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SCH4U- Atomic Structure Review 1) Describe the contributions of the following scientists to our current understanding of atomic structure and the quantum theory: a) Bohr: energy levels b) DeBroglie: wave properties of matter c) Heisenberg: uncertainty principle d) Schrodinger: wave equation for orbitals e) Pauli: magnetic spin property 2) Based on the quantum mechanical model of the atom: a) What is an "orbital"'? A region in which an electron with a certain set of quantum numbers has a certain probability of being found. b) What is the maximum electron carrying capacity of the following energy levels: n=2, n=5 n=2: 8 n=5: 50 c) What orbital sub-levels would be found in the following energy levels: n=2, n=5 n=2: s, p (l=0,1) n=5: s,p,d,f,g (l=0,1,2,3,4) d) How would the number of electrons determined in 2(b) be distributed amongst the orbital sub-levels in 2(c) (i.e.: number of e- per sub level) n=2: 2 in s, 6 in p n=5: 2 in s, 6 in p, 10 in d, 14 in f, 18 in g 3) a) Write the ground state electron configuration of a neutral atom of nitrogen. 1s2 2s2 2p3 2p 2s 1s | || || | | b) Assign quantum numbers to each or the electrons in your nitrogen atom. n m ml ms 2p 2 1 1 +.5 2p 2 1 0 +.5 2p 2 1 -1 +.5 2s 2 0 0 -.5 2s 2 0 0 +.5 1s 1 0 0 -.5 1s 1 0 0 +.5 SCH4U- Bonding and Molecular Shapes Review 1) Use Lewis diagrams to draw the following molecules or ions. Once completed, use the VSEPR theory to predict the shape of each. For each molecule, decide whether it is polar or non-polar. a) PCl3 b) SiF4 c) CO32d) H2S e) PO43trig pyram tetrahed planar trig bent tetrahed 3 single 4 single 1 double 1 single 4 single 1 lone pair 2 single 2 lone pair (1 co-ord) polar non-polar non-polar non-polar non-polar charged charged f) O3 bent 1 single 1 double 1 lone pair non-polar (1 co-ord) g) CO straight 2 double non-polar h) NO31planar trig 1 double 1 single (1 co-ord) non-polar charged 2) Methane and fluoromethane are two very similar molecules. Despite their similarities, there is an 86/C difference between their boiling points. Which of these two substances would have the higher boiling point? Explain your answer. CH3F, has a higher boiling point because it is polar and has greater London forces. 3) The element iodine exists as solid crystals composed of I2 molecules. A chemist wishing to dissolve iodine has a choice of two solvents; water and tetrachloromethane. Which of these two solvents would be the best choice? Explain your answer. Tetrachloromethane, since I2 is non-polar and so is CCl4. 4) Consider a molecule of ethanal. a) What type of orbital hybridization is each atom in the molecule using for bonding? #1-carbon is 2sp2, 2p #2-carbon is 2sp3 oxygen is probably not hybridized, but cannot know because it has only one sigma bond. b) What geometric arrangement would each of these orbitals take? #1-carbon is trigonal planar #2-carbon is tetrahedral c) Describe the orbital overlap that accounts for each bond and specify the type of bond formed (sigma, pi) #1-carbon has a sigma bond to carbon 2, a sigma bond to oxygen, and a pi bond to oxygen d) What are the bond angles around each carbon atom in this molecule? #1-carbon: 120o #2-carbon: 109.5o 5) Consider the data in the following table for substances labeled 'A" to “G" ___________________________________________________________________________________________ Substance Melting point (/C) Boiling point (/C) Solubility in H2O Electrical Conductivity as; solid/liquid/aq. sol. A (CH3OH) mol -94 65 SOLUBLE NO/NO/NO B (quartz) netw 1610 2230 INSOLUBLE NO/NO/NA C (S) mol 119 445 INSOLUBLE NO/NO/NA D (Al) metal 660 2467 INSOLUBLE YES/YES/NA E (KCl) ionic 770 1500 Slightly SOLUBLE NO/YES/slightly F (CH4) mol. -182 -164 INSOLUBLE NO/NO/NA G (CaO) ionic 2614 2850 SOLUBLE NO/YES/YE5 ____________________________________________________________________________________________ Classify each substance as to the type of solid it represents: molecular, metallic, ionic, network Match the letter for each sample to one of the following substances (one letter per substance) and explain your reasoning: quartz, CaO, Al, methanol, Sulfur, methane, KCl SCH4U- Organic Chemistry Review 1) Draw the structural formula for the following compounds: a) N,N-dimethyl-2-pentamine b) 3-ethylcyclohexane should be ethylcyclohexane c) methylpropanal d) 2-chloro-3-methyl-2-hexene e) 2-ethoxypropane f) propylbenzoate g) N-ethylpropanamide ethyl propanamide h) 2,3-difluorobutanoic acid 2) Name the compounds pictured here: a) 2,5-dimethyl-2-heptene b) 1-ethyl-3-methylbenzene c) N-ethyl-N-methylaminopropane d) 2-chloro-3-methylpentanoic acid f) 2,3-dimethylcyclopentane g) 2-methyl-3-heptyne h) 4-methyl-2-hexanone i) N-methylethyl propanamide j) ethoxypropane k) ethyl butanoate 3) Complete the following chemical reactions by naming and drawing the molecular formula the organic product(s). Assume all required conditions are available for each reaction, a) 3-methyl-1-butyne + 2Cl2 ---> 1,1,2,2-tetrachlorobutane b) methane + excess O2 ---> CO2 + 2H2O c) 2-methyl-3-pentanol + KMnO4 ---> 2-methyl-3-pentanone d) methylpropene + HBr ---> 2-bromomethylpropane e) 2-methylbutanal + KMnO4 ---> 2-methylbutanoic acid f) cyclopentene + H2O (w/ H2SO4) ---> 1-hydroxycyclopentane g) dimethyl-l,3-propandiol + KMnO4 ---> dimethylpropanedioic acid h) ethanoic acid + phenol ---> phenyl ethanoate 4) Consider the following organic compounds: All exhibit London Dispersion Forces (LDF) Highest Boiling Point proponoic acid LDF, dipole-dipole, H-bonding soluble 2-propanol LDF, dipole-dipole, H-bonding soluble propanone LDF, dipole-dipole soluble 2-chloropropane LDF and weak dipole-dipole (delta EN is 0.5) slightly soluble propane LDF only insoluble Lowest Boiling Point a) For each compound, identify the type of intermolecular forces acting between molecules. b) Rate the solubility of each compound in water as: insoluble, slightly soluble or very soluble c) Arrange the compounds in order of increasing boiling point. Generally speaking, order of increasing boiling point: hydrocarbons, ethers, alkyl halides, aldehydes, ketones, alcohols, carboxylic acids the more polar functional groups, the higher the boiling point and solubility the greater the molar mass, the higher the boiling point the more central the functional group, the greater the solubility and the lower boiling point (all other things equal). SCH4U- Energy and Rates Review 1) When 13.4g of ammonium chloride dissolve into 2.00x102 g of water the temperature changes from 20.0/C to 15.3/C. Determine the molar enthalpy of solution of ammonium chloride and write a balanced thermochemical equation for the reaction. Sketch an enthalpy diagram for this reaction. (Ans: +16.7kJ/mol) molar mass of NH4Cl mN = 14.01+(3x1.01)+35.45 = 52.49 g/mol moles of NH4Cl n = m/mN = 13.4 g / 52.39 g/mol = 0.256 mol enthalpy change for this process = msurrCp)T qsurr = 2.00x102g x 4.18 J/oC x (15.3 oC - 20.0 oC) = -3929.2 J = -3.93 kJ )H = qsys = -qsurr = 3.93 kJ )Hrxn = )H/n = 3.93 kJ / 0.256 mol = 15.36 kJ/mol 2) The following chemical reaction occurs in a car battery: Pb(s) + PbO2(s) + 2H2SO4(aq) –> 2PbSO4(s) + 2H2O(l) Given the following enthalpy of formation values, calculate the enthalpy change for the above reaction as written (also known as the enthalpy of reaction) PbO2(s) H2SO4(aq) PbSO4(s) H2O(l) )H o f -277.0 kJ/mol -909.3 kJ/mol -902.1 kJ/mol -285.5 kJ/mol )H = E (n x )Hof)prod - E (n x )Hof)react = ((2 x -902.1) + (2 x -285.5)) - ((1 x 0)+(1 x -277.0)+(2 x -909.3)) = (-1804.2 + -571.0) - (0 + -277.0 + -1818.6) = -2375.2 - -2095.6 = -279.6 kJ b) If it takes 39.5 kJ to start your car, what mass of lead (II) sulphate is produced when starting the car using the battery reaction described in 'a'? Since there are two moles of lead (II) sulphate produced in this reaction, the molar enthalpy for this reaction with respect to Lead (II) sulphate is: )Hrxn = )H/n = -279.6 kJ / 2 mol = -139.8 kJ/mol To calculate the number of moles of Lead (II) sulfate produced for a specific enthalpy change, use the equation: )H = n )Hrxn n = )H/)Hrxn = -39.5 kJ / -139.8 kJ/mol (-39.5 kJ since the battery needs to release -39.5 kJ to the car’s starter motor) = 0.282 mol molar mass of Lead (II) sulfate mN = 207.20 + 32.06 + (4 x 16.00) = 303.26 g/mol mass of Lead (II) sulfate produced m = n x mN = 0.282 mol x 303.26 g/mol = 85.6 g 3) The combustion of ethane is illustrated by the following equation: 2C2H6(g) + 7O2(g) –> 4CO2(g) + 6H2O(g) a) Show how the following equations can be combined using Hess’ Law to determine the enthalpy of combustion of Ethane. (Ans: -3028 kJ) (HS says -2973.8 kJ) )H1 = -1409 kJ 1) C2H4(g) + 3O2(g) –> 2CO2(g) + 2H2O(g) )H2 = -136.7 kJ 2) C2H4(g) + H2(g) –> C2H6(g) 3) H2(g) + ½O2(g) –> H2O(g) )H3 = -214.6 kJ 2 x 1) = 4) 2C2H4(g) + 6O2(g) –> 4CO2(g) + 4H2O(g) -2 x 2) = 5) 2C2H6(g) –> 2C2H4(g) + 2H2(g) 2 x 3) = 6) 2H2(g) + O2(g) –> 2H2O(g) 4) + 5) + 6) = )H4 = -2818 kJ )H5 = +273.4 kJ )H6 = -429.2 kJ 2C2H4(g) + 6O2(g) + 2C2H6(g) + 2H2(g) + O2(g) –> 4CO2(g) + 4H2O(g) + 2C2H4(g) + 2H2(g) + 2H2O(g) 2C2H6(g) + 7O2(g) –> 4CO2(g) + 6H2O(g) )H = )H4 +)H5 +)H6 = -2818 + 273.4 - 429.2 =-2974 kJ b) If you wish to burn ethane as fuel to heat 550.0 g of water from 12.5 /C to 100.0 /C, how many moles of ethane will be require? (Ans: 0.135 moles) Given that the )H from the equation above is for 2 moles of ethane, the molar enthalpy of combustion for ethane is: )Hrxn = )H/n = -2974 kJ / 2 mol = -1487 kJ/mol To heat the water (we will call it the surroundings), the energy needed is calculated by the calorimetry equation qsurr = m Cp )T = 550.0 g x 4.18 J/g oC x (100 oC - 12.5 oC) = 201163 J = 201 kJ )H = -qsurr = -201 kJ To calculate moles of ethane required, use the equation: n = )H / )Hrxn ()Hrxn = )H/n rearranged) = -201 kJ / -1487 kJ/mol = 0.135 mol 4) From your text on pages. 414-416 answer 13: a) rate doubles, b) rate drops by half, c) no change (reactants are not gases). 14: a) a series of steps, since there are 6 moles of reactants, see back of text for b) 15: a) first, b) first, c) second, d) r=k[HI][O2], e) k = r/([HI][O2]) = 0.0042/(0.01x0.01) = 42, f) 2 g) the coefficients do not match the exponents in the rate law expression 16: a) 2NO(g) + 2H2(g) —> N2(g) + 2H2O(g) b) Anything that is not in the overall balanced equation. c) r = k[NO(g)]2 18: Consider the physical states and past experience with precipitation reactions; reaction ‘a’ seems likely. SCH4U- Introduction to Equilibrium page 457 1: a) left, b) right, c) no change, d) left, e) right 2: a) F2 line spikes up followed by F2 and Br2 lines sloping down and the BrF line sloping up b) HOCl spikes down followed by sloping up while the other two lines slop down In both cases, all the lines become horizontal again. 3: a) It would speed it up. b) No effect. 4: the equilibrium reaction is exothermic as written and will shift right is temperature is decreased. page 465 1: N2O4 <––> 2NO2 K=[NO2]2/[N2O4] = 0.87 2 a) Q = 2.15 /5.30 = 0.87 Q=K, thus the system is at equilibrium b) Q = 1.552/0.80 = 3.0 Q<K, thus the equilibrium will shift right 2: Q = 1.5 x 3.2 / 2.0 = 2.4 Q<K, thus it is not at equilibrium page 481 1: N2 + 3H2 —> 2NH3; Q = (2.2x10 -4)2 / ((4.0x10 -3)3 (2.0x10 -4)); compare Q to K SCH4U -Solubility Equilibrium Review 1) Consider the following salts: AgI, AgNO3, AgCl, Ag2CO3, Ag2CrO4 a) List these salts in order of increasing solubility. If X=solubility (mol/L) then for a salt with the formula AnBm Ksp = (nX)n(mX)m The trick is to know when a subscript is part of a polyatomic ion (has no effect on the Ksp expression) and when the subscript is telling you how many ions are in the formula. This is important because two salts with the same Ksp can have very different solubilities if the exponents in the Ksp expressions are different. Salt AgI AgCl Ag2CO3 Ag2CrO4 AgNO3 Ksp 8.5x10 -17 1.8x10 -10 8.5x10 -12 1.1x10 -11 very large Solubility Product Expression Ksp = x2 Ksp = x2 Ksp = (2x)2 x = 4x3 Ksp = (2x)2 x = 4x3 Ksp = x2 x (Ksp)½ = 9.2x10 -9 <— lowest solubility (Ksp)½ = 1.3x10 -5 (Ksp/4)1/3 = 2.0x10 -4 (Ksp/4)1/3 = 2.2x10 -4 (Ksp)½ = large b)Write the chemical equilibrium equation for the formation of an aqueous solution of the least soluble silver salt. AgI(s) <—> Ag+(aq) + I -(aq) Ksp = [Ag+(aq)][I -(aq)] c) Determine the solubility in mol/L and in g/L for the least soluble silver salt in pure water. (Ans: 9.2X10-9 mol/L and 2.2X10-6 g/L) molar mass of AgI is mN = 107.87 g/mol + 126.90 g/mol = 234.77 g/mol molar solubility = 9.2x10 -9 mol/L mass solubility = CN x mN = 9.2x10 -9 mol/L x 234.77 g/mol = 2.2x10-6 g/L d)Determine the solubility in mol/L for the least soluble silver salt in pure water in a 0.025 M solution containing a sodium salt with the same negative ion. (Ans: 3.5X10-15 mol/L) Ksp = (x)(x+0.025) = 0.025x on the understanding that x is really small compared to 0.025 x = Ksp / 0.025 = 3.4x10 -15 mol/L 2) You are given a 100.0 mL sample of 0.0010 M silver nitrate and 100.0 mL of solution made by dissolving 6.4X10-4 g of sodium carbonate (not sodium bicarbonate) in enough water to make 100.0 mL of solution. If the two solutions are mixed at 25/C, will a precipitate of silver carbonate form? (Ans; Q = 8.0x10-12 ) Sodium carbonate solution molar mass of sodium carbonate (Na2CO3) mN = (2x22.99) + 12.01 + (3x16.00) = 105.99 g/mol Cm = 6.4x10 -4 g / 0.1000 L = 6.4x10 -3 g/L Cn = Cm / mN = 6.4x10 -3 g/L / 105.99 g/mol = 5.9x10 -5 M Silver nitrate solution Cn = 0.0010 M These two concentrations are premix concentrations. We need post-mix, or initial, concentrations (which will be half premix, since the volume is doubled on mixing). Initial concentrations are Ci = (Cpm x Vpm)/Vi AgNO3 0.0010 x 0.1 L / 0.2 L 0.0005 Na2CO3 5.9x10 -5 x 0.1 L / 0.2 L 2.95x10 -5 Q = [Ag+]2[CO32-] = (0.0005)2 x 2.95x10 -5 = 7.4 x 10 -12 Ksp = 8.5x10 -12 Q<Ksp, therefore no precipitate occurs SCH4U - Acid/Base Equilibrium Review 1) Determine the pH of the following: (Ans: a: pH 12.4 a) 0.025M KOH solution. b) 0.075M HNO3 solution b: pH 1.1) Both substances are strong electrolytes. a) [OH-] = [KOH] = 0.025 M; pOH = -log[OH-] = -log(0.025) = 1.6 pH = (14 - pOH) = 14 - 1.6 = 12.4 a) [H+] = [HNO3] = 0.075 M; pH = -log[H+] = -log(0.075) = 1.1 2) Nicotinic acid is a weak monoprotic organic add with the formula C6H4NCOOH. Question: what sort of acid to you think nicotinic acid is? If you said "carboxylic acid," give your self a gold star! a) Write the equilibrium equation for nicotinic acid in water. Ka = [H+][A-]/[HA] (A- being C6H4NCOO-) b) If a 0.050M solution of nicotinic acid has a pH of 3.1, determine the Ka of this acid. (Ans: 1.3x10-5) HA <---> H+ + AI 0.050 0 0 Ka = [H+][A-]/[HA] C -x +x +x = x2 / (0.050 - x) E 0.050 - x x x = (7.94 x 10-4)2 / (0.050 - 7.94 x 10-4) = 1.28x10 -5 + -pH -3.1 -4 But, we know that [H ]eq = 10 = 10 = 7.94 x 10 3) Consider the following salts: KNO3, LiC6H5COO, N2H5Cl, LiCN, NH4Cl a) Group the salts in to the categories: acidic, basic, neutral. acidic: (conjugates of a weak base present) N2H5Cl, NH4Cl basic: (conjugates of a weak acid present) LiC6H5COO, LiCN neutral: (no weak conjugates present) KNO3 b) Write an equilibrium Ka or Kb expression for all the non-neutral salts in the list. N2H5+(aq) + Cl -(aq) + H20(l) <---> N2H4(aq) + H30+(aq) + Cl -(aq) Ka = [N2H4(aq)][H30+(aq)]/[N2H5+(aq)] NH4+(aq) + Cl -(aq) + H20(l) <---> NH3(aq) + H30+(aq) + Cl -(aq) Ka = [NH3(aq)][H30+(aq)]/[NH4+(aq)] Li+(aq) + C6H5COO -(aq) + H20(l) <---> Li+(aq) + C6H5COOH (aq) + OH -(aq) Kb = [C6H5COOH (aq)][0H -(aq)]/[C6H5COO -(aq)] Li+(aq) + CN -(aq) + H20(l) <---> Li+(aq) + HCN (aq) + OH -(aq) Kb = [HCN(aq)][0H -(aq)]/[CN -(aq)] 4) What volume of 0.0150M HNO3, is needed to titrate 30.0 mL of 0.0250M NaOH to the equivalence point? Since both the titrant and base are “monoprotic”, the number of moles of titrant (nt) must equal the moles of base (ns) ns = CsVs = 0.0250 M x 0.0300L = 0.000750 mol nt = ns = 0.000750 mol Vt = nt/Ct = 0.000750 mol / 0.0150 M = 0.050 L = 50.0 mL 5) Hydrosulphuric acid (H2S) is a weak acid with a Ka of 9.1x10-9 H2S + H2O --> H3O+ + HSa) Determine the pH af a 0.075M solution of this acid. (Ans: pH 4.6) I 0.075 0 0 C -x +x +x E 0.075-x x x Ka = [H3O+][HS -] / [H2S] = x2 / (0.075-x) = x2 / 0.075 x = (Ka x 0.075)1/2 = (9.1x10 -9 x 0.075)1/2 = 2.6x10-5 M pH = -log(x) = 4.6 (0.075/Ka >> 100) b) What is the percent dissociation of this acid? (Ans: 0.035%) %dissoc. = 100% x [H+]eq/[H2S]i = 100% x 2.6x10-5 / 0.075 = 0.035% 6) Hydrocyanic acid is a weak acid with a Ka of 4.9x10-10 HCN + H2O --> H3O+ + CNa) What is the Kb of the cyanide ion? (Ans: Kb= 2.0x10-5) KaKb=10 -14 Kb = 10 -14 / 4.9x10-10 = 2.0x10 -5 b) Determine the pH of a 0.15M NaCN solution. (Ans: pH 11.2) CN - + H2O ---> HCN + OH Kb = x2/(0.15-x) = x2/0.15 (0.15/Kb >>100) I 0.15 0 0 x = (2.0x10 -5 x 0.15)½ = 0.0017 C -x +x +x pOH = -log(0.0017) = 2.8 E 0.15-x x x pH = 14 - pOH = 11.2