SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS TOPIC (1A): INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS1 1 Define (a) functional group. The atom or group of atoms that gives the organic compound its characteristic properties. (b) molecular formula. Formula that shows the number of atoms present in one molecule of compound (c) structural formula. Formula that shows the arrangement of atoms in a molecule 2 Complete the table below. (a) No. Group Functional Molecular Structural Formula Group Formula IUPAC NAME A Alkenes C=C C4H8 CH3CHCHCH3 2-Butene B Carbonyl –C=O C4H8O CH3COC2H5 Butanone C Alcohols –OH C3H8O CH3CH(OH)CH3 2-Propanol D Carboxylic acid –COOH C4H8O2 CH3CH2CH2CO2H Butanoic Acid E Halogenoalkane –X, –Cl C5H11Cl CH3CH2CHClCH2CH3 3Chloropentane (b) No. General Formula A Homologous series Alcohol CnH2n+2O Molecular Formula C4H10O B Cycloalkane CnH2n C5H10 C Alkene CnH2n C5H10 1 Structural Formula CH3CH2CH2CH2OH Or CH3CH2CH2(OH)CH3 CH2=CHCH2CH2CH3 SHE1325 3 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Write the expanded structural formula or full structural formula for each of the following condensed structural formula. (a) 2-pentene (b) H H H Br Br H C C C C C H H H H H H H H C C C H H H H (c) 2,2-dichloro-5-methylhexane H H Cl H H H H C C C C C C H Cl H H H C H H 1,2-dibromopropane (d) 3-bromo-2-pentene H H H H Br H H C C C C C H H H H H (d) 3-ethyl-2-methyl-1-hexene (e) 3-chloro-1-butene H H H H H H C H H H C C C C C C H H H H H C H H H H 2,2,4-trimethylheptane H H H H H C H H H H H C C C C C C C H C H C H H H H H H H H C C C H Cl H H C (f) H H H 2 H H C H SHE1325 4 5 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Write the IUPAC name for each of the following compounds. (a) CH3CH(CH3)CH2CH3 : 2-methylbutane (b) CH3C(CH3)ClCH2C(CH3)2CH3: (c) CH3(CH2)2CH(C2H5)NH2 : 4-chloro-2,2,4trimethylpentane (d) CH3CH2CH(C2H5COOH)CH3 : 3-hexylamine 4-methylhexanoic acid Draw five possible structural isomers of C5H10 which contain a double bond. Give the IUPAC names. H C H H H H H C C C C H H H H H H C C C C H H H C H H H H H A: 2-pentene H H H H C C C B: 1-pentene H H H C C H H C C C H C H H H C H H H H H H H C: 3-methyl-1-butene H H C C C C H H H D: 2-methyl-2-butene H C H H H H E: 2-methyl-1-butene (a) Identify two molecules which are positional isomers. A and B or C and D or C and E or D and E (b) Identify two molecules which are chain isomers. A and D or B and C or A and C or A and E or B and E (c) Identify one molecule which shows stereoisomerism and state the type of stereoisomerism shown. Draw and give the IUPAC name of the two stereoisomers. 3 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS 2-pentene shows geometrical isomerism CH3 CH2CH3 CH3 CH2CH3 Cis-2-pentene 6 trans-2-pentene Identify which of the following molecules display geometrical isomerism and, if they do, draw and name the two isomers: (a) 1-pentene No geometrical isomerism (b) 2-pentene H H C C CH2 CH3 CH3 2-methyl-1-butene No geometrical isomerism (d) 3-hexene C C CH2 H CH3 CH3 trans-3-hexene 7 CH3 H H H C CH2 trans-2-pentene (c) CH2 C H cis-2-pentene CH3 H CH3 C C CH2 CH2 CH3 cis-3-hexene Name the type of reaction or product in the following chemical reactions. (a) C2H4 (b) CH3CH2Cl (c) Propanal (d) 2-bromobutane C2H6 ADDITION CH3CH2OH Propanoic Acid 2–butene 4 SUBSTITUTION OXIDATION Elimination SHE1325 8 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Name all the functional groups, organic compound groups and the possible types of reaction in the compound below. Mark any chiral carbon with an (*) H H H C O C Cl H H No. Functional Group Type of reaction 1 C=O, carbonyl group Oxidation, addition 2 C–Cl, carbon-halogen Substitution 3 9 C C* reduction C = C,carbon-carbon Addition, oxidation double bond Complete the table below. Fission Reaction Chemical Equation (1 each) Homolytic fission in Cl2 Heterolytic fission in CH3Cl Examples (2 each) Electrophiles H+, NO2+, CH3+ Nucleophiles OH–, Cl–, H2O, NH3 Free Radicals H , CH3 , Cl , CH2Cl 5 or SHE1325 10 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Methane CH4 reacts with Cl2 in the presence of UV/sunlight. The reaction occurs through 3 stages. (a) Write the name of each stages and steps in each stage of the reaction. Step I : Initiation Cl–Cl Cl + Cl Step II : Propagation CH4 CH3 + + Cl Cl2 CH3 + HCl CH3Cl + Cl Step III : Termination (Show only 1 step) Or any other step between 2 radicals. CH3 + Cl CH3Cl CH3 + CH3 C2H6 (b) What is the name of the above mechanism? (Free)-radical substitution (c) Why is UV/sunlight important in the reaction? To break the Cl–Cl covalent bond. 11 State the type of reaction for halogenation of ethene with bromine in inert solvent. Show the mechanism of the reaction. Electrophilic addition/addition 6 SHE1325 12 13 14 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Write the reactions of complete combustion for the following alkanes and alkenes: (a) ethane C2H6 + 7/2O2 2CO2 + 3H2O (b) propane C3H8 + 5O2 3CO2 + 4H2O (c) ethene C2H4 + 3O2 2CO2 + 2H2O Cracking is used to break large hydrocarbon molecules into smaller molecules. (a) State the conditions for cracking reaction High temperature and pressure if without catalyst. Low T & P if with solid catalyst (aluminium oxide or silicon(IV) oxide). (b) A hydrocarbon, C15H32 is cracked to form 3 products. Ethene and propene are two of the products. What is the molecular formula for the other product? C10H22 (a) State Markovnikov’s Rule According to Markovnikov’s rule, the more positive reagent is added to the carbon of the double bond with higher number of hydrogen atoms. (b) The reaction between 1-butene and HBr give two products in unequal amounts. Identify the two products, state which is the major product. Explain in term of Markovnikov’s Rule why it is the major product. 2-bromobutane and 1-bromobutane. 2-bromobutane is the major product. According to Markovnikov’s rule, the more positive reagent (in this reaction is H+) is added to the carbon of the double bond with higher number of hydrogen atoms. Hence Br- is added to the carbon of the double bond with lower number of hydrogen atoms. 7 SHE1325 15 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Complete the following reaction by providing the major product or reagents needed. (a) H2 Pt Ans: H H (b) Br Br Br2 Ans: CH2Cl2 (c) OH H+ Ans: H2O 8 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS 16 Complete the chart below stating the reagents (R), conditions (C) and structural formulae where appropriate. 17 Reactions of halogenoalkanes (R–X) with CN– to produce nitrile (R–CN) compounds are important in organic chemistry. State two importance of this reaction. Increases the number of carbon atom in the compound. Nitriles are useful compounds that can be converted into amines (by reduction) and carboxylic acids (by hydrolysis). 18 The structures of two iodo-compounds, A and B are shown below. (a) Complete the table. Class of R–X (1o, 2o or 3o) Compound Systematic Name A CH3CI(CH3)2 2–iodo–2–methyl propane 3o 1–iodobutane 1o B CH3(CH2)3I (b) Both A and B undergoes hydrolysis reactions via nucleophilic substitution mechanism. What is the nucleophile of the reaction? OH- or H2O 9 SHE1325 19 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Show how the following conversions can be achieved. R: dilute acid (a) CH3(CH2)2Cl (b) Cl CH3–(CH2)2–CN CH3(CH2)2CO2H Br R: R: C: Br R: NaOH (ethanol) C; heat 20 R: Br2 in CCl4 Predict the products , reagents and conditions wherever appropriate for the following reactions : (a) (b) 1-Chloropropane NaOH in alcohol , boil CH3CH=CH2 I Bromocyclopentane I : NaOH in ethanol/ heat @ KOH in ethanol (c) Chloroethane NH3 , alcohol CH3CH2NH2 heat 21 Rank CH3CH2CH2CH2I, CH3CH2CH2CH2Br, CH3CH2CH2CH2F and CH3CH2CH2CH2Cl in ascending order of reactivity of haloalkane towards nucleophilic substitution. Explain your answer. CH3CH2CH2CH2I> CH3CH2CH2CH2Br> CH3CH2CH2CH2F C-X bond becomes stronger from I to F. 10 CH3CH2CH2CH2Cl > SHE1325 22 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Describe the iodomethane. SN2 reaction mechanism in the hydrolysis + INu- Electrophile transition state 11 product leaving group of SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS TOPIC 1 (B) : HYDROXY, CARBONYL, CARBOXYL AND AROMATIC COMPOUNDS HYDROXY COMPOUNDS 1 Name the following compounds according to IUPAC rules : 2 Write reaction equations that can produce ethanol from alkene, haloalkanes and aldehyde. State reagents and conditions in each reaction. (a) (b) (c) CH2=CH2 + H2O (g) CH3CH2OH Condition : H2SO4 CH3CH2Cl + NaOH in aqueous CH3CH2OH Condition : Heat CH3CHO + LiAlH4 in ether CH3CH2OH Condition : Room temperature. 12 SHE1325 3 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Below are isomers that have the molecular formula C4H10O (i) (ii) (iii) (iv) (a) (i) (ii) (iii) (iv) C3H7-CH2OH (CH3)2CH-CH2OH C2H5-CH(OH)CH3 (CH3)3COH State the class of each alcohol above. Primary Primary Secondary Tertiary (b) Name the reagent that can be used to identify the classes of the alcohol above. State the observation of the reaction for each class of alcohols. Lucas reagent : ZnCl2 + HCl (aq) 1° ROH : no ppt formed 2° ROH : white ppt formed within 5 minutes 3° ROH : white ppt formed immediately. (c) State the product of reaction between all of the isomers above with K2Cr2O7. (i) C3H7-COOH (ii) (CH3)2CH-COOH (iii) C2H5-CO-CH3 (iv) No reaction (d) One of the isomers gives a positive observation in iodoform test. Identify the isomer, write the equation and state the observation. C2H5-CH(OH)CH3 + I2 + NaOH CHI3 + C2H5-COONa + NaI + H2O Observation : yellow ppt formed with antiseptic smell. 4 Complete the following equation. (a) rt CH3CH2CH2OH + Na (s) 13 CH3CH2CH2ONa + H2 (g) SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS (b) O CH 3 CH 2 CH 2 CH 2 OH + CH 3 CH 2 CH COOH CH 3 CH 2 CH 2 CH 2 O C CH CH 2 CH 3 CH 3 CH 3 (c) OH CH3CH2CH2CCH 3 170°C + H2SO4 CH2 CH3CH2CH2C CH3 CH3 and C H 3C H 2C H C CH3 CH3 CARBONYL COMPOUNDS 1 Draw full structural formula of following compounds (a) 3-methyl-2-pentanone O CH3 H3C CH3 (b) Phenylethanal and phenylethanone 14 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS (c) Phenylethanone (d) 2,3-dimethylpentanal CH 3 CH 3 CH 2 CHCHCH CH 3 (e) O 4-methyl-3-hexanone CH3CH2CH(CH3)COCH2CH3 2 Fill in the empty boxes with products of reaction and the observations. Reagent A B CH3COCH3 CH3CH2CHO C CH3 C O 2,4-DNP CH3 CH3 C N CH2CH3 NO2 NH H NO2 NO2 C N NH NO2 CH3 C N NH Orange ppt Orange ppt No silver mirror CH3CH2COOH Silver mirror No silver mirror No brick red ppt CH3CH2COOH Brick red ppt No brick red ppt Tollen’s Fehlings NO 2 15 Orange ppt NO2 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS I2 + NaOH CH3COONa Yellow ppt & antiseptic smell ONa No Yellow ppt & antiseptic smell C O Yellow ppt & antiseptic smell 3 Show the mechanism involved in the reaction between propanone and HCN. O CH3 C CH3 + - CN cold NaOH - O CH3 + H C CH3 + OH cold NaOH CH3 C CH3 CN CN 4 Show how the following conversions can be carried out: (a) butanone to butane-1,2-diol O OH conc H2SO4 KMnO4/ OHL i A l H CH2 CHCH2CH3 CH3CCH2CH3 4 CH3CHCH2CH3 CH2( OH) CH( OH) CH2CH3 dry ether c ol d 0 160-180 C (b) 3-pentanone to 3-chloropentane O C H 3C H 2C C H 2C H 3 (c) dry ether K 2 C r2 O 7 /H OH LiAlH 4 C H 3C H 2C H C H 2C H 3 w arm + Cl C H 3C H 2C H C H 2C H 3 2-bromopropane to propanone Br NaOH(aq) OH CH3CHCH 3 heat CH3CHCH 3 16 K2Cr2O7 / H+ warm O CH3C CH3 SHE1325 5 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Compounds W and X have the molecular formula of C8H8O. W reacts with hot Fehling’s solution producing brick red precipitate. X gives negative result towards Fehling’s test but it reacts with LiAlH4 (dry ether) to form alcohol Y. Both X and Y react with iodine in NaOH solution to give a yellow precipitate with antiseptic smell. (a) 6 Determine the structural formula of W, X and Y. C H2 C HO O C C H3 OH C HC H3 W X Y Given the following compounds : G CH3COCH2CH2CH3 Q C6H5CH2COCH3 J C6H5CHO L C6H5CH2CH2CHO (a) Which of these compounds are aldehydes and which are ketones? Aldehydes – J and L (b) Ketones – G and Q Which of the above compounds can be transformed to the following compounds? Write the complete equation involved. (i) (ii) C6H5CH2CH2CH2Cl C6H5CH2C(OH)(CH3)COOH 17 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS (i) L LiAlH 4 / ether C6H5CH2CH2CHO PCl 5 C6H5CH2CH2CH2OH rt C6H5CH2CH2CH2Cl rt + POCl3 + HCl (ii) O Q CN HCN in alkaline C6H5CH2C(OH)CH C6H5CH2CCH3 H2SO4 (dilute) 3 cold boil COOH C6H5CH2C(OH)CH CARBOXYLIC ACID 1 Compound Y is given below : CH3CH(OH)CH2COOH (a) Write the IUPAC name for compound Y. 3-hydroxybutanoic acid (b) Complete the table below : Reagent Structural formula of organic Observation product (i) PCl5 (s) Cl White fume CH3CHCH 2COCl (ii) Na(s) + HCl(g) ONa CH3CHCH 2COONa + H2(g) 18 Gas bubbles of H2 3 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS (iii) LiAlH4 (dry) OH - CH3CH CH2CH2OH (iv) Na2CO3(aq) Gas bubbles of OH CO2 @ lime CH3CHCH 2COONa + CO2 solution turns cloudy (v) OH O CH3CH2CH2CH2NH2 - CH3CCH2CNHCH 2CH2CH2CH3 (vi) CH3CH2OH OH Sweet smell of O ester CH3CHCH 2COCH 2CH3 + H2O AROMATIC COMPOUNDS 1 Draw the structural formulae and give the IUPAC names for the isomers of C8H10 containing a benzene ring. CH3 CH3 CH3 CH3 1,2-dimethylbenzene 1,3-dimethylbenzene CH3 CH3 CH2CH3 19 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS 1,4-dimethylbenzene CH2CH3 ethylbenzene 2 Three reactions of methylbenzene can be summarized on the flow sheet as shown below; (a) State the reagents and conditions for step I and II I Reagents : Concentrated HNO3 Condition : Conc. H2SO4, T=50-60°C II Reagent : Br2 Condition : FeBr3, rt (b) Draw the structural formula of compound S CH 2 Br (c) Write the IUPAC nomenclature for compound Q 1, 4-nitromethylbenzene 20 SHE1325 3 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Show the conversions for the following compound; (state the reagent, conditions and intermediate compound involved) (a) COOH NO2 (b) Br NO2 NO2 Br Br2, FeBr3 HNO3, H2SO4 50 - 60°C warm 21 NO2 SHE1325 4 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Draw the mechanism of nitration of benzene. Step 1 : HO-SO3-H+ HO-NO2 H2O + NO2+ + HSO4Step 2 : Step 3 : 5 State the relative reactivity of benzene and methylbenzene in the nitration reaction. Explain. Relative reactivity : benzene < methylbenzene Methyl is activating groups which activate the benzene ring (by making it more partially negative) / methyl donating electron to the benzene ring, makes it more nucleophilic. 6 Benzene, methylbenzene and phenol form different products in the nitration reaction. Explain why this occurs. Benzene has no substituent, only one product formed. Methylbenzene has CH3 group which is ortho-para director. Phenol has OH group which is ortho-para director. Both will produce 2 products. 22 SHE1325 7 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Complete the reaction below : CH3 Cl CH3 AlCl 3/FeCl3 + + Cl2 Cl uv/sunlight CH2Cl 8 CH3 Complete the reaction below : 23 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS TOPIC 2 : CHEMICAL THERMODYNAMICS 1 Define the following terms: (a) Energy: the ability to do work (b) Heat, q: Energy transfer/exchange due to a difference in temperature between system and surroundings. (Heat flows from region of high temperature to lower temperature and does not belong to the system.) (c) Heats of rxn, qrxn: quantity of heat exchanged between a reaction (system) and its surroundings when the reaction occurs at a fixed temperature. (d) State function: A property of a system that is determined only by the present state/condition of the system and not by how the state is reached. e.g. P, T, V, E, H, S and G (e) Hess’s Law: The enthalpy change for an overall process is the sum of the enthalpy changes of the individual steps of the process. (f) Heat capacity: the measurable physical quantity of heat energy required to change the temperature of an object or body by a given amount (g) Specific heat capacity: the heat required to raise unit mass of a substance by unit temperature interval under specified conditions, such as constant pressure 2 Describe the transfer of heat and matter in an open, closed and isolated system. Heat and matter are exchanged in an open system. Only heat is exchanged in a closed system. No heat or matter is exchanged in an isolated system. 3 Consider following phenomenon. “ Calcium hydroxide solid was dissolved with hydrochloric acid in 250 ml beaker and stirred using glass rod.” Identify which is universe, surrounding and system. System: Neutralization reaction between calcium hydroxide and hydrochloric acid Surrounding: Beaker, glass rod, table, water 24 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Universe: Other than system and surrounding 4 Describe exothermic and endothermic reactions. An exothermic reaction releases heat to the surroundings such that the energy level of the reactant is higher than its product. A negative sign is used to indicate the drop in energy level. An endothermic reaction absorbs heat from the surroundings such that the energy level of the product is higher than its reactant. A positive sign is used to indicate the rise in energy level. 5 How much heat is released when 7.1 g of an alloy is cooled from 100°C to room temperature? Specific heat capacity of that alloy is 0.460 J/g.°C. ΔT = 25°C -100°C = -75°C q=mcΔT q= 7.1 g (0.460 J/ g.°C)( -75°C) = -244.95 J Hence, heat released is 244.95 J 6 25 mL of 1 M NaOH is placed into a calorimeter at 25.0 oC. 25 mL of 1.5 M HCl is added into the calorimeter at the same temperature. The temperature raised by 3.0 oC. Calculate the heat of reaction, qrxn and enthalpy of reaction, ΔHrxn. (Assume negligible heat loss. Density of water = 1.00 g/mL and Specific heat of water, c = 4.18 Jg-1C-1) qrxn = - qsoln = - (50 mL)(1 g/mL)( 4.18 Jg-1C-1)(3.0oC) = - 627 J 0.025L x 1 M = 0.025 mol NaOH reacts with 0.025 L x 1.5 M = 0.0375mol HCl. Thus, NaOH is the limiting reactant while HCl is in excess. ∆Hrxn = qrxn /mol water produced = - 627 J/0.025 mol = - 25 kJ/mol Check: Temperature rise indicates heat released by the system, thus exothermic reaction. 7 5 grams of potassium chloride is dissolved in 50 mL of water at 28.0 o C such that it reaches a final temperature of 32.0 oC. Calculate the heat of reaction, qrxn and enthalpy of reaction, ΔHrxn. (Assume negligible heat loss. Density of water = 1.00 g/mL and Specific heat of water, c = 4.18 Jg-1C-1) 25 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS qsoln = mcΔT = (50 mL)(1.00 g/mL)( 4.18 Jg-1C-1)(32.0 28.0)C = 836 J qrxn = - qsoln = - 836 J ΔHrxn = qrxn/mol of solute = -836 J/(5g/74.55 g/mol) = -12.5 kJ/mol 8 When a 1.23 g sample of propene, C2H4, is burned in a bomb calorimeter, the temperature of the surrounding bath rises by 18.22°C. Calculate the standard enthalpy of combustion of propene. (Heat capacity of bomb calorimeter = 2.12 kJ/°C) qrxn = -qcal = - (heat capacity x ΔT) = -(2.12 kJ/°C x 18.22°C) = -38.6264 kJ Moles of C2H4 = mass/molar mass = 1.23g / 28.06 g/mol = 0.0438 mol ΔH°c = -38.6264 kJ/0.0438 mol = -881.88 kJ/mol 9 10 Identify the type of standard enthalpies for the following equations: (a) Li(s) → Li(g) Standard enthalpy of atomization (b) NaCl(s) → Na+(g) + Cl-(g) Lattice energy (c) Cs(g) → Cs+(g) + e Ionization energy Calculate the enthalpy of formation of butane (C4H10) from the following data: Enthalpy of Combustion of graphite = -393.6kJmol-1 Enthalpy of Combustion of hydrogen = -285.9kJmol-1 Enthalpy of Combustion of butane = -2877.1kJmol-1 4C(s) + 5H2(g) → C4H10(g) ΔH°rxn = Σm ΔH°c (reactant) - Σn ΔH°c (product) = (4ΔH°c C + 5 ΔH°c H2) - ΔH°c C4H10 =-126.8 kJmol-1 26 SHE1325 12 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Given the following data, find the heat of formation for benzene using the ∆Hoc (C6H6(l)) = - 3267 kJ/mol ∆Hoc (C(s)) = - 394 kJ/mol ∆Hoc = - 286 kJ/mol (H2(g)) (a) algebraic method (b) energy cycle method (a) algebraic method 6 CO2(g) + 3 H2O(l) ⟶ C6H6(l) + 𝟏𝟓⁄𝟐O2 (g)∆Hoc=+3267 kJ/mol 6 C(s) + 6 O2(g)⟶ 6 CO2(g) ∆Hoc= 6(-394) kJ/mol 3 H2(g) + 𝟑⁄𝟐O2(g) ⟶3 H2O(l) ∆Hoc= 3(-286) kJ/mol 6 C(s) + 3 H2(g) ⟶C6H6(l) (b) ∆Horxn = +45 kJ/mol energy cycle method ∆Hof (C6H6 (l))= 6[∆Hoc (C(s))]+ 3[∆Hoc (H2(g))]- ∆Hoc (C6H6 (l)) = 6(-394) + 3(-286) + 3267 = + 45 kJ/mol 27 SHE1325 13 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS (a) Write a balanced equation for complete combustion of methanol, CH3OH. CH3OH (l) + 3/2 O2 (g) → CO2 (g) + 2 H2O (g) (b) Calculate the ΔH°f of methanol, using following data: ΔH°c of CH3OH ΔH°f of CO2 ΔH°f of H2O = -638.5 kJ = -393.5 kJ = -241.8 kJ ΔH°c or ΔH°rxn = Σm ΔH°f (product) – Σn ΔH°f (reactant) -638.5 kJ = (ΔH°f CO2 + 2 ΔH°f H2O) - ΔH°f CH3OH ΔH°f CH3OH = -238.6 kJ/mol (c) Calculate the ΔH°f of methanol, using energy cycle, based on the data given at 7(b) ΔH°f C(s) + 2 H2(g) + ½ O2(g) + O2 (-393.5) + O2 2(-241.8) CH3OH(l) + 3/2 O2 (-638.5) CO2(g) + 2 H2O(l) ΔH°f = (-393.5) + 2(-241.8) + (638.5) = -238.6 kJ/mol 28 SHE1325 14 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS The complete combustion of propane is represented by the following reaction, CH3CH2CH3 (g) + 5 O2 (g) → 3 CO2 (g) + 4H2O (l) (a) Calculate the ΔH°rxn using enthalpy of formation. Enthalpy of formation of propane= −104.7 kJ/mol Enthalpy of formation of carbon dioxide= -393.5 kJ/mol Enthalpy of formation of water= -285.8 kJ/mol ΔH°rxn = Σm ΔH°f (product) - Σn ΔH°f (reactant) = (3ΔH°f CO2 + 4 ΔH°f H2O) - (ΔH°f CH3OH + 0) = -2219 kJ/mol (b) Calculate the ΔH°rxn using average bond enthalpy data: Bond C-H C-C O=O C=O H-O (c) ΔH°BE (kJ/mol) +413 +347 +498 +805 +464 Bonds broken : 8(C-H) = 8(+413) = +3304 : 2(C-C) = 2(+347) = +694 : 5(O=O) = 5(+498) = +2490 Bonds formed : 3(2)(C=O) = 6(+805) = +4830 : 4(2)(O-H) = 8(+464) = +3712 ΔH°rxn ΔH°BE(bond broken) - ΔH°BE(bonds formed) (3304+694+2490) – (4830+3712) -2054 kJ/mol = = = Explain why the enthalpy combustion of propane calculated from ΔH°f is different than using ΔH°BE. Enthalpy combustion of propane calculated from ΔH°BE is less accurate because using average data. 29 SHE1325 15 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Construct Born-Haber cycle for reaction of formation of RbCl(s) by using the data given: Equation RbCl(s) →Rb+(g) + Cl-(g) ΔH° (kJ/mol) 696 +403 Rb(g) → Rb+(g) + e- +86 Rb(s) → Rb(g) Cl2(g) → 2 Cl(g) +242 Cl(g) + e- → Cl-(g) -349 =-696 ∆Hof= (d) Calculate the lattice enthalpy ΔH°rxn= ΔH°atm Rb + 1/2ΔH°atm Cl + ΔH°IE Rb + ΔH°EA Cl + (-ΔH°LE) = -435 kJ/mol 30 SHE1325 16 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Using the data given below, construct a Born-Haber cycle (enthalpy diagram) for the formation of CsF and calculate the lattice enthalpy of CsF. Label all heats of reaction involved and its values. Enthalpy of atomization of Cs (s) First ionization energy of Cs (g) Enthalpy of atomization of F2 (g) Electron affinity of F (g) Enthalpy of formation of CsF (s) ∆Hof CsF (s) = = = = = + 76.5 kJ/mol + 375.7 kJ/mol + 79.4 kJ/mol - 328.2 kJ/mol - 553.5 kJ/mol = ∆Hoatom Cs(s) + ∆HoIE1 Cs(g) + ∆Hoatom F2 + ∆HoEA F(g)+ (- ∆Holattice CsF (s)) - 553.5 kJ/mol = (76.5+375.7+79.4+ -328.2 – x ) kJ/mol x = +756.9 kJ/mol 31 SHE1325 17 (a) INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Define standard lattice enthalpy The enthalpy change when 1 mole of solid ionic compound breaks into its gaseous ions under standard states. (b) Explain how the following factors affect lattice energy. (i) ionic radius In a lattice, each ion is surrounded by a number of ions of opposite charge, resulting in strong forces of attraction. the smaller the radius, means the greater attraction, therefore the greater the lattice energy. (ii) ionic charge The higher the charge on an ion the stronger the attractive force that will result in an ionic lattice. the greater the lattice energy. 18 Write an equation to show the relationship between lattice enthalpy, enthalpy change of hydration and enthalpy change of solution. ΔH°solution= ΔH°lattice + ΔH° hydration 19 Spontaneity of a reaction is driven by two factors. State the factors. ∆H and ∆S 32 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS TOPIC 3 : CHEMICAL EQUILIBRIUM 1 2 Describe the following terms. (a) Reversible reaction (b) Dynamic equilibrium (a) A reversible reaction is a reaction that is able to proceed in the forward and reverse directions (b) A state of balance in which the rate of forward reaction is equal to the rate of reverse reaction and the concentrations of reactants and products remain constant with time Write the Kc and Kp expression for the following reactions. (a) CO(g) + Cl2(g) ⇌ COCl2(g) (b) SnO2(s) + 2 H2(g) ⇌ Sn(s) + 2 H2O(g) (c) NaHCO3 (s) ⇌ NaOH (s) + CO2 (g) (d) CH3COOH(l) + CH3CH2OH(l) ⇌ CH3COOCH2CH3(l) + H2O(l) a Kc Kc = [COCl2] / [CO][Cl2] Kp Kp = PCOCl2 / (PCO)(PCl2) b Kc = [H2O]2 / [H2]2 Kp = (PH2O)2 / (PH2)2 c Kc = [ CO2] Kp = d Kc= [CH3COOCH2CH3][ H2O]/ [CH3COOH][ CH3CH2OH] 33 (PCO2) No Kp SHE1325 3 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS The following reactions yield Nitrogen monoxide gas. Based on the Kc values provided, explain which reaction would yield the greater amount of Nitrogen monoxide. Reaction I Reaction II N2(g) + O2(g) ⇌ 2 NO(g) 2 NOBr(g) ⇌ 2 NO(g) + Br2(g) Kc = 4.8 x 10-31 Kc = 0.50 Reaction II because it has a large Kc which indicates a high yield of the products NO(g) and Br2(g) 4 Given the following information, calculate the Kc for the desired reactions. (a) Br2(g) + Cl2(g) ⇌ 2 BrCl(g) Kc1=58.0 BrCl(g) ⇌ ½ Br2(g) + ½ Cl2(g) Kc2 = ? Kc2= [Br2]1/2 [Cl2]1/2/BrCl = (1/58)1/2 = 0.13 5 Given the following two equilibrium: Equilibrium PbI2(s) ⇌ Pb2+(aq) + 2 I-(aq) PbSO4(s) ⇌ Pb2+(aq) + SO42-(aq) (a) Write the equilibrium constant expression for each reaction. Kc1 Kc2 (b) Kc 8.7 × 10-9 1.8 × 10-8 = = [Pb2+][I-]2 [Pb2+][SO42-] Calculate the equilibrium constant for the following reaction. PbI2(s) + SO42-(aq) ⇌ PbSO4(s) + 2 I-(aq) Kc3 = = = = [I-]2 /[SO42-] K1/K2 8.7 × 10-9/1.8 × 10-8 0.483 34 SHE1325 6 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS (a) Calculate the value of Kp for the following equilibrium. 2 SO3(g) ⇌ 2 SO2(g) + O2(g) Kc = 0.02 at 1000K Kp = Kc(RT)n; n = 3-2 = 1 = 0.02 [(0.0821 Latm/molK) x 1000K] = 1.642 (b) Calculate the Kc for the following reaction. 2 NO(g) + Cl2(g) ⇌ 2 NOCl(g) Kp = 6.5 10-4 at 308 K Kc = Kp /(RT)n gas ; ngas = 2-3 = -1 = 6.5 10-4 / [(0.0821 L atm/mol K) x 308K]-1 = 0.0164 7 Hydrogen and Iodine gas react together to form hydrogen iodide gas. The equation for this reaction is H2(g) + I2(g) ⇌ 2 HI(g) The equilibrium constant for this reaction is 7.1 x 102 at 25 °C. The initial concentration of gases are [H2]0 = 0.81 M [I2]0 = 0.44 M [HI]0 = 0.58 M In whicht direction will the reaction proceed to reach equilibrium? Calculate Q Q Q Q Q = = = = [HI]02/[H2]0·[I2]0 (0.58 M)2/(0.81 M)(0.44 M) 0.34/.35 0.94 Compare Q to K K = 7.1 x 102 Q = 0.94 Q < K, therefore the reaction will shift to the right. The reaction will shift to the right to produce more hydrogen iodide gas to reach equilibrium. 35 SHE1325 8 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS 0.50 moles of N2 gas is mixed with 0.86 moles of O2 gas in a 2.00 L tank at 2000 K. The two gasses react to form nitric oxide gas by the reaction N2(g) + O2(g) ⇌ 2 NO(g). Calculate the equilibrium concentrations of each gas. (Kc for the reaction is 4.1 x 10-4 at 2000 K) Conc (M) I N2(g) 0.50 mol/2.00L =0.25 M -x 0.25 M - x C E O2(g) 2 NO(g) 0.86 mol/2.00 L = 0.43 M -x 0.43 M - x 0 +2x 2x K = [NO]2/[N2][O2] [2x]2/(0.25-x)(0.43-x)= 4.1 x 10-4 assume X is negligible, 0.25-x≈ 0.25 and 0.43-x≈ 0.43 [2x]2/(0.25)(0.43)= 4.1 x 10-4 x = 3.32 x 10-3 Therefore [N2]= 0.25 M, [O2]= 0.43 M, [NO]=2x = 6.64 x 10-3 M 9 Nitrogen dioxide decomposes according to the following reaction. 2 NO2(g) ⇌ 2 NO(g) + O2(g) Kp = 4.48 x 10-13 0.55 atm of NO2 is introduced in a container and allowed to reach equilibrium. Calculate the equilibrium partial pressures of NO, O2 and NO2 gases. Atm I C E 2 NO2(g) 0.55 - 2x 0.55 – 2x 2 NO(g) 0 +2x 2x O2(g) 0 +x x Kp = 4.48 x 10-13 = (P NO)2(P O2) / (PNO2)2 = (2x)2 (x) / (0.55 – 2x)2 36 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Assume x is small since 0.55/4.48 x 10-13 >> 400, then (0.55 – 2x)2 ≈ (0.55)2 4x3 / (0.55)2 = 4.48 x 10-13 ; x = 3.24 x 10-5 P NO = 2x = 2(3.24 x 10-5) = 6.48 x 10-5 atm P O2 = x = 3.24 x 10-5atm P NO2 = 0.55 – 2x = 0.55 atm 10 Consider a reaction at equilibrium. Initially, the concentrations are 4.00 M CH4, 8.00 M H2S, 4.00 M CS2 and 8.00 M H2. At equilibrium, [CS2] = 2.44 M. Calculate the equilibrium concentration of H2. CH4(g) + 2 H2S(g) ⇌ CS2(g) + 4 H2(g) M I C E CH4(g) 4.00 +x 2 H2S(g) 8.00 +2x CS2(g) 4.00 -x 4.00-x = 2.44 4 H2(g) 8.00 -4x 8.00 – 4x x can only have a positive value. Thus, CS2 is a reactant. 4.00 - x = 2.44 ; x = 1.56 M [H2] = 8.00 – 4x = 1.76 M 11 The enthalpy change for the following reaction at standard conditions is +92.2 kJ. 2 NH3(g) ⇌ N2(g) + 3H2(g) Kp = 6.25 at 250C State the effect (direction) of the equilibrium and the amount of nitrogen (increase, decrease or unchange). (a) Adding more NH3(g) Shift to the right , increase N2 (b) Removing some H2(g) Shift to the right, increase N2 (c) Decreasing the temperature of system Shift to the right, decrease N2 (d) Decreasing the volume of the container Pressure increase, therefore Shift to the left, decrease N2 37 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS (e) Adding Krypton gas No effect, N2 remain unchanged (f) Adding catalyst No effect because catalyst will only increase the rate of reaction (g) What is the effect of temperature increase towards the value of Kp? Magnitude Kp increases. TOPIC 4 : ACID-BASE EQUILIBRIA 1 2 (a) Define acid and base, based on Bronsted-Lowry theory. Acid-a substance that can donate a proton to another substance Base-a substance that can accept a proton to another substance (b) Determine whether these species are acids and/ or bases according to Bronsted- Lowry theory: (a) CO32(b) HBr (c) H 2O BL base BL acid BL acid and base HClO, CH3NH2 and H2SO3 are some weak acids and base. For each of them (a) classify as a Bronsted-Lowry acid or base. (b) write the equation for dissociation in water. (c) identify all conjugate acid-base pairs. (d) write the Ka or Kb expression. (a) HClO is a BL acid HClO (aq) + H2O (l) ⇌ ClO- (aq) + H3O+ (aq) Conjugate acid-base pairs: HClO/ClO- and H3O+/H2O Ka = [ClO-][ H3O+]/[ HClO] (b) CH3NH2 is a BL base CH3NH2 (aq) + H2O (l) ⇌ CH3NH3+ (aq) + OH- (aq) 38 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Conjugate acid-base pairs: CH3NH3+/CH3NH2 and H2O/OHKb = [CH3NH3+][OH-]/[CH3NH2] (c) H2SO3 is a BL acid (diprotic) H2SO3 (aq) + H2O (l) ⇌ HSO3- (aq) + H3O+ (aq) 1st dissociation Conjugate acid-base pairs: H2SO3/HSO3- and H3O+/H2O Ka = [HSO3-][H3O+]/[H2SO3] HSO3- (aq) + H2O (l) ⇌ SO32- (aq) + H3O+ (aq) 2nd dissociation Conjugate acid-base pairs: HSO3-/SO32- and H3O+/H2O Ka = [SO32-][H3O+]/[ HSO3-] 3 Explain why hydrochloric acid is a strong acid while HF is a weak acid. HCl is a strong acid because of its complete dissociation/ionization in water, while HF is a weak acid because of its partial dissociation/ionization in water. This is supported by the large Ka value for HCl and small value (6.3 x 10-4) for HF. Ka value shows the extent of acid dissociation whereby the larger the Ka value, the greater the dissociation and the stronger the acid. 4 Complete the following equations and list down all conjugate acid-base pairs that are present : (a) NH3 (aq)+ CH3COOH (aq) ⇌ NH4+ (aq) + CH3COO- (aq) NH4+/NH3 and CH3COOH/CH3COO- (b) H2PO4- (aq) + CO32- (aq) ⇌ HPO42- (aq) + HCO3- (aq) H2PO4-/HPO42- and HCO3-/CO32- 39 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS (c) H2PO4- (aq) + HF (aq) ⇌ H3PO4 (aq) + F- (aq) H3PO4/H2PO4- and HF/F- 5 Determine if the following solutions are acidic or basic. Explain. (a) A solution with [OH-] = 6.3 x 10-12 A solution is basic when [OH-] > 1 x 10-7. Thus, this solution is acidic since [OH-] < 1 x 10-7. (b) A solution with [H3O+] = 2.4 x 10-4 A solution is acidic when [H3O+] > 1 x 10-7. Thus, this solution is acidic since [H3O+] > 1 x 10-7. (c) A solution with pH of 7.1 A solution is basic when pH > 7.0. Thus, this solution is basic. (d) A solution with pOH of 13.5 A solution with pOH = 13.5 has pH = 0.5. This solution is acidic since pH < 7.0 6 Which of the following is more acidic? Explain. (a) A 0.1M acid solution with Ka = 5.5 x 10-9 or 0.1M acid solution with Ka = 3.4 x 10-6 (b) 0.1M acid solution with pKa=7.38 or 0.1M acid solution with pKa=4.56 (c) An acid solution with concentration of 0.2M or 0.4M (d) 0.1M of a weak acid or 0.1M of a strong acid (e) 0.1M of a weak acid or 0.1M of weak base (a) The solution with Ka = 3.4 x 10-6 because it has greater Ka (b) The solution with pKa=4.56 because it has lower pKa 40 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS (greater Ka) (c) The solution with concentration of 0.4M because it is more concentrated (more H3O+ is present) (d) The strong acid because it dissociates completely in water (more H3O+ is present) (e) The weak acid because a solution of weak acid contain more hydronium ion than a solution of weak base 7 Calculate the Ka for the following solutions: (a) 0.05M HA with [A-] = 8.2 x 10-9 M (M) HA H2O A- H3O+ I 0.05 - 0 0 C -x - +x +x E 0.05-x - x x From the table, x = [A-] = [H3O+] = 8.2 x 10-9 Ka = [A-][ H3O+]/[HA] = x2/(0.05-x) = 1.34 x 10-15 (b) 1.6M HF, given that pH of the solution is 1.48. (M) HF H2O F- H3O+ I 1.6 - 0 0 C -x - +x +x E 1.6-x - x x Because pH = 1.48, [H3O+] = x = 10-pH = 3.29 x 10-2 Ka = [F-][ H3O+]/[HF] = x2/(1.6-x) = 6.9 x 10-4 [Theoretical Ka = 6.8 x 10-4] 41 SHE1325 8 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Calculate the [H3O+], [OH-], pH and pOH for the following solutions. For (c) and (d), calculate the pKa and pKb as well. (a) 1M HNO3 (b) 1M LiOH (c) 2.5 M CH3COOH [Ka = 1.8 x 10-5] (d) 5M NH3 [Kb = 1.8 x 10-5] (a) 1M HNO3 makes [H3O+] = 1M ; pH = - log [H3O+] = 0 [OH-] = (1 x 10-14)/1 = 1 x 10-14 ; pOH= - log [OH-] = 14 OR pOH = 14 – pH = 14 ; [OH-] = 10-pOH = 1 x 10-14 (b) 1M LiOH makes [OH-] = 1M ; pOH = - log [OH-] = 0 [H3O+] = 1 x 10-14/1 = 1 x 10-14 ; pH = - log [H3O ] = 14 + OR pH = 14 – pOH = 14 ; [H3O+] = 10-pH = 1 x 10-14 (c) 2.5 M CH3COOH (M) CH3COOH H2O CH3COO- H3O+ I 2.5 - 0 0 C -x - +x +x E 2.5-x - x x Ka = 1.8 x 10-5 = x2/(2.5-x) ≈ x2/2.5 ; can be simplified since 2.5/Ka >>400 Thus, x = 6.71 x 10-3 = [H3O+] ; pH = - log [H3O ] = 2.17 + [OH-] = (1 x 10-14)/6.71 x 10-3 = 1.49 x 10-12 pOH= - log [OH-] = 11.827 42 ; SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS OR pOH = 14 – pH = 11.827 ; [OH-] = 10-pOH = 1.49 x 10-12 pKa = - log (1.8 x 10-5) = 4.74 ; pKb = 14 - pKa = 9.26 (d) 5M NH3 (M) NH3 H2O NH4+ OH- I 5 - 0 0 C -x - +x +x E 5-x - x x Kb = 1.8 x 10-5 = x2/(5-x) ≈ x2/5 ; can be simplified since 5/Ka >>400 Thus, x = 9.49 x 10-3 = [OH-] ; pOH = - log [OH-] = 2.02 [H3O+] = (1 x 10-14)/ 9.49 x 10-3 = 1.05 x 10-12 ; pH= - log [H3O+] = 11.98 OR pH = 14 – pOH = 11.98 [H3O ] = 10 + -pH = 1.05 x 10 pKb = - log (1.8 x 10-5) = 4.74 9 ; -12 ; pKa = 14 - pKa = 9.26 Determine and explain if the following salt is acidic, basic or neutral. Arrange these salts in the order of increasing pH. (a) NH4Br (b) CH3COOK (c) Zn(NO2)2 [Ka (Zn2+) = 1.0 x 10-9 ; Ka (HNO2) = 7.1 x 10-4] (a) [Kb (NH3) = 1.8 x 10-5] Acidic salt. NH4+ is acidic because it is derived from a weak base while Br- is neutral because it is derived from a strong acid. NH4+ will hydrolyze to form H3O+ 43 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS (b) Basic salt. CH3COO- is basic because it is derived from a weak acid while K+ is neutral because it is derived from a strong base. CH3COO- will hydrolyze to form OH- (c) Acidic salt. Ka (Zn2+) > Kb (NO2-) Order of increasing pH: Zn(NO2)2 < NH4Br < CH3COOK 10 For each of the following, write the relevant acid-base equilibrium equation and predict if pH increases or decreases with each change. Explain your prediction. (a) KCN is added into HCN (b) NH4Cl is added into NH3 (c) NaOH is added into LiOH (d) NaCl is added into HCl (a) HCN (aq) + H2O (l) ⇌ CN- (aq) + H3O+ (aq) Addition of CN- will shift equilibrium to the left (reactant side) and produce more HCN which leads to lesser H3O+. Thus, pH will increase as acidity is reduced. (b) NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq) Addition of NH4+ will shift equilibrium to the left (reactant side) and produce more NH3 which leads to lesser OH-. Thus, pH will decrease as basicity is reduced. (c) Both NaOH and LiOH will dissociate completely in water. Thus, addition of NaOH only adds more OH-, increasing the pH of solution. (d) 11 NaCl is a neutral salt. Thus, the addition of NaCl has no effect whatsoever on the pH of solution. Define buffer. A solution of weak acid and its conjugate base or weak base and its conjugate acid that can resist pH changes when a small amount of strong acid or base is added. 44 SHE1325 12 13 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Predict the pH at equivalence point when the following titration is carried out: (e) CH3COOH with LiOH Basic (f) Ca(OH)2 with HClO3 Neutral (g) Sr(OH)2 with CH3COOH Basic (h) HClO4 with NH3 Acidic A 1.0 L buffer solution is made up of 0.5M HCOOH and 0.4M HCOONa. [Ka = 1.8 x 10-4] (a) Calculate the initial pH (b) Calculate the pH if 0.01 mol of KOH is added into the buffer solution (c) Calculate the pH if 0.01 mol of HCl is added into the buffer solution (a) pH = pKa + log [A-]/[HA] = - log (1.8 x 10-4) + log (0.4)/(0.5) = 3.648 (b) 0.01 mol of OH- will react with 0.01 mol of HCOOH to form 0.01mol of HCOO-. pH = pKa + log [A-]/[HA] = - log (1.8 x 10-4) + log (0.4+0.01 )/(0.5-0.01) = 3.667 (c) 0.01 mol of H+ will react with 0.01 mol HCOO- to form 0.01 mol HCOOH pH = pKa + log [A-]/[HA] = - log (1.8 x 10-4) + log (0.4-0.01)/(0.5+0.01) = 3.628 45 SHE1325 14 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Listed below is some acid, base and salt. Acids (Ka) Bases (Kb) Salts HCOOH (1.8 x 10-4) CH3NH2 (4.4 x 10-4) NaF HF (6.8 x 10-4) C6H5NH2 (4.0 x 10-10) CH3NH3Br (a) How many pairs of buffer can you make from this list? (b) Suggest the pH in which each buffer works best. (a) Two. HF and NaF. CH3NH2 and CH3NH3Br (b) Best at pH = pKa. Thus, HF is best at pH = - log Ka = 3.17 Thus, CH3NH2 is best at pH = - log Kw/Kb = 10.64 15 20 mL of 0.30M HF is titrated with 0.10 M KOH. [Ka (HF) = 6.8 x 10-4] (a) Predict whether pH at the equivalence point will be acidic, basic or neutral. (b) Determine a suitable indicator for this reaction. (c) Calculate the initial pH. (d) Calculate the pH after 10 mL of KOH is added (e) Calculate the pH after 60 mL of KOH is added (f) Calculate the pH after 70 mL of KOH is added (g) Sketch the titration pH curve based on your calculations in c, d, e and f. (a) Basic pH due to the presence of F- ion which will hydrolyze to form OH- (b) Phenolphthalein 46 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS (c) Initial pH (M) HF H2O F- H3O+ I 0.30 - 0 0 C -x - +x +x F 0.30 - x - x x 0.30M/Ka >> 400, so the Ka expression can be simplified into Ka = x2/0.30 ; x = [H3O+] = 1.43 x 10-2 ; pH = - log 1.43 x 10-2 = 1.85 (d) 10 mL KOH (mol) HF KOH KF H2O I 0.020Lx0.30M 0.010Lx0.10M 0 - = 6 x 10-3 = 1 x 10-3 C - 1 x 10-3 - 1 x 10-3 + 1 x 10- - 3 F 5 x 10-3 0 1 x 10-3 - From the table, you can see that this is prior to equivalence point. Also, acid and its conjugate base are present. pH = pKa + log [A-]/[HA] = - log (6.8 x 10-4) + log (1 x 10-3)/(5 x 10-3) = 2.47 47 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS (e) 60 mL KOH This is at the equivalence point whereby 6 x 10-3 mol of KOH reacts with 6 x 10-3 mol of HF to form 6 x 10-3 mol of KF. KF will hydrolyze in water as such: (mol) F- H2O HF OH- I 6 x 10-3 - 0 0 C -x - +x +x F 6 x 10-3 - x - x x Kb = Kw/Ka = 1.47 x 10-11 = x2/(6 x 10-3-x) ≈ x2/(6 x 10-3) ; can be simplified since 0.30/Ka >>400 x = [OH-] = 2.97 x 10-7 ; pH = 14 – pOH = 14 – (- log 2.97 x 10-7) = 8.12 (f) 70 mL KOH This is beyond the equivalence point whereby an excess of 0.010 L x 0.10M = 1 x 10-3 mol KOH is present. [OH-] = (1 x 10-3 mol)/(0.020+0.070)L = 0.0111 MH = 14 – pOH = 14 – (- log 0.0111) = 12.05 (g) Sketch of pH-volume curve Titration of HF with KOH 14 12 10 8 pH 6 4 2 0 0 20 40 48 60 Volume of KOH (mL) 80 SHE1325 16 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS 30 mL of 0.15M NH3 is titrated with 0.10 M HBr. [Kb (NH3) = 1.8 x 10-5] (a) Predict whether pH at the equivalence point will be acidic, basic or neutral. (b) Determine a suitable indicator for this reaction. (c) Calculate the initial pH. (d) Calculate the pH after 10 mL of HBr is added (e) Calculate the pH after 45 mL of HBr is added (f) Calculate the pH after 60 mL of HBr is added (g) Sketch the titration pH curve based on your calculations in c, d, e and f. (a) Acidic pH due to the presence of NH4+ ion which will hydrolyze to form H3O+ (b) Methyl red (c) Initial pH (M) NH3 H2O NH4+ OH- I 0.15 - 0 0 C -x - +x +x F 0.15 - x - x x 0.15M/Kb >> 400, so the Kb expression can be simplified into Kb = x2/0.15 ; x = [OH-] = 1.64 x 10-3 ; pH = 14 - pOH = 14 – (-log 1.64 x 10-3) = 11.21 49 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS (d) 10 mL HBr (mol) NH3 HBr NH4+ Br- I 0.030Lx0.15M 0.010Lx0.10M 0 - = 4.5 x 10-3 = 1 x 10-3 C - 1 x 10-3 - 1 x 10-3 + 1 x 10- - 3 F 3.5 x 10-3 0 1 x 10-3 - From the table, you can see that this is prior to equivalence point. Also, base and its conjugate acid are present. pH = pKa + log [A-]/[HA] = - log [(1 x 10-14)/(1.8 x 10-5)] + log [(3.5 x 10-3)/(1 x 10-3)] = 9.80 (e) 45 mL HBr This is at the equivalence point whereby 4.5 x 10-3 mol of HBr reacts with 4.5 x 10-3 mol of NH3 to form 4.5 x 10-3 mol of NH4Br. NH4Br will hydrolyze in water as such: (mol) NH4+ I 4.5 x 10- H2O NH3 H3O+ - 0 0 3 C -x - +x +x F 4.5 x 10-3 - x - x x Ka = Kw/Kb = 5.56 x 10-10 = x2/(4.5 x 10-3-x) ≈ x2/(4.5 x 10-3) ; can be simplified since 0.15/Ka >>400 x = [H3O+] = 1.58 x 10-6 pH = - log 1.58 x 10-6 = 5.80 50 ; SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS (f) 60 mL HBr This is beyond the equivalence point whereby an excess of 0.015 L x 0.10M = 1.5 x 10-3 mol HBr is present. [H3O+] = (1.5 x 10-3 mol)/(0.045+0.060)L = 0.0143 M pH = - log 0.0143 = 1.84 (g) Sketch of pH-volume curve 12 Titration of NH3 with HBr 10 8 pH 6 4 2 0 0 10 20 30 40 50 Volume of HBr (mL) 51 60 70 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS TOPIC 5 : ELECTROCHEMISTRY 1 Balance the following redox equations by using the half-reaction method. (a) Cr2O72- + C2O42- Cr3+ + CO2 [acidic] Cr2O72- + 14 H+ + 3C2O42- 2Cr3+ + 6CO2 +7H2O (b) Mn2+ + H2O2 MnO2 + H2O [basic] Mn2+ + H2O2 +2OH- MnO2 + 2H2O (c) Bi(OH)3 + SnO22- SnO32- + Bi [basic] 2Bi(OH)3 + 3SnO22- 3SnO32- + 2Bi + 3H2O 2 State two differences between voltaic cell and electrolytic cell. Voltaic cell 1. Chemical energy to electrical energy 2. Spontaneous chemical reaction occur 3. Anode – negative terminal Cathode – positive terminal 3 Electrolytic cell 1. Electrical energy to chemical energy 2. Non-spontaneous reaction is forced to occur 3. Anode – positive terminal Cathode – negative terminal Consider the cell notation below ZnZn2+(aq, 1M) Fe3+(aq, 1M), Fe2+(aq, 1M)Pt (a) Write the overall cell reaction. Anode : Zn (s) Zn2+ (aq) + 2ē Cathode : 2Fe3+ E0red = -0.763 (aq) + 2ē 2Fe2+ (aq) E0red = +0.771 Overall: Zn (s) + 2Fe3+ (aq) Zn2+ (aq) + 2Fe2+ (aq) (b) Sketch a cell. Label the anode and cathode, salt bridge and direction of electron flow. 52 SHE1325 (b) INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Calculate the E0cell. E0cell = E0red (cathode) - E0red (anode) = 0.771 – (-0.763) = 1.534 V (c) Calculate the Ecell if [Zn2+]=0.01 M, [Fe3+]=0.01 M and [Fe2+]=1.5 M. Ecell = E0cell– 0.0592 log Q n = 1.534 – 0.0592 log [Zn2+] [Fe2+]2 2 [Fe3+]2 = 1.534 - 0.0592 log (0.01) (1.5)2 2 (0.01)2 = 1.464 V (d) State two purpose of salt bridge. What might it contains? - (i) It prevents the electrolytes in two half-cells from mixing (ii) It maintains ionic balance in the cell Inert electrolyte eg. KNO3,KCl 53 SHE1325 4 (a) INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Define the term Standard Hydrogen Electrode (SHE). Standard electrode potential of an element is defined as the potential difference between SHE half-cell and a half-cell of the element in a solution of its ions at 1M, 298 K and 1 atm. (b) 5 Draw labeled diagrams that could be used to measure the standard electrode potential of Cu. Given a selected standard reduction potentials (298 K) V2+ (aq) + 2 e → V (s) Pb2+ (aq) + 2 e → Pb (s) l2 (s) + 2 e → 2 l- (aq) Cl2 (g) + 2 e → 2 Cl- (aq) (a) E° E° E° E° = = = = -1.18V -0.13V +0.54V +1.36V Determine: (i) (ii) (iii) (iv) (v) (vi) the strongest reducing agent V (s) the ion/s or element/s could be reduced by Pb Cl2 (g), I2 (g) the ion/s or element/s could be oxidized by Pb2+ V (s) the ion/s or element/s could be reduced by V Cl2 (g), I2 (g), Pb2+ (aq) the ion/s or element/s could be oxidized by V2+ none Arrange the appropriate species in the order of increasing oxidizing strength V2+< Pb2+< I2 < Cl2 54 SHE1325 6 7 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Calculate whether the following reactions would occur spontaneously in aqueous solution at 25°C. Assume that the initial concentrations of dissolved species are 1.0 M. (a) Ca(s) + Cd2+ (aq) Ca2+(aq) + Cd(s) E° = E°red (cathode) - E°red (anode) = -0.40-(-2.76) = 2.36 V (spontaneous) (b) 2Br- (aq) + Sn2+(aq) Br2(l) + Sn(s) E° = E°red (cathode) - E°red (anode) = -0.14-1.07 = -1.21 V (nonspontaneous) (a) Calculate the standard potential of a cell consisting of Zn/Zn2+ half cell and the Standard Hydrogen Electrode. Zn (s) + 2H+ Zn2++ H2 (g) E° = E°red (cathode) – E°red (anode) =0.00 V – (-0.76 V) = 0.76V (b) What is the cell potential (Ecell) if the concentration of Zn2+ is 0.45 M, pressure of H2 is 2.0 atm and the concentration of H+ is 1.8 M. E = E° - (0.0592 / n) log Q = 0.76 V – (0.0592/2) log [(0.45 x 2.0)/(1.82) = 0.78 V 8 Consider the following galvanic cell set up at 298 K: Al(s) l Al3+(aq) ll Ni2+(aq)l Ni(s) (a) Calculate Eocell. Eocell= -0.25V – (-1.66V) = +1.41V 55 SHE1325 (b) INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Calculate Ecell when water is added into Ni2+(aq)l Ni(s) half cell such as Ni2+(aq) decreasing ten times. Explain how does cell potential change. 𝟎.𝟎𝟓𝟗𝟐 𝟏 ) 𝐥𝐨𝐠 𝟏 𝟔 ( )𝟑 Ecell=1.41-( =+1.38V. Cell potential decreases. 𝟏𝟎 9 (a) Sketch a labeled diagram of electrolysis of diluted NaCl solution using inert electrode. (b) Write the reactions at anode and cathode and the observations. Anode Reaction Observation Cathode Reaction Observation : : 2H2O (l) O2 (g) + 4H+ + 4eBubbles of gas evolved. : : 2H2O (l) + 2e- H2 (g) + 2OHBubbles of gas evolved 56 SHE1325 (c) 10 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Write the reactions at anode and cathode when concentrated NaCl is used as electrolyte. Explain. Anode Reaction Observation : : 2Cl-(aq) Cl2(g) + 2eBubbles of yellow-green gas evolved. Cathode Reaction Observation : : 2H2O (l) + 2e- H2 (g) + 2OHBubbles of gas evolved Plastic object can be chromium-plated by coating them with a thin layer of graphite paste first and then placing them in a bath of aqueous chromium (III) sulphate and electroplating. (a) Draw a fully labeled diagram of the electrical circuit for the electroplating. Plastic object coated with graphite (b) (a) Cr2(SO4)3(aq) Write the half reactions at both the electrodes. Anode Cathode 11 chromium or graphite : Cr(s) Cr3+(aq) + 3e: Cr3+(aq) + 3e- Cr(s) Write all the possible half reactions occur at the anode and cathode when the following substances are electrolysed using inert electrodes:(i) AgNO3(aq) Anode: NO3- cannot be oxidized. : 2H2O(l) O2(g) + 4H+(aq) + 4eCathode: Ag+(aq) + e- Ag(l) : 2H2O + 2e- H2(g) + 2OH-(aq) 57 SHE1325 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS (ii) Molten NaBr Anode: 2Br-(l) Br2(g) + 2eCathode: Na+(l) + e- Na(l) (iii) K2SO4(aq) Anode: SO42- cannot be oxidized. : 2H2O(l) O2(g) + 4H+(aq) + 4eCathode: K+(aq) + e- K(l) : 2H2O + 2e- H2(g) + 2OH-(aq) (iv) Concentrated MgCl2(aq) Anode: Cl-(aq) Cl2(g) + 2e: 2H2O(l) O2(g) + 4H+(aq) + 4eCathode: Mg2+(aq) + 2e- Mg(s) : 2H2O + 2e- H2(g) + 2OH-(aq). (b) What are the products formed at the anode and cathode in(a)(i)-(iv)? Give your reasoning. (i) Anode: oxygen gas because only water can undergo oxidation. No competition Cathode: silver molten because of its more positive reduction potential (ii) Anode: bromine gas Cathode: sodium molten Only sodium ion and bromide ion are present. (iii) Anode: oxygen gas because only water can undergo oxidation. No competition. Cathode: hydrogen gas because water has a more positive reduction potential. (iv) Anode: chlorine gas because of its high concentration. The formation of oxygen needs an ‘overvoltage’. Cathode: hydrogen gas because Mg has less positive reduction potential. 58 SHE1325 12 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS (a) Sketch a labeled diagram of electrolysis of purification of copper. (b) Describe the industrial processes of purification of copper. The impure copper from a smelter is cast into a block to form the positive anode. The cathode is made of previously purified copper. These are dipped into an electrolyte of copper(II) sulphate solution. When the d.c electrical current is passed through the solution electrolysis takes place. The copper anode dissolves forming blue copper(II) ions Cu2+. These positive ions are attracted to the negative cathode and become copper atoms. The mass of copper dissolving at the anode exactly equals the mass of copper deposited on the cathode. The concentration of the copper(II) sulphate remains constant. Any impurities present in the impure copper anode fall to the bottom of the electrolysis cell tank. This 'anode sludge' is not completely mineral waste, it can contain valuable metals such as silver. 59 SHE1325 13 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS Consider the electrolysis of molten barium chloride. (a) Write the half- reactions. Anode : 2Cl-(aq) Cl2(g) + 2eCathode : Ba2+ (aq) +2e- Ba (s) (b) 14 How many grams of barium metal can be produced by supplying 0.50 A for 30 min? Q = It = 0.50 x 30 x 60 = 900 C Mass of I2 = 900 C x 1 mol e/96500 C x 1 mol Ba/2 mol e x 137.3 g/1 mol Ba = 0.62 g When an aqueous solution of KI is electrolyzed, the following halfreaction occurs. Anode : 2I-(aq) I2(aq) + 2e Cathode : 2H2O (l) + 2e H2 (g) + 2OHIn the electrolysis, a current of 8.52 x 10-3 A flows through the cell for 2 hours. Calculate (a) The amount of charge required to produce 1 mole of I2. Q = It = 8.52 x 10-3 x 2 x 60 x60 = 61.344 C (b) Calculate the mass of iodine and hydrogen produced. Mass of I2 = 61.344 C x 1 mol e/96500 C x 1 mol I2/2 mol e x253.8g/1 mol I2 = 0.079 g Mass of H2 61.344 C x 1 mol e/96500 C x 1 mol I2/2 mol e x 2.016g/ 1 mol H2 = 6.408 x 10-4 g 60 SHE1325 15 INTRODUCTION TO ORGANIC CHEMISTRY, HYDROCARBON AND HALOGEN COMPOUNDS A constant current deposits 365 mg of Ag in 216 min from aqueous silver nitrate solution. Calculate the current used. Q =365 mg Ag x 1 g Ag/1000 mg Ag x 107.9 g Ag/ 1 g Ag x 1 mol e/1 mol Ag x 96500 C/1mol e = 326.4 C A = Q/t = 326.4/12960 = 0.0252 A. 61