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CONTENT TRAINING

This training aims to enhance learner performance in multiple choice questions related to physical properties, reaction rates, and chemical equilibrium. It includes analysis of past exam papers and emphasizes the importance of the Examination Guidelines (EG). Productive pedagogics methods are utilized for increased learning and feedback. Specific topics such as recognition of conjugate acid-base pairs and hydrolysis reactions are covered.

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CONTENT TRAINING

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  1. CONTENT TRAINING PHYSICAL SCIENCES P2 (CHEMISTRY) Nov 2014 and 2015 • DON FRANCIS Pacaltsdorp SS 21 May 2016

  2. MCQ’s Today the purpose of the Content Training is to improve learner performance in the following questions asked in the November 2015 question paper: Multiple Choice Questions 2, Physical Property Relationships Reaction Rate Chemical Equilibrium EQUILIBRIUM GRAPHS HAVE BEEN ADDED Analysis-subject-2013

  3. MCQ’s The most important document that should be consulted in order to prepare successfully for the 2016 PHYSICAL SCIENCES FINAL EXAMINATION is the EXAMINATION GUIDELINES (EG) dated 2014. Analysis-subject-2013

  4. STATISTICAL MODERATION OF SBA EXAM GUIDELINE (EG) EG is more important than the Textbook It tells learners everything they should know for their examination. Each learner should be given a copy of the EG ONLY statements of laws, principles and definitions in the EG should be studied for the Final Exam. DO NOT USE statements of laws, principles and definitions FROM OTHER SOURCES Analysis-subject-2013

  5. STATISTICAL MODERATION OF SBA • PRODUCTIVE PEDAGOGICS • Use teaching methods that increase learning, such as • Reduce teaching time. Increase learning time • Increase time to read, remember, calculate, interpret, analyse, draw, solve problems, … • Better to have learners writing answers on black board on a daily basis (than to teach whole period daily) • Increase diagnosis • Increase remediation – correct mistakes • REMEMBER: FEEDBACK TO LEARNERS INCREASES LEARNING THE MOST (HSRC) • KNOW WHAT IS WORKING AND WHAT IS NOT • WORKING DAILY Analysis-subject-2013

  6. MCQ’s RECOGNITION OF A CONJUGATE ACID-BASE PAIR: There is ONLY 1 proton (H+) the difference between the Acid and the Base i.e. they DIFFER by 1 proton (H+) ONLY. They are found on OPPOSITE SIDES of the arrow (SINGLE or DOUBLE arrows) in the equation. 1,1 Consider the reaction represented by the balanced equation below. H3PO4(aq) + H2O(ℓ) ⇌ H3O+(aq) + H2PO4-(aq) Which ONE of the following is a conjugate acid-base pair? H3O+(aq) and H2O(ℓ) H3PO4(aq) and H2O(ℓ) H3PO4(aq) and H3O+(aq) H3O+(aq) and H2PO4-(aq) A B C D Analysis-subject-2013

  7. MCQ’s DEFINITION OF HYDROLYSIS REACTION: A reaction between an ALKYL HALIDE and a (dilute aqueous) strong BASE such as NaOH or KOH, or H2Oto form an ALCOHOL. 1.2 Which ONE of the following is a product formed during the hydrolysis of bromoethane? NOTES Reaction conditions: Mild heat is required for the alcohol to form. REASON: Alkyl groups are electron repelling groups, making it easier to remove the Br - 3o > 2o > 1o A B C D Water Ethene Ethanol Bromine Analysis-subject-2013

  8. MCQ’s Empirical formula is the SIMPLEST formula i.e. the number of the different atoms in this formula are relatively prime to each other i.e. their HCF(Highest Common Factor) is 1. 1.3 Which ONE of the following is the EMPIRICAL FORMULA of 1,2-dichloroethane? NOTES The molecular formula of 1,2-dichloroethane Is C2H4Cℓ2 2, 4 and 2 are not relatively prime because their HCF is 2. If 2, 4 and 2 are divided by their HCF viz. 2 respectively, then the numbers resulting viz. 1,2 and 1 are relatively prime i.e. their HCF is 1. B is the answer. CHCℓ CH2Cℓ CHCℓ2 C2H4Cℓ2 A B C D Analysis-subject-2013

  9. MCQ’s Technique to answer 1.4: There are 2 changes that can be made that will favour the forward reaction (a) Add more of one reactant (b) Decrease the amount or concentration of a product or Remove a product. None of the reactants are distracters. The amount or concentration of a product must be decreased or a product must be removed. D is the answer – OH- removes H+ from the products. 1.4 The reaction represented by the balanced equation below reaches equilibrium in a closed container. Cℓ2(g) + H2O(ℓ) ⇌ Cℓ─(aq) + CℓO─(aq) + 2H+(aq) Which ONE of the following reagents will favour the forward reaction when added? NOTES H2O(ℓ) does not appear in the Kc equation for the reaction but an increase in the mol of any product increases the mol of H2O(ℓ). If H2O is in the gaseous phase, it is written in the Kc equation . All will be normal. This is also applicable to pure solids in heterogeneous equilibriums such as CaCO3(s) and CaO(s) in example below. Example: CaCO3(s) ⇌ CaO(s) + CO2(g) i.e. Increase or decrease in [CO2], increases or decreases n(CaCO3) Hydrogen Sodium chloride Hydrogen chloride Sodium hydroxide A B C D Analysis-subject-2013

  10. MCQ’s • ELABORATING ON THE STATEMENTS MADE IN THE PREVIOUS SLIDE • Pure liquids such as H20(ℓ) and pure solids such as CaCO3(s) do not appear in the Kc equation but the n(H20) and the n(CaCO3) change in accordance with LCP. • Example 1 • Cℓ2(g) + H2O(ℓ) ⇌ Cℓ─(aq) + CℓO─(aq) + 2H+(aq) • H20(ℓ) is not written in the Kc equation but if the n(H+) is increased then, according to LCP, n(Cℓ─) and n(CℓO─) decrease, and n(Cℓ2) and n(H2O) increase. The opposite occurs, according to LCP, when the n(H+) is decreased i.e. n(Cℓ2) and n(H2O) decrease and n(Cℓ─) and n(CℓO─) increase. • Example 2 • CaCO3(s) ⇌ CaO(s) + CO2(g) • CaCO3(s) and CaO(s) arenot written in the Kc equation but if the n(CO2) is increased then, according to LCP, n(CaO) decreases, and n(CaCO3) increases. The opposite occurs, according to LCP, when the n(CO2) is • decreased i.e. n(CaCO3) decreases and n(CaO) increases Analysis-subject-2013

  11. MCQs Refer to EG, page 23. NOTES: In cell notation: A comma “,” is used to separate two substances in the SAME phase. A vertical line “|” is used to separate two substances in different phases. In the case where an electrode is a gas or a liquid, an inert electrode such as Pt or C is used. Read from Left to Right: Aℓ is oxidised to Aℓ3+ and Co3+ is reduced to Co2+. 1.5 The following half-reactions take place in a galvanic cell: Co3+ + e- ⇌ Co2+ Aℓ3++ 3e- ⇌ Aℓ Which ONE of the following is the cell notation for this cell? Aℓ ∣ Aℓ3+ ∥ Co3+, Co2+ Aℓ ∣ Aℓ3+ ∥ Co3+, Co2+ ∣ Pt Aℓ ∣ Aℓ3+ ∥ Co2+, Co3+ ∣ Pt Pt ∣ Co2+ , Co3+ ∥ Aℓ3+ ∣ Aℓ A B C D • Study Tips: CELL NOTATION • You only need to know what is written on ONE SIDE of (the salt bridge). • Suggestion: Use OAR which means Oxidising Agent is written on Right of • Then the Reducing Agent will be written on the Left of Analysis-subject-2013

  12. MCQs 2015 TECHNIQUE for 1.6: Reducing Agents and Oxidising Agents are found on the LHS of an equation. The stem of the question gives you the reactants. The reducing agent is  one of them. D can thus be eliminated because it is a product. From Table 4B in the data sheets, potassium ions and chlorine gas are oxidising agents. A and D can also be eliminated.  Answer is C 1.6 Chlorine gas (Cℓ2) is bubbled through a potassium iodide solution (KI). The reducing agent in this reaction is: Reaction taking place: Cℓ2 + 2KI → 2KCℓ + I2 OA and RA are ALWAYS found on LHS of the arrow. A B C D Potassium ions Chlorine gas Iodide ions Chloride ions • DEFINITIONS • Reduction is a GAIN in electrons OR Reduction is a DECREASE in OXIDATION NUMBER. • Oxidation is a LOSS of electrons OR OXIDATION is an INCREASE in OXIDATION NUMBER • Reducing agent is a substance that REDUCES another substance but it itself undergoes OXIDATION • Oxidising agent is a substance that OXIDISES another substance but it itself undergoes REDUCTION NO2(g) 2NO(g) + O2(g) In this reversible redox reaction Reducing and Oxidising agents occur on both sides of the equation. Analysis-subject-2013

  13. MCQs 2015 This question is similar to Q1.6 in the 2015 question paper. The difference is that the equation is provided, making it easier. 1.7 Consider the reaction represented by the balanced equation below: Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) In the above reaction, Cu(s) is the ... • The same technique to answer Q1.6 can be applied to Q1.7: • Reducing agent (Cu) and oxidising agent (Ag+) are found on the LHS of the equation. oxidising agent and is reduced. oxidising agent and is oxidised. reducing agent and is reduced. reducing agent and is oxidised. A B C D Analysis-subject-2013

  14. PHYSICAL PROPERTY RELATIONSHIPS DEFINITIONS Independent variable: The variable that is changed in the experiment/investigation Dependent variable: The variable that is measured in the experiment/investigation An investigative question: A relationship between the independent and dependent variables. Hypothesis: Is a prediction about the relationship between the independent and dependent variables When is an experiment/investigation fair? When it has only one independent variable. Framework for formulating investigative questions: What is the relationship between the independent and dependent variables? (In this order) Analysis-subject-2013

  15. PHYSICAL PROPERTY RELATIONSHIPS Refer to EG, page 17. There are 12 physical property relationships that should be studied. 9 are provided in Table 1 below. The trend is shown horizontally from Left to Right. Table 1: Dependent Variables (What is measured ) Independent Variables (What you change) Analysis-subject-2013

  16. PHYSICAL PROPERTY RELATIONSHIPS Refer to EG, page 17. There are 12 physical property relationships that should be studied. The remaining 3 are shown below. The trend is shown vertically from Top to Bottom. Dependent Variables (What is measured ) Independent Variables (What you change) Boiling point increases Vapour pressure decreases Melting point increases Analysis-subject-2013

  17. PHYSICAL PROPERTY RELATIONSHIPS Relationship between boiling point, melting point and vapour pressure, and Type of functional group. First we need to arrange the functional groups in a particular order e.g. in increasing order according to the strengths of their intermolecular forces. This is done below and is based on the principle that you are comparing functional groups in molecules of comparable molecular mass. Alkenes < Alkanes < Alkynes < CHO < CO < COO < OH < RX < COOH where RX is an alkyl halide and X = I. Before we do examples we take a deeper look at intermolecular forces. Acknowledgements: M. Human Analysis-subject-2013

  18. PHYSICAL PROPERTY RELATIONSHIPS Hydrogen bonds: Hydrogen bonding occurs between molecules in the following homologous series: Alcohols and Carboxylic Acids. Since hydrogen bonds are stronger than London/Dispersion forces, the boiling and melting points of alcohols and carboxylic acids are higher than those of alkanes of comparable molecular mass. NOTES: In alcohols that contain only one –OH group there is only one site and in carboxylic acids there are two sites for hydrogen bonding to occur. Comparing the strengths of hydrogen bonds: The trend in the strengths of hydrogen bonds from strongest to weakest is: H – F > H – O > H – N The differences are due to electro-negativity differences between the atoms in each hydrogen bond. H – O – R O H – O R – O R – C C – R H O – H O H-bonding in alcohols H-bonding in carboxylic acids Carboxylic acids have higher boiling points than alcohols of comparable molecular mass. Analysis-subject-2013

  19. PHYSICAL PROPERTY RELATIONSHIPS RELATIVE STRENGTHS OF DIFFERENT BONDING FORCES In the table below the dissociation energies (energy needed to break bonds) of three types of bonds are compared: Analysis-subject-2013

  20. PHYSICAL PROPERTY RELATIONSHIPS UNDERSTANDING INTERMOLECULAR FORCES Intermolecular forces: These are forces that occur between molecules and they are Van der Waals forces. In organic chemistry intermolecular forces are attractive forces between the molecules that make up a substance. For the purpose of understanding physical property relationships, it is more useful to view boiling and melting point as: “The energy required to separate a substance (liquid or solid) into the individual molecules that it is composed of.” The higher the boiling and melting points are, the stronger the intermolecular forces between particles. Which Van der Waals forces affect the physical properties of organic compounds? Dipole – dipole forces and London forces which are also called Dispersion forces (or Momentary (instantaneous) dipoles) affect physical properties of organic compounds. Analysis-subject-2013

  21. PHYSICAL PROPERTY RELATIONSHIPS London (Dispersion or Momentary) forces: These are the intermolecular forces that affect the physical properties of organic molecules the most because they occur between all organic molecules. When two molecules approach their electrons repel each other creating momentary dipoles that attract each other. When the electrons move again they create opposite charges that are different but still attract. In this way London forces can act on molecules forever. Shape and size of molecules can increase or decrease the effect of London forces: Long thin molecules develop bigger (stronger) London forces than short fat ones because they can lie close together, maximising the number of electrons for London forces. Atoms (e.g. Halogens) with larger molecular mass have more electrons that can travel through a larger radius and thus create bigger (stronger) London forces than small atoms. Analysis-subject-2013

  22. PHYSICAL PROPERTY RELATIONSHIPS Explaining relationships in Table 1. Mr (H-COOH) =MrCH3CH2OH = 46 Analysis-subject-2013

  23. PHYSICAL PROPERTY RELATIONSHIPS EXAMPLE to illustrate the relationship between strength of intermolecular forces and boiling point. Question 1 A laboratory technician is supplied with three unlabelled bottles containing an alcohol, an aldehyde and an alkane respectively of comparable molecular mass. She takes a sample from each bottle and labels them P, Q and R. In order to identify each sample, she determines the boiling point of each under the same conditions. The results are shown in the table below. Explaining Physical Property relationships: STEP 1: Describe the molecular structure (straight or branched chain or number of H-bonds or Dispersion or Dipole-Dipole forces, surface area, etc.) STEP2: State the strength of the IMF (Weak or Strong) STEP 3: State how much energy is needed to overcome the IMF (More or Less) Analysis-subject-2013

  24. PHYSICAL PROPERTY RELATIONSHIPS 1.1 For this investigation, write down the: 1.1.1 Independent variable (1) 1.1.2 Dependent variable (1) 1.2 From the question’s stem, write down a phrase that shows that this investigation is a fair test. (1) 1.3 Which sample (P, Q or R) is the: 1.3.1 Alkane (1) 1.3.2 Alcohol (1) 1.3.3 Refer to boiling point and the type of intermolecular forces present between alcohol molecules to give a reason for the answer in question 1.3.2. (2) 1.4 The alkane is identified as pentane. Will the boiling point of hexane be HIGHER THAN or LOWER THAN that of pentane? Refer to MOLECULAR STRUCTURE, INTERMOLECULAR FORCES and ENERGY needed to explain the answer. (4) Analysis-subject-2013

  25. PHYSICAL PROPERTY RELATIONSHIPS • 1.1.1 strength of intermolecular forces (1) • 1.1.2 Boiling point (1) • 1.2 “comparable molecular mass” or “under the same conditions” (1) • 1.3.1 Q (1) • 1.3.2 R (1) • 1.3.3 highest boiling point or boiling point is 118 oC (1) • H-bonding occurs in alcohol molecules (1) • 1.4 Higher than (1) • Longer (straight) chain length or Larger surface area (1) • increase in the strength of the intermolecular forces (1) or • increase in the number of dispersion or London forces • More energy is required to overcome the inter- • molecular forces (1) or More energy is required to separate the alkane • into its individual molecules Analysis-subject-2013

  26. PHYSICAL PROPERTY RELATIONSHIPS Explaining relationships in Table 1: Analysis-subject-2013

  27. PHYSICAL PROPERTY RELATIONSHIPS Example to illustrate the relationship between chain length and boiling point. Question 2 During a practical investigation the boiling points of the first six straight-chain ALKANES were determined and the results were recorded in the table below. Analysis-subject-2013

  28. PHYSICAL PROPERTY RELATIONSHIPS • 2.1 Write down the general formula of the alkanes. (1) • Refer to the table to answer question 2.2 and 2.3. • 2.2 For this investigation, write down the: • 2.2.1 Independent variable (1) • 2.2.2 Dependent variable (1) • 2.2.3 Conclusion that can be drawn from the results in the table. (2) • 2.3 Write down the name of an alkane that is a liquid at 25 oC. (1) • 2.4 State an investigative question for this investigation. (2) • 2.5 Will the boiling points of the structural isomers of hexane be HIGHER • THAN, LOWER THAN or EQUAL TO that of hexane? Refer to MOLECULAR • STRUCTURE, INTERMOLECULAR FORCES and ENERGY NEEDED to explain • the answer. (4) Q2.5: Follow the 3 STEPS to explain physical property relationships to obtain the answer Analysis-subject-2013

  29. PHYSICAL PROPERTY RELATIONSHIPS 2.1 CnH2n+2 (1) 2.2.1 Chain length (1) 2.2.2 Boiling point (1) 2.2.3 As chain length increases (1) boiling point increases(1) 2.3 pentane or hexane (1) 2.4 What is the relationship between chain length (1) and boiling point (1) 2.5 Lower (1) They are branched chains (1) or there is decrease in surface area. Weaker intermolecular forces (1) or Number of dispersion / London forces decreases/is less (1) Less energy is required to overcome intermolecular forces (1) Analysis-subject-2013

  30. PHYSICAL PROPERTY RELATIONSHIPS Explaining relationships in Table 1: Analysis-subject-2013

  31. PHYSICAL PROPERTY RELATIONSHIPS Example to illustrate the relationship between branched chains and boiling point. Question 3 Three hydrocarbons (A, B and C) with molecular formula C5H12are used to investigate the effect of BRANCHING on the BOILING POINTS of hydrocarbons. The results obtained are shown in the table below. Analysis-subject-2013

  32. PHYSICAL PROPERTY RELATIONSHIPS 3.1 Write down for this investigation the: 3.1.1 Dependent variable (1) 3.1.2 Independent variable (1) 3.1.3 Controlled variable (1) 3.2 Are these hydrocarbons saturated or unsaturated? Explain the answer. (3) 3.3 One of the hydrocarbons (A, B or C) has a straight chain with no branches. Write down the following: 3.3.1 The letter (A, B or C) that represents this hydrocarbon (1) 3.3.2 Its IUPAC name (2) 3.4 Consider hydrocarbon C and its boiling point. 3.4.1 Write down the structural formula of hydrocarbon C. (2) 3.4.2 Explain why hydrocarbon C has the lowest boiling point. In your explanation, refer to its structure, intermolecular forces and the energy involved. (3) 3.5 Which ONE of hydrocarbons (A, B or C) has the highest vapour pressure? Refer to the data in the table to give a reason for the answer. (3) Q3.4.2: Follow the 3 STEPS to explain physical property relationships to explain the answer, Analysis-subject-2013

  33. PHYSICAL PROPERTY RELATIONSHIPS 3.1.1 boiling point (1) 3.1.2 branched chains (1) 3.1.3 molecular mass (1) 3.2 saturated (1) have general formula C5H12 (1) have only carbon-carbon single bonds (1) 3.4.1 (2) 3.3.1 A (1) 3.3.2 pentane (2) 3.4.2 it has the smallest surface area (1) or It is spherical It has the weakest intermolecular forces (1) or It has the east number of dispersion / London forces decreases Least energy required to overcome the intermolecular forces (1) 3.5 C (1) It has the lowest boiling point (2) Or its boiling point is 10 oC Q3.4.2: Follow the 3 STEPS to explain physical property relationships to explain the answer, Analysis-subject-2013

  34. PHYSICAL PROPERTY RELATIONSHIPS Relationship between boiling point, melting point and vapour pressure, and Type of functional group. First we need to arrange the functional groups in a particular order e.g. in increasing order according to the strengths of their intermolecular forces. This is done below and is based on the principle that you are comparing functional groups in molecules of comparable molecular mass. Alkenes < Alkanes < Alkynes < CHO < CO < COO < OH < RX < COOH where RX is an alkyl halide and X = I. Before we do examples we take a deeper look at intermolecular forces. Acknowledgements: M. Human Analysis-subject-2013

  35. PHYSICAL PROPERTY RELATIONSHIPS Hydrogen bonds: Hydrogen bonding occurs between molecules in the following homologous series: Alcohols and Carboxylic Acids. Since hydrogen bonds are stronger than London/Dispersion forces, the boiling and melting points of alcohols and carboxylic acids are higher than those of alkanes of comparable molecular mass. NOTES: In alcohols that contain only one –OH group there is only one site and in carboxylic acids there are two sites for hydrogen bonding to occur. Comparing the strengths of hydrogen bonds: The trend in the strengths of hydrogen bonds from strongest to weakest is: H – F > H – O > H – N The differences are due to electro-negativity differences between the atoms in each hydrogen bond. H – O – R O H – O R – O R – C C – R H O – H O H-bonding in alcohols H-bonding in carboxylic acids Carboxylic acids have higher boiling points than alcohols of comparable molecular mass. Analysis-subject-2013

  36. PHYSICAL PROPERTY RELATIONSHIPS Definition: A molecule or bond is a dipole if a partial positive and a partial negative charge are separated by some distance. Dipole – dipole forces: Dipoles occur in alkyl halide molecules because they have polar bonds e.g. in CH3CH2Cl the C – Cl bond is polar because Cl being more electro-negative than C attracts the shared electron pair more to its side. This creates a partial positive charge on the C atom and a partial negative charge on the Cl atom: - . An example of other homologous series that contain bonds that are dipoles are aldehydes and ketones (C=O is a dipole). Analysis-subject-2013

  37. PHYSICAL PROPERTY RELATIONSHIPS RELATIVE STRENGTHS OF DIFFERENT BONDING FORCES In the table below the dissociation energies (energy needed to break bonds) of three types of bonds are compared: Analysis-subject-2013

  38. PHYSICAL PROPERTY RELATIONSHIPS Refer to EG, page 17. There are 12 physical property relationships that should be studied. The remaining 3 are shown below. The trend is shown vertically from Top to Bottom. Dependent Variables (What is measured ) Independent Variables (What you change) Boiling point increases Vapour pressure decreases Melting point increases Analysis-subject-2013

  39. PHYSICAL PROPERTY RELATIONSHIPS • Example to illustrate the relationship between type of functional group and vapour pressure. • Question 4 • 1.1 Define the term functional group of an organic compound. • 1.2 Four compounds of comparable molecular mass are used to investigate • the effect of functional groups on vapour pressure. The results obtained • are shown in the table below. Analysis-subject-2013

  40. PHYSICAL PROPERTY RELATIONSHIPS 1.2.1 For this investigation write down: (a) The independent variable (b) The dependent variable (c ) Two variables that must be controlled (d) An investigative question 1.2.2 Which ONE of the compounds (A, B, C or D) in the table has the: (a) Highest boiling point (Refer to the vapour pressures in the table to give a reason for the answer.) (b) Weakest intermolecular forces 1.2.3 Refer to the type of intermolecular forces to explain the difference between the vapour pressure of compound A and compound B. 1.2.4 Briefly explain the difference in vapour pressure between compound C and D Analysis-subject-2013

  41. PHYSICAL PROPERTY RELATIONSHIPS 1.1 A bond/an atom/or a group of atoms that determines the physical and chemical properties of organic compounds. See Pg. 16 EG. 1.2.1 (a) type of functional group (b) vapour pressure (c ) Molecular mass, temperature at which vapour pressure is measured (d) What is the relationship between type of functional group and vapour pressure 1.2.2 (a) D because it has the lowest vapour pressure (b) A 1.2.3 A has dispersion or London forces between its molecules. B has dipole- dipole forces between its molecules. Dipole-dipole forces are stronger than dispersion or London forces 1.2.4 C has one H-bond between its molecules but D has two. The IMF in C are weaker than in D. C has a lower vapour pressure than D Analysis-subject-2013

  42. CHEMICAL CHANGE: REACTION RATE EXOTHERMIC AND ENDOTHERMIC REACTIONS ENTHALPY (H)Is the heat content of a system OR The sum of the internal energy (U) of a system and the energy available to do work on the surroundings (pV) As an equation, enthalpy is written as: H = U + pV (1) Although it is difficult to measure enthalpy, a change in enthalpy (OR an increase or decrease in enthalpy) at constant pressure, p, can be measured. As an equation, “Change in Enthalpy” is written as: ∆H =∆U + p∆V (p is constant) (2) A change in enthalpy is called the Heat of Reaction (∆H) !st Law of Thermodynamics 2nd Law of Thermodynamics: Heat moves from a hot object to a cold object and not vice versa. (NOT EXAMINABLE) Analysis-subject-2013

  43. CHEMICAL CHANGE: REACTION RATE • ACTIVATION ENERGY Is the minimum energy required for a reaction • to occur. • H or HEAT OF REACTION Is the energy released or absorbed during a • chemical reaction at constant pressure • As an equation it is written as: • ∆H =∆U + p∆V (p is constant) • ∆H has more than one meaning in chemistry: • H is the DIFFERENCE between the energy of the products and the energy of the reactants in a potential energy diagram(Graph of Ep versus Course of Reaction) • ∆H is the heat of reaction • ∆H is the change in enthalpy • ∆H =∆U + p∆V (p is constant) • ∆H = Hfinal – Hinitial = Hf - Hi All definitions, laws and principles should be taken from the EG Definition: ACTIVATED COMPLEX Iis the unstable transition state from reactants to products. Analysis-subject-2013

  44. CHEMICAL CHANGE: REACTION RATE • ENDOTHERMIC REACTION Is a reaction in which energy is absorbed • (taken in) from the surroundings. • Notes: • The following observations are made duringendothermic reactions: • The reaction mixture gets cold (its temperature decreases) • ∆H > 0 (In an Ep (potential energy) diagram, Hf > Hi OR Epf> Epi) • Any one of the bulleted observations can be used to explain why a reaction is endothermic. Potential Energy (kJ) • Potential energy diagram OR • Graph of Potential energy versus Course of reaction for an ENDOTHERMIC REACTION. Hf> Hi  Hf - Hi = ∆H > 0 Ea Products ∆H > 0 Reactants Course of Reaction Analysis-subject-2013

  45. CHEMICAL CHANGE: REACTION RATE • EXOTHERMIC REACTION Is a reaction in which energy is released (given off) • to the surroundings. • Notes: • The following observations are made duringexothermic reactions: • The reaction mixture gets hot (its temperature increases) • ∆H < 0 (In an Ep (potential energy) diagram, Hf< Hi OR Epf < Epi) • Any one of the bulleted observations can be used to explain why a reaction is exothermic. • Potential energy diagram OR • Graph of Potential energy versus Course of reaction for an EXOTHERMIC REACTION. • Hf < Hi  Hf - Hi = ∆H < 0 Analysis-subject-2013

  46. CHEMICAL CHANGE: REACTION RATE (POSITIVE )CATALYST It isa substance that increases the rate of a chemical reaction without itself undergoing a permanent change. A (positive) CATALYSTincreases the rate of a reaction by making the reaction proceed along an alternative path with lower activation energy. The net activation energy for the reaction therefore decreases. It does not change the value of ∆H Potential Energy (kJ) Lowering of Ea Products ∆H > 0 Reactants Course of Reaction Analysis-subject-2013

  47. CHEMICAL CHANGE: REACTION RATE RATE OF A CHEMICAL REACTIONIt is the rate of change in the concentration of one of the reactants or one of the products in a chemical reaction. The following formulae can be used to calculate the AVERAGE RATE of a reaction: Average reaction rate = and the unit of measurement will be mol.dm-3.s-1 OR Average reaction rate = and the SI unit of measurement will be kg.s-1 OR Average reaction rate = and the SI unit of measurement will be m3.s-1 Analysis-subject-2013

  48. CHEMICAL CHANGE: REACTION RATE In the case where a change cannot be measured directly but only seen, such as the disappearance of a cross as the result of the formation of a precipitate, the following equation should be used: Reaction rate = and the SI unit of measurement will be s-1 Consider the reaction taking place in a Zn-Cu galvanic cell: Cu2+ + Zn → Cu + Zn2+ During this reaction the mass of both the Cu2+ and Zn decrease, whereas the mass of Cu and Zn2+increase. Determine the reaction rate on the LHS and RHS of the arrow: LHS: Reaction rate = = < 0 (Because mf < mi , mf - mi < 0 ) RHS: Reaction rate = = > 0 (Because mf > mi , mf - mi > 0 Dilemma: LHS Reaction Rate is negative but positive on the RHS This dilemma also occurs with [Cu2+]and {Zn2+] i.e. concentration Analysis-subject-2013

  49. PROPORTION REACTION RATE IS ALWAYS POSITIVE (According to IUPAC’s Golden Book) The average reaction rate on the LHS is made positive by taking the negative of its value. EXAMPLE In a Zn-Cu galvanic cell, the [Cu2+] changes from 1 mol.dm-3 to 0,9 mol.dm-3 in 100 seconds. Calculate the average rate of this reaction. ANSWER Reaction Rateav = - = - = 0,001 mol.dm-3 s-1 or 1x10-3mol.dm-3s-1 Analysis-subject-2013

  50. CHEMICAL CHANGE: REACTION RATE • The collision theory is a model that explains reaction rate as the result of particles colliding with a certain minimum energy. • NOTES: ONLY EFFECTIVE COLLISIONS RESULT IN A CHEMICAL REACTION. The following two factors determine whether a collision will be effective or not: • The molecules must have sufficient kinetic energy • The molecules must be correctly orientated. • According to the Collision Theory: • Reaction Rate  x x In place of this, DBE uses “effective collisions”. Analysis-subject-2013

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