H2SO4) = 98g/ mol Mr (hydrogen) = 2g/ mol Calculate the unknown moles of NaOH. moles = concentration x volume n (NaOH) = CV = (50x10-3 L) x 0.1mol/ L = 0.005 mol Therefore the percentage of H = (2/ 98) x100% =2.04 % Manual 15 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 UNIT 2: ACIDS AND BASES 2.1 Instructions Ref. No SO2 AC1-6 Resources Learning materials CCFO2-4, 6, 7 SO2 AC1-4 CCFO1-7 Samples of household acids and bases Litmus paper Learning Methodology Read through Unit 2 of the learning materials. Make notes of things you do not understand and/ or need more information on. As an introduction to Unit 2, collect samples of household acids and bases, such as lemons, vinegar, soap and other household products and bring it to the first lecture. SO2 AC1-4 Classroom/ Laboratory Attend a lecture on: CCFO1-7 Materials and equipment for the experiments the properties of acids and bases strong and weak acids the ionisation of water and the concept of pH. Computer and projector with animations of the experiments and/ or PowerPoint slides of the experiments Manual Workbook Assessment N/ a N/ a Act. 5 Ass. 1 During the lecture, you will observe and/ or participate in experiments/ animations of the experiments to help you understand the concept of acids and basis. Ex. 3 Act. 6 16 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Ref. No Resources Learning Methodology SO2 AC5, 6 Classroom Attend a lecture on how to: CCFO1-7 Facilitator balance and explain a range of acid reactions by using accepted chemical techniques and notation perform standardisation calculations. SO2 AC1-6 CCFO1-7 Learning materials and workbook PoE Facilitator/ SME Manual Revise the work that you have done up to this point. Make sure that you have completed the CCFO checklist and obtained the required evidence for your PoE. If there is anything that you do not understand, ask your facilitator. 17 Act. 7 Workbook Assessment Ex. 4 Ass. 1 CCFOs CCFOs Act. 8 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 2.2 Introduction It was not until a few hundred years ago that it was discovered that vinegar and other sour tasting foods have this taste because they are acids. The word acid comes from the Latin term ‘acere’, which means sour. 2.2.1 Properties of acids and bases Characteristics of acids: releases hydrogen ions [H+] into water solution corrodes active metals turns a blue litmus paper red tastes sour. Characteristics of bases: generate hydroxide ion [OH-] into water solution turns red litmus paper blue has a soapy ('slippery') feel has a bitter taste. It would be too dangerous to taste a liquid in order to find out if it is a base or an acid. That is why chemists use indicators. These indicators change colour when added to acids or bases. 2.2.2 Definitions for acids and bases Bases are also known as alkalis and are the chemical opposite of acids. There are many definitions for acids and bases; in this section we will define them by three well-known definitions. Arrhenius's definition Bronsted-Lowery's definition Lewis's definition 2.2.2.1 Arrhenius’s definition Arrhenius's equation for an acid and a base is as follows: Acid + Base ↔ Water + Salt Manual 18 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 The equation above is used to explain a reaction called neutralisation. Figure 3: Acids turn litmus paper red and bases turn litmus paper blue. It is important to note that litmus paper is the most commonly used and oldest indicator of acids and bases. 2.2.2.2 Bronsted-Lowery's definition An acid is a material that donates protons [H+] and a base is a material that accepts protons. The following equation is normally used to illustrate this process: acid + base ↔ acid + base Example HCl + H2O ↔ HзO+ + ClEach base has a conjugate acid and each acid has it own conjugate base, which only differs with a proton. From the example HCl is an acid, its conjugate base is Cl - and H2O is the base with HзO+ as its conjugate acid. The table below gives a few examples of acids and bases with their conjugates. Manual 19 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Acids Conjugate base Base Conjugate acid HCl Cl- OH- H2O HNOз NOз- NHз NH4+ H2SO4 HSO4- COз²- HCOз- HSO4- SO4²- HCOз- H2COз Table 2: Acids and bases with their conjugates 2.2.2.3 Lewis's definition An acid is a material that accepts an electron pair and a base donates an electron pair. When using this definition, an acid can also be called an electrophile and a base a nucleophile. This makes it possible for many more reactions to be considered as acidbase reactions even if they do not occur in solution. 2.2.3 Strong and weak acids Since an acid is a proton donor when dissolved in water, a hydrogen ion (proton) is transferred to a water molecule. This reaction produces a hydronium ion (H3O+ or H+) and a negative ion. This is generally represented by the equations below: HX + H2O ↔= HзO+ + X- or HX ↔ H+ + X- At a certain time virtually 100% of the acid will have reacted to produce these products. This acid gives the maximum concentration of [H+] and is considered to be a strong acid. A strong acid is an acid that dissociates completely or nearly completely in an aqueous solution. For a strong acid, the concentration of HзO+ will be equal to [X-], meaning that after equilibrium has been reached the concentration of the acid HX will be close to zero. Table 5 shows the common acids that are almost hundred percent ionised, for example when hydrogen chloride dissolves in water. Manual 20 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Chemical formula and names of acids HBr Hydrobromic acid HCl Hydrochloric acid HI Hydroiodic acid HNOз Nitric acid HClO4 Perchloric acid Table 3: Common strong acids A weak acid is one that does not completely ionise in water. The degree to which the acid dissolves is also measured by the equilibrium constant Ka When the equilibrium constant is relatively large then the acid is strong and when equilibrium constant is small then the acid is weak. A very small equilibrium constant means a very weak acid. Consider the dissociation of a weak acid, namely hypochlorous acid: HOCl + H2O ↔= HзO+ + OClWhen formulating the expression for the equilibrium constant for the reaction: Ka = [HзO+] x [OCl-]/ [HOCl] x [H2O] The molar concentration of water is always assumed to be 1, therefore: Ka = [HзO+] x [OCl-]/ [HOCl] The numerical value of the ionisation constant gives the measure of the acid strength. The larger the Ka value, the stronger the acid will be. The Ka for strong acids is in the vicinity of infinity, implying that the equilibrium lies far to the right. Do not confuse strong and weak acids or bases with concentrated and diluted acids or bases. An acid or base is not necessarily stronger when it is present in a high concentration. A diluted acid or base does not make that acid or base weaker. Manual 21 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Acid Formula Ionisation constant Ka Hydrochloric HCL 107 Sulfurous H2SO3 2.4 ×10-6 Phosphoric H3PO4 7.5×10-3 Ammonia NH3 1.8 x10-5 Hydrogen Sulfate ion HSO4- 1.1×10-2 Hypochlorous HCLO 3.5 x 10-8 Table 4: The ionisation constants of common acids 2.2.4 Water ionisation and pH Water can act as a base or an acid and as such is termed amphitropic. Amphitropic elements can either accept or donate a proton. The equation below, called the auto ionisation of water, has the equilibrium constant Kw: H2O + H2O ↔HзO+ + OHKw = [HзO+] x [OH-] The dissociation constant (Kw), also known as the ion product, has the very small value of 10-14 at 25 ºC and gives us the concentrations of hydronium and hydroxide ions in pure water, acidic and basic solutions. The table below indicates the range of hydronium and hydroxide ion concentrations in various solutions. Solution Hydronium ion molar concentration [HзO+] Hydroxide ion molar concentration [H-] Neutral 1x10-7 1x10-7 Acidic > 1x10-7 < 1x10-7 Basic < 1x10-7 > 1x10-7 Table 5: Hydronium and hydroxide ions concentrations in various solutions Another handy way of writing concentrations in terms of pH, which corresponds to the concentration of hydronium ions in a solution, is: pH = - log [H+] Manual 22 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 The pH scale measures the acidity or alkalinity of a solution. Taking the exponent of the HзO+ concentration and removing the negative sign will determine the pH of the concentration. This gives you the pH of a solution. The pH scale ranges from 0 to 14, where 7 is considered neutral ([HзO+] = [OH-]), less than 7 is acidic while more than 7 basic. The further away from 7 you are on the pH scale, the more acidic or basic the solution. Therefore, a low pH means that the substance is more acidic, while a high pH means that the substance is basic. The table below contains examples of our everyday acids and bases and their pH values. [HзO+] pH [OH-] Example 1 X 100 0 1 X 10-14 Battery acid 1 X 10-1 1 1 X 10-13 Stomach acid 1 X 10-2 2 1 X 10-12 Lemon juice 1 X 10-3 3 1 X 10-11 Grapefruit 1 X 10-4 4 1 X 10-10 Tomato juice 1 X 10-5 5 1 X 10-9 Black coffee 1 X 10-6 6 1 X 10-8 Saliva 1 X 10-7 7 1 X 10-7 Pure water 1 X 10-8 8 1 X 10-6 Sea water 1 X 10-9 9 1 X 10-5 Baking Soda 1 X 10-10 10 1 X 10-4 Milk of magnesia 1 X 10-11 11 1 X 10-3 Ammonia solution 1 X 10-12 12 1 X 10-2 Soapy water 1 X 10-13 13 1 X 10-1 Bleaches 1 X 10-14 14 1 X 100 Caustic soda Table 6: pH values of common acids and bases Manual 23 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 From the theory above it can be concluded that the lower the pH value, the more acidic the medium is and therefore the higher the value of the Ka will be. 2.2.5 Acid reactions Acids reacts with many compounds, too many to look at in one experiment. In this section we will discuss only five basic reactions. These reactions are discussed below. 2.2.5.1 Acid and metal reactions Reactions with acids would normally involve a proton transfer (H+), but when an acid and a metal react an electron transfer takes place. The hydrogen from the acid is displaced by the metal, releasing hydrogen gas and producing a salt: Acid + Metal ↔ A salt + Hydrogen Example H2SO4 + Mg ↔ MgSO4 + H2 2HCl + Mg ↔MgCl2 + H2Hydrogen causes bubbling during the reaction. It can be detected by holding a burning match close to the mouth of the gas collection tube, which will cause an explosive sound. 2.2.5.2 Acid and carbonate reactions When an acid reacts with a carbonate it produces water, salt and carbon dioxide. Acid + Carbonate ↔ A salt + Water + Carbon dioxide Example 2HCl + Na2CO3 ==> 2NaCl + H2O + CO2 2HNO3 + CaCO3 ==> Ca(NO3)2 + H2O + CO2 Carbon dioxide also causes bubbling. It can be detected using lime water, which will turn clear lime water milky. Manual 24 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 2.2.5.3 Acid and metal oxides reactions Metal oxides are also bases. When they react with acids they form a salt and water, generally: Acid + Metal oxide ↔ A salt + Water Example H2SO4 + CuO ↔CuSO4+ H2O 2HNO3 + CuO ↔Cu(NO3)2+ H2O 2.2.5.4 Acid and metal hydroxide reactions Metal hydroxide is also a special type of a base, because when it reacts with a base it also produces a salt and water. Acid + Metal hydroxide ↔ A salt + Water Example H2SO4 + 2NaOH ↔Na2SO4 + 2H2O NaOH + 3HNO3 ↔NaNO3 + 2H2O 2.2.5.5 Acid and alkali reactions Not all the bases are water soluble; those that are soluble in water are called alkalis. When an acid reacts with an alkali it produces a salt and water. Acid + Alkali ↔ A salt + Water Example HCl + NaOH ↔ NaCl + H2O H2SO4 + 2KOH ↔K2SO4 + H2O This is also called a neutralisation process because an alkali is a special type of a base. Manual 25 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 2.2.6 Acid-base titration calculations Titration is a method that is commonly used to determine the concentration of an unknown liquid by comparing it with a known liquid. Acid-base titration involves the progressive addition of a base of known concentration from the burette to the unknown volume of the acid in the conical flask. This is depicted in the next figure. The volume of the acid and base is precisely measured. This process continues until the equivalence point is reached. Equivalence point, which is shown by an indicator, is the point at which the moles of H+ are equal to the moles of OH-. Figure 4: Titration setup This is the basis of all titration calculations. The following example illustrates how titration calculations are performed. Example It was found that 50 mL of 0.10M NaOH neutralised 25.0mL of hydrochloric acid. Determine the concentration of the acid used. 1 Write the balanced chemical equation for the reaction. NaOH (aq) + HCl (aq) ↔ NaCl (aq) + H2O(l) 2 Calculate the unknown moles of NaOH. moles = concentration x volume n (NaOH) = Manual CV 26 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 3 = (50x10-3 L) x 0.1mol/L = 0.005 mol From the balanced equation, find the mole ratio. It is clear that the reactants are consumed on a 1:1 mole ratio. 4 Find the moles of the acid. NaOH : HCl = 1:1 At equivalence point n(NaOH) = n(HCl) = 0.005 mol 5 Calculate the concentration of HCl. n= 0.005 mol, V = 25.0 x 10-3L C= n/ VC C(HCl) = 0.005 / 25.0 x 10-3 = 0.2 mol/ L Example In an experiment 100mL of 0.2mol-1 L NaOH neutralised 50mL of sulphuric acid. Determine the concentration of the acid. 1 Write the balanced chemical equation for the reaction. 2NaOH(aq) + H2SO4(aq) ↔ Na2SO4(aq) + 2H2O(l) 2 Calculate the unknown moles of NaOH. moles = concentration x volume n(NaOH) 3 = CV = (100x10-3 L) x (0.2mol/ L) = 0.02 mol From the balanced equation, find the mole ratio. The reactants are consumed on a 2:1 mole ratio. Manual 27 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 4 Find the moles of the acid. NaOH : H2SO4= 2:1 At equivalence point n(H2SO4) = (0.5) n(NaOH) = 0.01 mol. 5 Calculate the concentration of H2SO4. V = 50. x 10-3L n = 0.01 mol, C = n/ V C(H2SO4) = 0.01/ 50 x 10-3 = 0.2 mol/ L Example In an experiment, it was found that 10mL of 0.025 mol-1 Ba (OH)2 neutralised 35 mL of nitric acid. Determine the concentration of the acid. 1 Write the balanced chemical equation for the reaction. Ba(OH)2(aq) + 2HNO3(aq) ↔ Ba(NO3)2(aq) + 2H2O(l) 2 Calculate the unknown moles of Ba (OH)2. moles = concentration x volume n(Ba(OH)2 = 3 CV = (10x10-3 L) x 0.025 mol/ L = 2.5 x 10-4 mol Calculate the moles of the acid. Ba(OH)2:HNO3 = 1:2 At equivalence point n(HNO3) = 2 x n(Ba(OH)2 =5 x 10-4mol 4 Calculate the concentration of HNO3. n =5 x 10-4mol; V = 35 x 10-3L C = n/ V C(HNO3) =5 x 10-4/ 35 x 10-3 Manual 28 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 = 0.0143 mol/ L 2.3 Experiment The following experiment will be conducted in the chemistry laboratory so that you can see how the concentration of an unknown acid solution can be determined. Observe a demonstration of and/ or conduct the following experiment. Purpose To determine the concentration of an unknown acid solution. Apparatus and materials pipette bulb burette clamp 50 ml graduated cylinder 100 ml graduated cylinder 125 ml Erlenmeyer flask 20 ml pipette retort stand spatula wash bottle unknown acid solution 0.1 M NaOH solution. Correct PPE, including eye protection should be worn at all Safety hazards and precautions times throughout the experiment. If you spill the base or the acid solution on your skin, rinse with water immediately. Procedure Rinse the 125 ml flask and 20 ml pipette with distilled water. Rinse the pipette using a small volume of the unknown acid solution. Manual Pipette 20ml of the unknown acid solution into the 125 ml flask. 29 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Add about 20 ml of distilled water and a few drops of phenolphthalein. Record your observation of the colour of the solution. Clean the burette with distilled water and rinse it with a small volume of the NaOH solution. Fasten the burette to the retort stand. Pour the NaOH solution in a burette, making sure that the stop cock is closed. Place the 125 ml flask below it as shown in the sketch. Perform the titration by slowly adding small amounts (1 ml) of the NaOH solution into the unknown acid solution by opening the burette to release the solution. Be careful not to add too much at one time. Record your observation of the colour of the solution as you add more base. When your solution has reached the equivalence point, the solution should have a pale pink colour. Record the amount of base you added to the acid and then perform the calculations in the workbook to determine the acid concentration. Perform the experiment again to make sure you get same results. Sketch Manual 30 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 UNIT 3: OXIDATION-REDUCTION (REDOX) REACTIONS 3.1 Instructions Ref. No SO3 AC1-5 Resources Learning materials CCFO2-4, 6, 7 Learning Methodology Read through Unit 3 of the learning materials. You can either work individually or in pairs. Make notes of things you do not understand and/ or need more information on. Workbook Assessment N/ a N/ a Ex. 5 Ass. 1 Act. 9 SO3 AC1-3 Classroom Attend a lecture on: CCFO1-7 Facilitator the principles of reduction and oxidation reactions the principles of standard electrode potentials. Experiment -Clean plate of zinc, beakers, zinc sulphate solution, copper plate, copper sulphate salt bridge, cotton wool, voltmeter. Computer and projector with animations of the experiments and/ or PowerPoint slides of the experiments Manual During the lecture, you will observe and/ or participate in experiments/ animations of the experiments to help you understand the concept of reduction-oxidation reactions. 31 Act. 10 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Ref. No Resources Learning Methodology SO3 AC4, 5 Classroom Attend a lecture on how to: CCFO1-7 Facilitator balance Redox reactions by using the oxidation states of the elements. make predictions based on information obtained from a redox table. SO3 AC1-5 CCFO1-7 Learning materials and workbook PoE Facilitator/ SME Manual Revise the work that you have done up to this point. Make sure that you have completed the CCFO checklist and obtained the required evidence for your PoE. If there is anything that you do not understand, ask your facilitator. 32 Act. 11 Workbook Assessment Ex. 6 Ass. 1 CCFOs CCFOs Act. 12 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 3.2 Introduction Reduction-oxidation (redox) reactions are a group of reactions that are concerned with the transfer of electrons between compounds. These reactions always go together. A reduction reaction will not happen without an oxidation reaction happening simultaneously in chemical equivalent quantities. When oxidation occurs, electrons are lost and electrons are gained when reduction occurs. Hint: (LEO-loss of electrons is oxidation, GER- gain of electrons is reduction). Each reaction on its own is called a half reaction. A substance that is oxidised loses electrons and is called a reducing agent while the one which is reduced gains electrons and is called an oxidising agent. For example, the reaction between magnesium and oxygen atoms may be broken down into two half reactions: This process represents an oxidation half reaction with magnesium as a reducing agent. This reaction represents a reduction half reaction with oxygen as an oxidising agent. Figure 4: Corrosion of metals is a redox reaction that occurs naturally. 3.3 Oxidation number The change in the oxidation number of a material provides a proper definition of oxidation and reduction. The oxidation number is the charge an atom has if bonding electrons are transferred completely to the atom with greater affinity for them in a particular situation. Manual 33 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 The ratio by which the atoms will combine when forming compounds can be predicted by these oxidation numbers. It means that the oxidation number is the charge an atom has if it is in a compound composed of ions. In the case of oxidation, the oxidation number will increase. The oxidation number will decrease when reduction occurs. 3.3.1 Rules for assigning oxidation numbers The oxidation number of elements in their natural form is zero. For example, the oxidation number of hydrogen in H2 or oxygen in O2 is 0. An element in a monatomic ion has an oxidation number equal to the charge of the ion. For example, aluminium and oxygen in Al2O3 have oxidation numbers of +3 and -2 respectively. The ions in this case are Al3+ and O2-. Some elements have the same oxidation number in almost all their compounds. The 1A metals always have an oxidation number of +1 while 2A elements always have an oxidation number of +2 in their compounds. The most electronegative element, fluorine always has an oxidation number of -1 in all its compounds. Oxygen is assigned -2 as its oxidation number. Hydrogen, in its compound with metals is assigned -1 while with nonmetals it is assigned +1. The sum of the oxidation numbers of atoms in a neutral species equals zero in an ion that equals the charge on that species. We can use this principle to find oxidation number of elements in polyatomic substance. Application of this principle is illustrated by the following examples. Example 1 What is the oxidation number of manganese in MnO4-? For MnO4-, oxygen’s oxidation number is taken as -2 and it should be realized that the sum is -1 Therefore, 4 x (-2) +ox no (Mn) = -1 ox no (Mn) = +7 2 What is the oxidation number of sulphur in sulphuric acid? In H2SO4, oxygen’s oxidation number is taken as -2, hydrogen as +1 and the sum is 0. Therefore, Manual 4 (-2) +2(+1) + ox no (S) = 0 34 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 oxidation no (S) = +6 3.4 Balancing redox reactions There are many methods used to balance redox reactions, however in this section we will describe the use of the oxidation number method to balance these reactions. 3.4.1 Oxidation number method The oxidation number method is illustrated by the examples below. Example Mn2+ is oxidised by NaBiO3 in acidic conditions according to this equation: Mn2+ + BiO3- + H+ → MnO4- + Bi3+ + H2O 1 2 Determine the oxidation number of all elements on each side of the equation. Ox no of reactants Ox no of products Mn +2 +7 oxidised Bi +5 +3 reduced O -2 -2 - H +1 +1 - From this, one can deduce that manganese was oxidised by freeing five electrons and bismuth was reduced by accepting two electrons. To make an increase in oxidation number equal a decrease in oxidation number, there must be five Bi atoms reduced for two Mn atoms oxidised: 2Mn2+ + 5BiO3- + H+ → MnO4- + Bi3+ + H2O 3 From here the rest can be balanced by inspection and it is easy to do, giving: 2Mn2+ + 5BiO3- + 14H+ →2MnO4- + 5Bi3+ + 7H2O Manual 35 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Example In an acidic medium, potassium permanganate oxidises chlorine ions to chlorine gas by this chemical equation: MnO4- + Cl- + H+→Mn2+ + Cl2 + H2O 1 What is the oxidation number of manganese in MnO4-? For MnO4-, oxygen’s oxidation number is taken as -2 and it should be realized that the sum is -1. Therefore, 4 x (-2) +ox no (Mn) = -1 ox no (Mn) = +7 Example In an acidic medium, potassium permanganate oxidises chlorine ions to chlorine gas by this chemical equation: MnO4- + Cl- + H+ → Mn2+ + Cl2 + H2O 1 2 Ox no of reactants Ox no of products Mn +7 +2 reduced Cl -1 0 oxidised O -2 -2 - H +1 +1 - In this case, the oxidation number of manganese decreases by five units while chlorine's oxidation number increases by one unit. To make an increase in oxidation number equal a decrease in oxidation number, there must be five Cl atoms oxidised per Mn reduced. Thus: MnO4- + 5Cl- + H+ → Mn2+ + Cl2 + H2O 3 Balance the rest by inspection, yielding: MnO4- + 5Cl- + H+ → Mn2+ + 5/2 Cl2 + H2O 4 Manual Or multiply through by a factor of two to eliminate fractional coefficients: 36 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 2MnO4- + 10Cl- + 16H+ → 2Mn2+ + 5Cl2 + 8H2O Example Consider this equation and balance it using the oxidation number method. HNO3(aq)+H3AsO3(aq)→NO(g)+H3AsO4(aq)+H2O(l) 1 From the equation it looks like it will be difficult to do a balance on oxygen and hydrogen. Let’s rather do a balance on nitrogen and arsenic atoms. 3 Ox no of reactants Ox no of products N +5 +2 oxidised As +3 +5 reduced From this, one can deduce that nitrogen was oxidised by losing 3 electrons and arsenic was reduced by gaining two electrons in the process. To make an increase in oxidation number equal a decrease in oxidation number, there must be three As atoms reduced for every two N atoms oxidised: 2HNO3(aq)+3H3AsO3(aq)→ NO(g)+H3AsO4(aq)+H2O(l) 4 From here one can balance this equation by inspection, yielding: 2HNO3(aq)+3H3AsO3(aq)→NO(g)+3H3AsO4(aq)+H2O(l) 5 Balancing redox reactions using half equations. An easy way of balancing redox reactions occurring in an aqueous solution involves breaking up the equation into two half equations (reduction half reaction and oxidation half reaction). The two reactions are balanced separately and then combined to get the overall equation. There must be no net change in the number of electrons in the overall equation. In each half reaction: Balance all other atoms except O and H. O is balanced by adding H2O. H is balanced by adding H+. charges are balanced using electrons. Manual 37 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Example Consider a reaction that occurs between chlorine and potassium permanganate: MnO4-(aq) + Cl- (aq) →Mn2+ (aq)+ Cl2 (g) 1 Separate the equation into its half equations and work on them separately. Half equation for reduction: MnO4-(aq) → Mn2+ (aq). 2 From the equation we have the same number of manganese atoms on both sides, but we have four oxygen atoms in excess. To balance the oxygen, we add an appropriate number of H2O molecules; in this case we add four H2O molecules to the right of the equation: MnO4-(aq) → Mn2+ (aq) + 4H2O. 3 We add eight H+ ions to the left of the equation to balance hydrogen atoms that are in excess on the right side of the half equation: MnO4-(aq) + 8H+ (aq) → Mn2+ (aq)+ 4H2O. 4 Finally the charge must be balanced; we have +2 on the right and +7 on the left. This +7 on the left comes from (+1x8) -1. Thus we add five electrons on the left to balance the charge: MnO4-(aq) + 8H+ (aq) + 5e- → + Mn2+ (aq) + 4H2O 5 equation 3.1 Next, we balance the oxidation reaction: Cl- (aq) → Cl2 (g). 6 Lastly, we balance the charge; we have -2 on the left. Thus we add -2 on the right: 2Cl- (aq) → Cl2 (g) + 2e- 7 equation 3.2 We have two chlorine atoms on the right, thus we balance the atoms on the left by multiplying the left side of the equation by two: 2Cl- (aq) → Cl2 (g). 8 Now, combine the two balanced half equations in such a way that the electron gain equals the electron loss. So, we multiply equation 3.1 by 2 and equation 3.2 by 5. 10Cl- (aq) →2Cl2 (g) + 10e2MnO4- (aq) + 16H+ (aq) + 10e-→ 2Mn2+ (aq)+ 8H2O 10Cl- (aq) + 2MnO4- (aq) + 16H+ (aq)→2Cl2 (g) +2Mn2+ (aq) + 8H2O Manual 38 equation 3.3 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 9 Equation 3.3 is our final equation. From the above example, one would wonder why H2O and H+ molecules are used to balance oxygen and nitrogen. It is done because they are the only substances that contain O and H in large quantities in aqueous solutions. 3.5 Standard electrode potential Figure 5: example of regular batteries A normal battery can be used to explain half reactions. The battery is also called an electrochemical cell. In this type of cell, the production of electricity brings about a change in chemical reaction. There are two types of electrochemical cells, namely a Galvanic cell, where the reaction occurs spontaneously and an electrolytic cell where a nonspontaneous reaction occurs. These cells contain electrodes where oxidation and reduction occur. Oxidation occurs at the electrode called the anode and reduction occurs at the cathode. The cathode is a positive electrode while an anode is a negative electrode. A Galvanic cell is also called the voltaic cell and is commonly used because it is spontaneous. In this section we will discuss this type of cell in its simplest form. 3.5.1 Galvanic cell The following experiments will illustrate how a simple galvanic cell works. Experiment Conduct the following experiment. Purpose To illustrate how a simple galvanic cell works Apparatus and materials clean plate of zinc 2 Beakers zinc sulphate solution copper plate Manual 39 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Procedure copper sulphate solution cotton wool salt bridge. Place a clean plate of zinc in a beaker containing zinc sulphate solution and a copper plate in another beaker containing copper sulphate. Connect the two solutions with a salt bridge filled with a saturated solution of KCl or KNO3 and plug it with cotton wool to prevent the solution from running out. The functions of the salt bridge are: to complete the circuit to allow the movement of ions to ensure electrical neutrality and to prevent direct contact between the two electrode solutions (electrolytes). The copper plate is connected to the positive terminal of the voltmeter while the zinc is connected to the negative terminal. The electrical potential difference between the redox pairs is measured by the voltmeter. Observation The zinc electrode loses electrons and its mass decreases. Electrons flow from the zinc electrode through the voltmeter. A layer of copper deposits form on the copper electrode. Conclusions The zinc electrode is an anode because it loses electrons. Thus oxidation occurs at this electrode by this half reaction: Zn → Zn2+ + 2eThe electron is accepted by the Cu2+ ions in the solution: Cu2+ + 2e-→Cu Therefore, the copper electrode is a cathode because reduction occurs in this electrode. This is depicted in Figure 6. Figure 6: Simple galvanic cell Manual 40 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 3.6 Standard potentials The most important property of a voltaic cell to chemists is the voltage or the electrical potential. This potential measures the driving force behind the reaction taking place in the cell. The electrical potential is measured in volt (V). The potential of a single half cell cannot be measured directly by a voltage measuring device; however, when two half cells are connected to form a cell, the potential difference can be measured. In other words, the potential difference between the anode and the cathode is measured. A reduction half cell has a higher reduction potential than the half cell where oxidation occurs, because it has a higher tendency to acquire electrons. The reduction potential of a half cell measures the tendency of a half cell to occur as reduction. Cell potential is the difference between the reduction potential of the two half cells at hand: E ocell = E ocathode- E oanode E ocell, the standard cell potential, is the measure of the cell potential under standard conditions, which are: for gases, a pressure of 1 atmosphere a temperature of 25 oC an ion concentration in the half cell is 1M. As mentioned previously, the half cell potentials cannot be measured, so scientists have chosen a reference electrode. This electrode is the standard hydrogen electrode and was assigned a value of 0.00V. All other potential measurements can be made against this electrode potential. The table below gives the standard reduction potentials of different chemicals in water solution at 25 oC. Manual Half-Reaction E 0(V) Li+(aq) + e- → Li(s) -3.05 K+(aq) + e- → K(s) -2.93 Ba2+(aq) + 2e- → Ba(s) -2.90 Sr2+(aq) + 2e- → Sr(s) -2.89 Ca2+(aq) + 2e- → Ca(s) -2.87 Na+(aq) + e- → Na(s) -2.71 41 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Mg2+(aq) + 2e- →Mg(s) -2.37 Be2+(aq) + 2e- → Be(s) -1.85 Al3+(aq) + 3e- → Al(s) -1.66 Mn2+(aq) + 2e- → Mn(s) -1.18 2H2O + 2e- → H2(g) + 2OH-(aq) -0.83 Zn2+(aq) + 2e- →Zn(s) -0.76 Cr3+(aq) + 3e- → Cr(s) -0.74 Fe2+(aq) + 2e- → Fe(s) -0.44 Cd2+(aq) + 2e- → Cd(s) -0.40 PbSO4(s) + 2e- → Pb(s) + SO42-(aq) -0.31 Co2+(aq) + 2e- → Co(s) -0.28 Ni2+(aq) + 2e- → Ni(s) -0.25 Sn2+(aq) + 2e- → Sn(s) -0.14 Pb2+(aq) + 2e- →Pb(s) -0.13 2H+(aq) + 2e- →H2(g) 0.00 Sn4+(aq) + 2e- →Sn2+(aq) +0.13 Cu2+(aq) + e- →Cu+(aq) +0.13 SO42-(aq) + 4H+(aq) + 2e- → SO2(g) + 2H2O +0.20 AgCl(s) + e- →Ag(s) + Cl-(aq) +0.22 Cu2+(aq) + 2e- →Cu(s) +0.34 O2(g) + 2H2 + 4e- →4OH-(aq) +0.40 I2(s) + 2e- →2I-(aq) +0.53 MnO4-(aq) + 2H2O + 3e- →MnO2(s) + 4OH-(aq) +0.59 O2(g) + 2H+(aq) + 2e- →H2O2(aq) +0.68 Fe3+(aq) + e- →Fe2+(aq) +0.77 Manual 42 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Ag+(aq) + e- →Ag(s) +0.80 Hg22+(aq) + 2e- →2Hg(l) +0.85 2Hg2+(aq) + 2e- →Hg22+(aq) +0.92 NO3-(aq) + 4H+(aq) + 3e- →NO(g) + 2H2O +0.96 Br2(l) + 2e- →2Br-(aq) +1.07 O2(g) + 4H+(aq) + 4e- →2H2O +1.23 MnO2(s) + 4H+(aq) + 2e- →Mn2+(aq) + 2H2O +1.23 Cr2O72-(aq) + 14H+(aq) + 6e- →2Cr3+(aq) + 7H2O +1.33 Cl2(g) + 2e- →2Cl-(aq) +1.36 Au3+(aq) + 3e- →Au(s) +1.50 MnO4-(aq) + 8H+(aq) + 5e- →Mn2+(aq) + 4H2O +1.51 Ce4+(aq) + e- →Ce3+(aq) +1.61 PbO2(s) + 4H+(aq) + SO42-(aq) + 2e- →PbSO4(s) + 2H2O +1.70 H2O2(aq) + 2H+(aq) + 2e- →2H2O +1.77 Co3+(aq) + e- →Co2+(aq) +1.82 O3(g) + 2H+(aq) + 2e- → O2(g) + H2O +2.07 F2(g) + e- →F-(aq) +2.87 Table 7: Reduction potentials at 25ºC Whether the reaction will be oxidised or reduced is measured by the potentials listed in Table 7. The standard reduction potentials make it possible for you to predict whether or not the reaction will take place. The half reaction with a more positive potential always undergoes reduction. Thus, the other half reaction undergoes oxidation. The more negative the potential, the more difficult it is to bring about the half reaction. One can determine if a redox reaction will take place under certain conditions by calculating the voltage of the redox reaction. If it is positive, it implies that the reaction is spontaneous. If the calculated voltage is negative it means that the reaction cannot occur by itself. Manual 43 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Example 1 Is the following reaction spontaneous? Hg + Mg2+ → Hg2+ + Mg reduction: Mg2+ + 2e-→ Mg E0 = -2.37 V oxidation: Hg →Hg2+ + 2e- E0 = 0.79 V E ocell = E ocathode - E oanode = -2.37 V- (+0.79) = -3.16 V 2 Since the standard electrode potential is negative, this redox reaction is not spontaneous. Thus energy will have to be applied for this reaction to take place. In Table 7 the strength of reducing agents increases from the top to the bottom in the table, while the strength of oxidising agents increases from the bottom to the top in the table. 3.7 Experiment The following experiment will be conducted in the chemistry laboratory so that you can investigate a simple redox reaction. Experiment Observe a demonstration of and/ or conduct the following experiment. Purpose To investigate a simple redox reaction Apparatus and materials aqueous copper sulphate zinc metal 2 test tubes 2 tweezers 25 ml graduated cylinder small beaker Manual 44 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Safety hazards and precautions Procedure Correct PPE, including eye protection should be worn at all times throughout the experiment. Obtain a small piece of zinc metal and record its appearance specifically. Carefully place the zinc into the test tube so that it rests at the bottom. Use the cylinder to measure 15 ml of CuSO4 solution and put it into the small beaker. Also record your observation of the colour of the CuSO4 solution. Add 5 ml of the CuSO4 solution into the test tube and allow it to rest. Record your observations during this period and again at the end of the reaction. Manual 45 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 UNIT 4: CHEMICAL REACTION RATES 4.1 Instructions Ref. No SO4 AC1-4 Resources Learning materials CCFO2-4, 6, 7 Learning Methodology Read through Unit 4 of the learning materials. You can either work individually or in pairs. Make notes of things you do not understand and/ or need more information on. Workbook Assessment N/ a N/ a Act. 13 SO4 AC1-4 Classroom/ Laboratory Attend a lecture on the: CCFO1-7 Facilitator principles of chemical reaction rates Material and equipment for experiments: factors affecting chemical reaction rates principles of endothermic and exothermic reactions Computer and projector with animations of the experiments and/ or PowerPoint slides of the experiments industry applications demonstrating the role of chemical reaction rates in industry. During the lecture, you will observe and/ or participate in experiments/ animations of the experiments to help you understand the concept of endothermic and exothermic reactions. Ex. 7 Act. 14 Ass. 1 Ex. 8 Manual 46 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Ref. No Resources SO4 AC4 On-site/ Classroom CCFO1-7 Facilitator/ SME Computer with internet access Learning Methodology CCFO1-7 Learning materials and workbook PoE Facilitator/ SME Manual Assessment Ex. 9 Ass. 1 CCFOs CCFOs In pairs, do research on the application and effect of chemical reaction rates in industry and collect examples of the application of chemical reaction rates in industry. Present your findings to the class. Act. 15 Multimedia for presentation SO4 AC1-4 Workbook Revise the work that you have done up to this point. Make sure that you have completed the CCFO checklist and obtained the required evidence for your PoE. If there is anything that you do not understand, ask your facilitator. 47 Act. 16 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 4.1.1 Chemical reaction rates A chemical reaction is a process in which chemical substances interact and conversion takes place at a certain rate. The chemical substances in a reaction are referred to as reactants. There are two types of chemical reactions namely endothermic and exothermic. An endothermic reaction cannot occur spontaneously and thus require energy to proceed. An exothermic reaction can occur spontaneously and releases energy in the form of heat, light, or sound. The rate of chemical reactions is defined as the change in concentration of a substance per time unit. The substance can be a reactant or the product. The reaction rate can be best explained by the collisions theory, which states that a reaction can occur when two atoms, ions or molecules collide with enough energy in a given time to form one or more products. The more collisions in the system, the more combinations of molecules will occur and the faster the reaction will happen. The reaction rate of this reaction is higher. Consider the following reaction: A + B →C If one measures the concentration of A at a certain time interval, one typically sees a trend as shown in the graph below. Figure 7: Rate of reaction curve This graph shows the concentration decreasing with time. Since this is not a straight line, the slope (rate) of this graph changes with time, indicating that the reaction rate is not constant. The rate is therefore related to the time the concentration measurements were made. Manual 48 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 4.1.2 Factors affecting the rate of reaction There are several factors that can influence the chemical reaction rate. Generally, factors that increase the number of collisions will increase the reaction rate, while those decreasing the collisions will decrease the reaction rate. These factors are discussed below: 4.1.2.1 Temperature The increase in temperature will increase the kinetic energy in the reaction, which makes the motion of particles more chaotic. Due to this, the number of collisions per unit time will increase at higher temperatures. 4.1.2.2 Pressure It is a known fact that atoms of a gas are very far apart. For two compounds to react there must be contact between their molecules. By increasing the pressure, you force the molecules to make contact with one another, which increase the frequency of collisions between the particles. An increase in pressure also causes the temperature to rise. Thus an increase in pressure of a gas results in an increase in chemical reaction rate. 4.1.2.3 Concentration In a given volume, the number of reacting particles affects the reaction rate. An increase in concentration is in essence an increase in reacting particles. The higher the concentration, the more collisions occur and therefore the faster the reaction rate will be. 4.1.2.4 Catalyst A catalyst is a substance that speeds up the reaction rate without being consumed in the reaction. Catalysts are widely used in the chemical industry to increase the reaction rate for economical reasons. The presence of a catalyst decreases the energy required for the reaction and therefore increasing the effective collisions. 4.1.2.5 Reaction surface area If one of the reacting substances is a solid, its surface will affect the rate of chemical reaction. A larger surface area will result in an increase of collisions between particles. It is important to note that powders have a larger surface area than solids. Manual 49 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 4.1.2.6 Reaction Energetics For a chemical reaction to occur chemical bonds must be formed and others must be broken. Energy is required to break these bonds. Some reactions require energy to occur while others release energy in the process. Energy can be released in terms of heat, light or sound. Reactions that release energy are called exothermic reactions. The energy released by these reactions is taken up in the environment. This may heat up the surroundings if energy is released in form of heat. Exothermic reactions are mostly spontaneous and explosive. As a general rule, an increase in temperature results in an increase in chemical reaction rate. Once an exothermic reaction starts, the reaction rate quickly increases due to the heat that is produced. Care must be taken when dealing with such reactions. The combustion of methane is an example of an exothermic reaction: CH4 (g) + 2O2 (g) →CO2 (g) + 2H2O (g) Generally exothermic reactions are written as: Reactants →Products + energy In an endothermic reaction, energy has to be supplied to the system continually in order for the reaction to occur. Photosynthesis is an example of an endothermic reaction. Plants use energy from sunlight to convert water and carbon dioxide to produce oxygen and glucose. This reaction is provided below: Sunlight + 6CO2 (g) + H2O (l) = C6H12O6(aq) + 6O2(g) In a general form, endothermic reactions are written as follows: Reactants + energy →Products 4.1.2.7 Industrial applications of chemical reaction rates The most important thing in industry is time, since time is money. The faster the reaction rate of chemical reactions is, the more economic it is. It is important to know how long the reaction will take to produce the desired products. This is why catalysts are used in the chemical industry to increase reaction rates. Reaction rates are studied for health and safety reasons. Fine powders have a large surface area, which increases the reaction rate. This could cause an explosion if it is not controlled well. For example, methane gas and fine dust particles in mines are all potentially dangerous. Manual Thus knowledge of the ignition temperature and explosion 50 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 threshold concentration of reactions helps us to design systems with minimal operational risks. Reaction rates are also used to design reactors. 4.2 Experiment The following experiment will be conducted in the chemistry laboratory so that you can investigate the factors that affect chemical reaction rates. Experiment Observe a demonstration of and/ or conduct the following experiment. Purpose To investigate the factors that affect chemical reaction rates Apparatus and materials 8 x 0.1 g Magnesium strips 50 ml 1 M HCl solution 50 ml 2 M HCl solution 50 ml glass syringe 8 x test tube with rubber stoppers with holes at the top elastic tube stopwatch. Correct PPE, including eye protection should be worn at all Safety hazards and precautions times throughout the experiment. In case of skin contact with HCl, flush the affected area with lukewarm water for 15 minutes and seek medical attention. Procedure Construct the system as shown in the illustration. Pour 10ml of the 1 M HCl solution into the test tube. Add one 0.1 Mg strip to the edge of the test tube. Plug the system and shake it to make the Mg strip contact the HCl solution, start measuring the time with the stopwatch immediately. Repeat the experiment with the magnesium folded many time to reduce the surface area. Manual 51 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Repeat the experiment with the Mg strip in the 2 M HCl solution. Repeat the experiment with the Mg strip folded many times in the 2M HCl solution. Record the exact time each reaction took to complete. Sketch (please redraw) Manual 52 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 UNIT 5: CHEMICAL EQUILIBRIUM 5.1 Instructions Ref. No SO5 AC1-5 Resources Learning materials CCFO2-4, 6, 7 Learning Methodology Read through Unit 5 of the learning materials. You can either work individually or in pairs. Make notes of things you do not understand and/ or need more information on. Workbook Assessment N/ a N/ a Act. 17 SO5 AC1-4 Classroom/ Laboratory Attend a lecture on: CCFO1-7 Facilitator The principles of phase equilibrium. Material and equipment for experiments: The principles of chemical equilibrium. The factors affecting chemical equilibrium. Equilibrium calculations involving concentrations and the equilibrium constant. Computer and projector with animations of the experiments and/ or PowerPoint slides of the experiments During the lecture, you will observe and/ or participate in experiments/ animations of the experiments to help you understand the concept of chemical equilibrium. Ex. 10 Act. 18 Ex. 11 Manual 53 US 244241 Rev.1 – Sep 08 Sparrow Consulting © September 08 Ref. No Resources SO5 AC5 On-site/ Classroom CCFO1-7 Facilitator/ SME Computer with internet access Learning Methodology CCFO1-7 Learning materials and workbook PoE Facilitator/ SME Manual Assessment Ex. 12 Ass. 1 CCFOs CCFOs In pairs, do research on industry applications demonstrating the role of chemical equilibrium using practical industry examples. Present your findings to the class. Act. 19 Multimedia for presentation SO5 AC1-5 Workbook Revise the work that you have done up to this point. Make sure that you have completed the CCFO checklist and obtained the required evidence for your PoE. If there is anything that you do not understand, ask your facilitator. 54 Act. 20 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 5.1.1 Equilibrium Equilibrium is a fundamental concept in chemistry, so in this section we will explore the application of equilibrium to chemical reactions. In many cases, reactions do not reach completion, in other words there are still reactants left at the end of the reaction. In some cases, the reactants are favoured over products and vice versa. These are called reversible reactions. In such reactions, the conversion of reactants to products and of products to reactants occurs at the same time. Reversible reactions are indicated by double arrows. As the reaction starts there are no products, therefore the reaction rate of the reverse reaction is zero. The rate of the forward reaction decreases as the concentration of reactants decrease, while the products build up. Since the concentration of the products increases, the rate of the reverse reaction increases. Eventually the rate of reactants converting to products will equal the rate of which products are converting to reactants. This phenomenon is known as the chemical equilibrium. Chemical equilibrium is reached when the rate of the forward reaction equals the reverse reaction. At chemical equilibrium, the concentration of both products and reactants stop changing with time but are not equal. For example, making of sulphur trioxide is a reversible reaction: 2SO2 (g) + O2 (g) ↔ SO3 (g) Equilibrium implies that for every one molecule of oxygen, two molecules sulphur dioxide that react two sulphur trioxide molecules will dissociate. Both the forward and reverse reactions are illustrated in Figure 8. The graph shows that initially there was no SO3; hence the reverse reaction could not proceed. As the concentration of SO 3 increases, the reverse reaction rate increases while the forward rate decreases. This continues until equilibrium is reached, when both reaction rates are equal. Manual 55 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Figure 8: Equilibrium of a trioxide producing reaction 5.1.2 Equilibrium Constant Chemists use equilibrium constants instead of percentages to indicate the position of equilibrium in a chemical reaction. Equilibrium constants relate to the concentration of reactants and products at equilibrium at a given temperature. Consider the hypothetical reaction, where A reacts with B to produce C and D at equilibrium. aA + bB ↔ cC + dD The equilibrium constant: Keq = [C] c x [D] d / [A] a x [B] b The important feature of the equilibrium expression is that: K is dimensionless. Concentrations of reactants always appear on the numerator. Concentrations of products always appear on the denominator. Concentrations are always raised to power of their stoichiometric coefficients. Square brackets indicate the amount of a material in moles per litre. It is important to note the following: Equilibrium is temperature dependent. When the temperature changes, the constant changes as well. If the reagent is solid, its concentration is taken as 1 and does not appear on the equilibrium constant expression. Manual 56 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 For equilibrium reaction: C(s) + CO2 (g) ↔ 2CO (g) Keq = [CO]2 / [CO2] Concentrations of solvents of dilute solutions are also taken as 1 and are omitted from the Keq expression. For the reaction; H2SO4 (aq) + H2O (l) ↔ HSO4-(aq) + H3O+ (aq) Keq = [HSO4-] x [H3O+] / [H2SO4] Example 1 What is the equilibrium constant expression for the following reaction? 4R + ½ S ↔ 3T + U Keq = [R] 4 x [S] ½ / [T] 3x [U] Example 1 A litre of a gas mixture at 450 oC at equilibrium contains 0.5 mol of hydrogen gas, 0.5 mol of iodine and 3.55 mol of hydrogen iodide. Determine the equilibrium constant for the reaction t is the equilibrium constant expression for the following reaction: H2(g) + I2(g) ↔ 2HI(g) Keq = [HI]2 / [H2] x [I2] = (3.55 mol/l)2/ (0.5 mol/ l)x(0.5 mol/ l) = 50.41 Example 1 One mol of hydrogen gas reacts with 1 mole iodine in a 1 litre flask at 450 oC. The equilibrium, 1.8 mol of hydrogen iodide is present at equilibrium. Calculate the Manual 57 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 equilibrium constant. From the reaction: H2(g) + I2(g) ↔ 2HI(g) 1.4 mol of HI implies that 0.7 mol of H2 and 0.7 mol of I2 were consumed in the process. Thus at equilibrium: mol I2 = mol H2 = 1 mol – 0.7mol = 0.3 mol left at equilibrium Therefore Keq = [HI]2/ [H2] x [I2] = (1.4)2/ (0.3)x(0.3) = 21.78 Example 1 At a certain temperature, bromine chloride decomposes to form chlorine and bromine and the equilibrium constant is 50. The equilibrium mixture contains 5mol Br2 and 5mol Cl2. How many moles of bromine chloride are present at equilibrium in a litre of a reactor? For the reaction: 2BrCl (g) ↔ Br2(g) + Cl2(g) Keq = [Br2] x [Cl2] / [Br]2 Let y represent concentration of bromide chloride at equilibrium and solve for y: 50 = (5 mol/ l) x (5 mol/l)/ (y)2 y = 0.707 mol/ l Since the volume of the flask is 1 litre, moles of bromine chloride is 0.707mol. Equilibrium constants give valuable information regarding the reactions. Amongst other things, they give an indication of whether the reaction favours the reactants or products at equilibrium. The equilibrium constant expression is always written as the ratio of products to reactants. Therefore, when the value of Keq is greater than one it simply means that there are more products than reactants at equilibrium. However, if Keq is less than one, it implies that there are more reactants than products at equilibrium. Therefore: Manual 58 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Keq >1 Products are favoured at equilibrium. Keq <1 The expression gives a fraction, reactants are favoured at equilibrium. 5.1.3 Factors affecting chemical equilibrium When a chemical system is in equilibrium, there is a delicate balance between reactants and products. When stress is applied to such a system it can disturb the equilibrium. The three most common ways in which equilibrium is disturbed are: changes in temperature changes in pressure changes in concentration. A change in any of the factors affecting the equilibrium conditions of a system will cause the system to change in such a way that it cancels or reduces the effect of the change. This is known as the Le Chatelier's Principle. 5.1.3.1 Effects of temperature changes All chemical reactions are accompanied by an energy release or uptake. An increase in temperature causes the equilibrium position to shift in to the half reaction that will absorb heat. For example, an exothermic reaction producing SO3: 2SO2 (g) + O2 (g) ↔ SO3 (g) + heat In this reaction, heat will be the by-product of the reaction. Suppose this reaction is at equilibrium at temperature T1 and the temperature is increased to T2. According to Le Chatelier, the net reaction will advance in the direction that will counteract this effect, thus the equilibrium position will shift to the left and the reactants will be favoured. For an endothermic reaction producing nitric oxide: Heat + N2+ O2 ↔ 2NO In this reaction, an increase in temperature will result in the equilibrium position shifting to the right, producing more nitric oxide. 5.1.3.2 Effects of pressure changes The effects of pressure are considered for gases only. For gases, an increase in pressure will result in a decrease in volume, because volume and pressure are inversely proportional. Manual 59 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 A change in pressure only affects the equilibrium with an unequal number of moles of products and reactants. The increasing pressure shifts the equilibrium position so that the least number of moles form. For example, in the reaction: N2 (g) + 3H2 (g) ↔ 2NH(g) If pressure increases, the equilibrium will shift to the right, because there are 4 moles on the left and only 2 on the right. 5.1.3.3 Effects of concentration changes Changing the concentration of either the reactants or the products at equilibrium will result in the reaction shifting to a new equilibrium position in order to counter the effect. Look at the following example: H2 + I2 ↔ 2HI If you increase the concentration of hydrogen gas, the equilibrium will shift to the right. If you remove some hydrogen gas, the equilibrium will shift to the left producing more reactants. Example 1 For the reaction: PCl5(g) + heat ↔ PCl3(g) + Cl2 (g) How will the following affect the equilibrium? Addition of PCl5 Increase in pressure Removal of heat Removal of Cl2 Addition of PCl5 will cause the equilibrium position to shift to the right, giving more products. An increase in pressure will shift the equilibrium position to the side with fewer molecules. Thus it will shift to the left, favouring the reactants. Removing heat will result in the equilibrium position shifting to the left, decreasing the products at equilibrium. Removal of Cl2 causes a shift to the right to produce more Cl2. Manual 60 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 5.1.3.4 Industry applications of chemical equilibrium To maximise profits in the process industry, the amount of products manufactured need to be as high as possible and the ‘leftover’ reactants as low as possible. Le Chatelier’s principle and the other principles of reaction kinetics can be used to design reactions that are economical and produce the highest amounts of product and the lowest amount of byproducts. 5.2 Experiment The following experiment will be conducted in the chemistry laboratory so that you can investigate shifting of equilibrium using Le Chatelier’s principle. Experiment Observe a demonstration of and/ or conduct the following experiment. Purpose To investigate shifting of equilibrium using Le Chatelier’s principle. Apparatus and materials test tubes nitrogen dioxide is prepared by your facilitator in a fume hood and placed in a test tube with a stopper on top Safety hazards and precautions ice and water bath hot water bath. The first part of the experiment should be conducted in a fume cupboard. Correct PPE, including eye protection should be worn at all times throughout the experiment. Procedure Record your observation of the colour of the gas in the test tube. Place the test tube filled with nitrogen dioxide in the ice and water bath for five to six minutes. Record your observations after this time. Place the test tube in hot water bath for the same amount of time and record your observations. Manual 61 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 ANNEXURE 1: RESOURCES Primary Resources http://www.wpbschoolhouse.btinternet.co.uk/ page07/ equilibria3.htm www.sasked.gov.sk.ca/ docs/ chemistry/ unit8chem30.html ANNEXURE 2: ICONS Checklist Demonstration Example Group work/ discussion Important Note/ Something to think about Lecture Multimedia Experiment Practical exercise/ assessment Safety & PPE Manual 62 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Self-study Site visit Written exercise This manual was developed by Sparrow Research and Industrial Consultants CC e-mail: [email protected] Tel: 012 – 460 9755 ANNEXURE 3: US 244075 SOUTH AFRICAN QUALIFICATIONS AUTHORITY REGISTERED UNIT STANDARD: Apply knowledge of chemical reactions in a processing environment. SAQA US ID UNIT STANDARD TITLE 244241 Apply knowledge of chemical reactions in a processing environment. SGB NAME REGISTERING PROVIDER Chemical Industries SGB FIELD SUBFIELD Field 06 - Manufacturing, Engineering and Technology Engineering and Related Design ABET BAND UNIT STANDARD TYPE NQF LEVEL CREDITS Undefined Regular 3 6 REGISTRATION STATUS REGISTRATION START DATE REGISTRATION END DATE SAQA DECISION NUMBER Registered 2007-06-27 2010-06-27 SAQA 0371/07 LAST DATE FOR ENROLMENT Manual LAST DATE FOR ACHIEVEMENT 63 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 2011-06-27 2014-06-27 This unit standard does not replace any other unit standard and is not replaced by any other unit standard. PURPOSE OF THE UNIT STANDARD Learners who demonstrate competence as described in the outcomes of this unit standard will be able to understand elementary chemistry and its application in the process industry. The qualifying learner is able to: Perform elementary chemical calculations. Demonstrate an understanding of acids and bases. Demonstrate an understanding of oxidation-reduction (redox) reactions and their industrial applications. Demonstrate understanding of chemical reaction rates. Demonstrate understanding of chemical equilibrium. LEARNING ASSUMED TO BE IN PLACE AND RECOGNITION OF PRIOR LEARNING National Certificate Chemical Operations NQF Level 2 or equivalent, including communication and mathematical literacy, NQF Level 2. UNIT STANDARD RANGE The typical context of this unit standard includes any processing environment, for example chemical, minerals or beverage processing. Range statements, which are applicable to the unit standard titles, specific outcomes and assessment criteria are found beneath the applicable assessment criteria. Specific Outcomes and Assessment Criteria: SPECIFIC OUTCOME 1 Perform elementary chemical calculations. ASSESSMENT CRITERIA ASSESSMENT CRITERION 1 The concepts of atomic number, mass number, atomic mass, mole and Avogadro constant are defined and explained in terms of scientific principles. ASSESSMENT CRITERION 2 The relative atomic masses for a range of elements are calculated with information obtained from the periodic table of elements. ASSESSMENT CRITERION 3 The relative formula masses for a range compounds are calculated with information obtained from the periodic table of elements. ASSESSMENT CRITERION 4 The term "isotope" is defined and explained in terms of an atomic model. ASSESSMENT CRITERION 5 Manual 64 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 The masses for a mole of different substances are calculated by using accepted scientific principles. ASSESSMENT CRITERION 6 The volumes for a mole of different gases are calculated by using accepted scientific principles. ASSESSMENT CRITERION 7 The mass and percentages of substances are calculated using chemical formulas and balanced chemical equations. SPECIFIC OUTCOME 2 Demonstrate an understanding of acids and bases. ASSESSMENT CRITERIA ASSESSMENT CRITERION 1 The properties of acids and bases are defined and explained using chemical terminology. ASSESSMENT CRITERION 2 Strong and weak acids are defined and explained in terms of the acid and base equilibrium constants. ASSESSMENT CRITERION 3 The ionisation of water and the concept of pH are explained in terms of the ionisation constant for water. ASSESSMENT CRITERION 4 A list of acids is made in order of descending strength by using the ionisation constant for water for each acid. ASSESSMENT CRITERION 5 A range of acid reactions are balanced and explained by using accepted chemical techniques and notation. ASSESSMENT CRITERION RANGE Included are reactions with metals, metal oxides, metal hydroxides, carbonate and alkalis. ASSESSMENT CRITERION 6 Standardisation calculations are performed using basic scientific principles. SPECIFIC OUTCOME 3 Demonstrate an understanding of oxidation-reduction (redox) reactions and their industrial applications. ASSESSMENT CRITERIA ASSESSMENT CRITERION 1 The principles of reduction and oxidation reactions are explained in terms of electron transfer principles. ASSESSMENT CRITERION 2 The principles of standard electrode potentials are explained with the assistance of a table with standard electrode potentials. ASSESSMENT CRITERION 3 Manual 65 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 A range of redox reactions are written down and balanced using scientific principles and notation. ASSESSMENT CRITERION 4 Redox reactions are balanced by using the oxidation states of the elements. ASSESSMENT CRITERION 5 A prediction can be made whether redox reactions will take place based on information obtained from a redox table. SPECIFIC OUTCOME 4 Demonstrate understanding of chemical reaction rates. ASSESSMENT CRITERIA ASSESSMENT CRITERION 1 The principles of chemical reaction rates are described in terms of scientific principles and terminology. ASSESSMENT CRITERION 2 The factors affecting chemical reaction rates are explained with the assistance of practical examples. ASSESSMENT CRITERION RANGE Factors include but are not limited to the nature of substances involved in the reaction, reaction surface area, concentrations, temperature and presence of a catalyst. ASSESSMENT CRITERION 3 The principles of endothermic and exothermic reactions are described in terms of scientific principles and terminology. ASSESSMENT CRITERION 4 Industry applications demonstrating the role of chemical reaction rates are described using practical industry examples. SPECIFIC OUTCOME 5 Demonstrate understanding of chemical equilibrium. ASSESSMENT CRITERIA ASSESSMENT CRITERION 1 The principles of phase equilibrium are described in terms of scientific principles and terminology. ASSESSMENT CRITERION 2 The principles of chemical equilibrium are described in terms of the equilibrium constant. ASSESSMENT CRITERION 3 The factors affecting chemical equilibrium are explained with the assistance of Le Chatelier`s Principle. ASSESSMENT CRITERION RANGE Factors include but are not limited to concentration, pressure and temperature. ASSESSMENT CRITERION 4 Manual 66 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Equilibrium calculations are performed involving concentrations and the equilibrium constant. ASSESSMENT CRITERION 5 Industry applications demonstrating the role of chemical equilibrium are described using practical industry examples. UNIT STANDARD ACCREDITATION AND MODERATION OPTIONS An assessor, accredited with a relevant NQF Level 3 or higher qualification, will assess the learner’s competency. Only an assessor with considerable first hand experience in process operations will assess the learner’s competency. Anyone assessing a learner or moderating the assessment of a learner against this qualification must be registered as an assessor with the relevant ETQA. Direct observation in simulated or actual work conditions is required. UNIT STANDARD ESSENTIAL EMBEDDED KNOWLEDGE The qualifying learner understands and is able to: Perform elementary chemical calculations. Demonstrate an understanding of acids and bases. Demonstrate an understanding of oxidation-reduction (redox) reactions and their industrial applications. Demonstrate understanding of chemical reaction rates. Demonstrate understanding of chemical equilibrium. UNIT STANDARD DEVELOPMENTAL OUTCOME N/ A UNIT STANDARD LINKAGES N/ A Critical Cross-field Outcomes (CCFO): 1. UNIT STANDARD CCFO IDENTIFYING Identify and solve problems in the following areas, which response displays that responsible decisions, using critical and creative thinking, have been made: Refer to all Specific Outcomes. 2. UNIT STANDARD CCFO ORGANISING The learner is able to organise and manage himself and his activities responsibly and effectively. Refer to all Specific Outcomes. 3. UNIT STANDARD CCFO COLLECTING Collect, analyse, organise and critically evaluate information by: Refer to all Specific Outcomes. Manual 67 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 4. UNIT STANDARD CCFO COMMUNICATING Communicate effectively by using mathematical and/or language skills in the modes of oral and/ or written presentations during: Refer to all Specific Outcomes. 5. UNIT STANDARD CCFO SCIENCE Use science and technology effectively and critically, showing responsibility towards the environment and health of others when: Refer to all Specific Outcomes. 6. UNIT STANDARD CCFO DEMONSTRATING Demonstrate an understanding of the world as a set of related systems. This is evident in: Refer to the following Specific Outcome/ s: Demonstrate understanding of chemical reaction rates. Demonstrate understanding of chemical equilibrium. 7. UNIT STANDARD CCFO CONTRIBUTING Contribute to the full personal development of each learner and the social and economic development of the society at large by: Refer to all Specific Outcomes. ANNEXURE 4: GLOSSARY You will often encounter words, acronyms or abbreviations that you do not understand. It is important that you familiarise yourself with industry-specific terms as soon as possible. Abbreviations, acronyms and terminology: Term Description A Atomic number The number of protons in the nucleus of an atom. C Corrosion To destroy a metal or an alloy gradually by a process called oxidation or by a chemical reaction Electrochemical cell A cell where the reaction occurs spontaneously and an electrolytic cell where a non-spontaneous reaction occurs. Endothermic reaction A reaction where energy has to be supplied to the system continually in order for the reaction to occur. Equilibrium constant Chemists use equilibrium constants instead of percentages to indicate the position of equilibrium in a chemical reaction. Equilibrium constants relate the concentration of reactants and products at equilibrium at a given temperature. E Manual 68 US 244241 Rev.1 – Sept 08 Sparrow Consulting © September 08 Isotopes Atoms that have the same atomic number but a different atomic mass. They have the same number of protons and electrons but different number of neutrons. Mass number The sum of the protons and neutrons in the nucleus. Oxidise A chemical reaction where oxygen is added or combined to a substance and an oxide is formed Oxidation number The charge an atom would have if bonding electrons were transferred completely to the atom with greater affinity for them in a particular situation. Rate of chemical reaction The change in concentration of a substance per time unit. Relative atomic mass The ratio of the absolute mass of an atom of that element to that of the atomic mass unit. Reduction-oxidation (redox) Reactions are a group of reactions that are concerned with transfer of electrons between compounds. I M O R Manual 69 US 244241