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LIMITING REACTANT • The limiting reactant is used up first • The limiting reactant determines the amount of product • Need balanced equation to proceed Limiting Reagents 2H2 + O2 10 H2 and 7 O2 Æ 2H2O 10 H2O and 2 O2 reaction stops when one of the reactants is depleted What if you had only 220.0 g Pb(NO3)2? How much PbI2 would precipitate? PERCENT YIELD actual yield Percent x 100% = Yield theoretical yield Calculation is just one more step beyond a standard stoichiometry calculation PERCENT YIELD EXAMPLE Silicon carbide (SiC) is made from sand (silicon dioxide, SiO2) and carbon at high T. CO is also formed. If 100.0 kg of sand are reacted and 55.0 kg SiC are formed, what is the percent yield? SiO2(s) + 3 C(s) → SiC(s) + 2 CO(g) Convert kg SiO2 to moles: 1 mol 100.0 kg x 103 g/kg x = 1664 mol SiO2 60.09 g moles SiO2 = moles SiC Convert moles SiC to kg: 1664 mol x 40.10 g/mol x 10-3 = 66.73 kg SiC % yield = 55.0 kg x 100% = 82.4% 66.73 kg TITRATION • Goal: find conc. of unknown • React solution of known conc. (std. soln.) with soln. of unknown conc. • Find equivalence point (vol.) add std. soln. soln. of unknown conc. Find volume at equivalence point (need way to signal the equiv. point such as indicator) ACIDS-BASE NEUTRALIZATION ACID + BASE → SALT + WATER HNO3 + KOH → KNO3 + H2O complete ionic equation H+(aq) + NO3–(aq) + K+(aq) + OH–(aq) → H2O(l) + K+(aq) + NO3–(aq) eliminate spectator ions net ionic equation H+(aq) + OH–(aq) → H2O(l) Strong Acids and Bases Table 4.2 Strong Acids Hydrochloric, Hydrobromic, Hydroiodic, Chloric, Perchloric, Nitric, Sulfuric, HCl HBr HI HClO3 HClO4 HNO3 HSO4 Strong Bases Group 1A metal hydroxides (LiOH, NaOH, KOH, RbOH, CsOH) Heavy Group IIA metal hydroxides (Ca(OH)2, Sr(OH)2, Ba(OH)2) TITRATION EXAMPLE A flask contains an unknown amount of HCl. This solution is titrated with 0.101 M NaOH. It takes 23.35 mL of NaOH to complete the reaction. How many grams of HCl were there? HCl + NaOH → NaCl + H2O (1) Find moles of NaOH used (23.35 x 10–3 L NaOH)(0.101 mol/L) = 2.36 x 10–3 mol (2) Find moles of HCl……same as NaOH (3) Find g of HCl (2.36 x 10–3 mol)(36.5 g/mol) = 86.1 x 10–3 g 86.1 mg HCl If we knew the volume of HCl soln., we could calc. the M of the HCl soln. TITRATING H2SO4 with NaOH H2SO4(aq) + 2 NaOH(aq) Æ Na2SO4(aq) + 2 H2O(l) net ionic equation 2 H+(aq) + 2 OH– (aq) → 2 H2O(l) Begin with 20 mL of 0.1 M H2SO4(aq), titrate with 0.1 M NaOH(aq) As NaOH(aq) is added, H+(aq) ions are used to make H2O(l). When enough NaOH is added, all of the H+ ions are consumed and the indicator turns pink. What volume of 0.1 M NaOH is required? What is final volume of solution? What is final conc. Of H+(aq) and OH–(aq)? What is final conc. of Na+(aq) and SO42–(aq) STOICHIOMETRY When chemical reactions involve gases, the balanced equation provides the number of moles of reactants and products. The ideal gas equation provides the link between number of moles and P, V, T of gases. How much gas is produced or consumed by a chemical reaction? find n and then V or P Air Bag 2 NaN3(s) → 2 Na + 3 N2(g) sodium azide (40 msec) gas in air bag How many L of N2 at 735 mm Hg and 26 °C are produced from 125 g NaN3? Moles NaN3 125 g 65.01 g/mol = 1.92 mol Moles N2 (1.92 mol NaN3)(3/2) = 2.88 mol N2 V= (2.88)(0.0821)(299) 0.967 = 73.1 L 29 NaN3 + 4 Fe2O3 + NaNO3 → 15 Na2O + 8 Fe + 44 N2 2/3 = 0.67 29/44 = 0.66 IDEAL GAS LAW EXAMPLE Calculate pressure change in cylinder of a car’s engine when 0.250 g of octane [C8H18(l)] is burned with a stoichiometric amount of O2 assuming complete combustion. Cylinder volume = 0.100 L Initial Temp = 80 oC; final Temp = 700 oC ENERGY KINETIC ENERGY • mechanical (moving mass ½ mv2) [ joule = kg m2/s2] • electrical (moving charge) • light (photons) • sound (molecules moving uniformly) • heat (molecules moving randomly) POTENTIAL ENERGY • mechanical (mass in a place where force can act) • chemical (bonds) • nuclear (binding energy) ENERGY CONVERSION Energy can be converted from one form to another When it is converted, the total energy remains constant Law of Conservation of Energy First Law of Thermodynamics All energy lost by a system under observation is gained by the surroundings During energy conversion, some heat is always produced First Law of Thermodynamics Conservation of Energy The energy of the universe is constant. ΔEuniverse = ΔEsystem + ΔEsurroundings = 0 Energy is neither created nor destroyed, only converted from one form to another. ΔEsystem = q + w q is heat gained or lost by the system w is work done by or on the system ENERGY CHANGES ΔE = Efinal – Einitial ΔE is a state function State function: a function whose value does not depend on pathway used to get to present state. System energy Surroundings Surroundings ΔE is – energy System ΔE is + ΔE = q + w work (work done to system +) heat (heat added to system +) ENTHALPY When changes occur at constant pressure ΔE = qp + wexpansion this is negligible ΔH = qp ΔH is the enthalpy change ΔH is the quantity of thermal energy transferred into a system at constant pressure Like energy, enthalpy is a state function ENTHALPY OF REACTION ΔH = H(products) – H(reactants) Endothermic ΔH > 0 Exothermic ΔH < 0 2 H2(g)+ O2(g) → 2 H2O(g) + heat ΔH = – 483.6 kJ Characteristics of Enthalpy (1) Enthalpy is an extensive property (2) ΔH for a reaction is equal in magnitude but opposite in sign to ΔH for reverse reaction (3) ΔH for a reaction depends on states of reactants and products (gas, liquid) THERMOCHEMICAL EQUATIONS A balanced chemical equation that also includes the enthalpy change. ΔHo delta H standard standard P (1 bar) & T (usually 25 °C) 2 H2 + O2 → 2 H2O ΔH = −483.6 kJ Is this reaction exothermic or endothermic? ________ How much heat is given off per mole of O2? _____ How much heat is given off per mole of H2? _____ What is ΔH for 2 H2O → O2 + 2 H2 ? _______ How many kJ of heat are needed to convert 9.0 g of H2O into H2 and O2 ? 2 H2O → O2 + 2 H2 H 2 HI(g) products ΔH = 53 kJ H2(g) + I2(g) reactants ENDOTHERMIC REACTION H CH4(g) and 2 O2(g) reactants ΔH = –890 kJ CO2(g) + 2 H2O(l) products EXOTHERMIC REACTION EXOTHERMIC REACTION DEMONSTRATION sodium sulfite and bleach SO32−(aq) + OCl−(aq) → SO42−(aq) + Cl−(aq) + heat Laundry bleach is 5% NaOCl This is a redox reaction S4+ → S6+ + 2 e− Cl1+ + 2 e− → Cl− Heat is evolved Exothermic ΔH is negative FLAMING COTTON Na2O2(s) + 2H2O(l) → 2Na+ + 2OH– + H2O2(aq) sodium peroxide 2H2O2(aq) → 2 H2O + O2 reaction produces heat have high conc. of O2 cotton has low ignition temperature ….so….. flame ENDOTHERMIC REACTION DEMONSTRATION Ba(OH)2•8H2O(s) + 2 NH4(SCN)(s) → Ba(SCN)2(aq) + 2 NH3(g) + 10 H2O(l) mix two solids highly endothermic reaction increase in entropy is driving force go from two ordered crystals to ions and gases in solution HEAT CAPACITY Experimental measurement of heat flow heat gained or lost q = C m ΔT mass temp change specific heat For H2O: 4.184 J g °C QUANTITATIVE CALORIMETRY EXAMPLE A calorimeter with 200 g H2O is used for a reaction. If T rises from 25.0 °C to 33.0 °C, how much energy is being released? heat capacity of H2O(l) = 4.184 J/°C• g q = C m ΔT 4.184 J °C g H2O (200 g H2O)(8.0 °C) = 6694.4 J or 6.7 kJ HESS’S LAW ΔH for a sum of steps is the same as for the overall process Reason: ΔH is a state function Analogy to location and distance HESS’S LAW EXAMPLE SMOG REACTION N2(g) + 2 O2(g) → 2 NO2(g) ΔH = ? Two steps N2(g) + O2(g) → 2 NO(g) 2 NO(g) + O2(g) → 2 NO2(g) N2(g) + 2 O2(g) → 2 NO2 (g) ΔH = 180 kJ ΔH = – 112 kJ ΔH = 68 kJ sum of ΔH values gives ΔH of net reaction ΔH OF FORMATION Hess’s law applied to combination reactions Elements combine Compound ΔHf heat of formation When all substances in standard states (T, P, state) then ΔHfo standard heat of formation ΔHfo for stable form of element = 0 ΔHFo to calculate ΔHRXNo energy elements C(s) + H2(g) + O2(g) -(ΔHFo)R reactants C6H12O6 + 6 O2 (ΔHFo)P ΔHRXNo products 6 CO2 + 6 H2O ΔHRXNo = (Δ HF)P – (Δ HF)R ΔHrxno ΔHrxno = Σn ΔHfo (products) – Σm ΔHfo (reactants) an application of Hess’s law EXAMPLE: What is ΔHrxno for the reaction ? ΔHfo values kJ/mol C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l) –277.7 0 2(–393.5) 3(–285.8) ΔHrxno = 2(–393.5) – 3(285.8) – (–277.7) = – 1366.7 kJ exothermic ΔHfo for ethanol 2 C (graphite) + 3 H2(g) + ½ O2(g) → C2H5OH(l) reactants all in standard states 1 bar 25 °C ΔHfo = – 277.7 kJ/mol produce 1 mol of C2H5OH ΔHrxno ΔHrxno = Σn ΔHfo (products) – Σm ΔHfo (reactants) an application of Hess’s law EXAMPLE: What is ΔHrxno for the reaction ? ΔHfo values kJ/mol C6H6(l) + 15/2 O2(g) → 6 CO2(g) + 3 H2O(l) 49.0 0 6(–393.5) 3(–285.8) ΔHrxno = 6(–393.5) + 3(–285.8) – (49.0) = – 3267.4 kJ exothermic ΔHfo for benzene 6 C (graphite) + 3 H2(g) → C6H6(l) reactants all in standard states 1 bar 25 °C ΔHfo = 49.0 kJ/mol produce 1 mol of C6H6(l) ΔHrxno Example Calculate ΔHRXNo for 2 C(s) + H2(g) Æ C2H2(g) From the following information C2H2(g) + 5/2 O2(g) Æ 2 CO2(g) + H2O(l) ΔHRXNo = -1299.6 kJ C(s) + O2(g) Æ CO2(g) ΔHRXNo = -393.5 kJ H2(g) + ½ O2(g) Æ H2O(l) ΔHRXNo = -285.8 kJ WHERE DOES THE ENERGY COME FROM? Bond breaking and formation BOND ENTHALPY energy to break one mole of bonds Cl2(g) → 2 Cl(g) ΔHo = 243 kJ Bond breaking is always endothermic Bond formation always exothermic ΔH for a reaction depends on the breaking and forming of bonds -- the numbers of bonds and their strengths (more in Chap. 8) COVALENT BONDS LENGTH, ENERGY BOND LENGTH • • bond length Larger atoms have longer bonds Multiple bonds are shorter: C–O 143 pm but C=O 122 pm BOND ENERGY Bond length & bond energy closely related As electron density (and number of bonds) between 2 atoms increases, the bond gets shorter and stronger. COVALENT BONDS LENGTH, ENERGY Covalent bonds are strong Overall stability of a molecule is related to the covalent bonds it contains Strength of bond = energy needed to break it BOND ENERGY Value is always positive -- it takes energy to break a covalent bond Value depends on identity of atoms involved and number of shared electron pairs H H C H H . . C . + 4 H. . ΔH = 1660 kJ/mol D(C–H) = 1660/4 = 415 kJ/mol COVALENT BOND ENERGIES C–H 416 kJ/mol C–Cl 327 kJ/mol C–Br 285 kJ/mol C–C 356 kJ/mol C=C 598 kJ/mol C≡C 813 kJ/mol Table 8.4 shorter bond length more electrons shared shorter bond length BOND ENERGIES AND ΔH The ΔH for a reaction can be estimated from bond energies ΔHrxn = ΔH(bonds broken) – ΔH(bonds formed) EXAMPLE Estimate ΔHrxn R C=C H R′ for + Br2 H Br Br R C C H H R′ BOND ENERGIES AND ΔH Br Br R C=C H R′ + Br2 R H break break form C C H H R′ form ΔHrxn = 1[C=C] + 1[Br–Br] – {1[C–C] + 2[C–Br]} 598 791 taken up 193 356 276 x2 908 given off ΔHrxn = 791 – 908 = – 117 kJ/mol exothermic reaction COMBUSION OF METHANE CH4 + 2 O2 → CO2 + 2 H2O C – H 413 kJ/mol O = O 495 kJ/mol C = O 799 kJ/mol O – H 463 kJ/mol CONSUME RELEASE 4 x 413 = 1652 2 x 495 = 990 2 x 799 = 1598 4 x 463 = 1852 2642 3450 – 808 kJ/mol expt value = – 802 kJ/mol OLD EXAM QUESTION The thermite reaction below is used for welding. What is the ΔHrxno for the reaction involving 1 mole of Al? 2 Al (s) + Fe2O3 (s) → Al2O3 (s) + 2 Fe (s) ΔHfo of Al2O3 (s) = – 1669.8 kJ/mol ΔHfo of Fe2O3 (s) = – 822.2 kJ/mol A. B. C. D. E. + 847.6 kJ – 847.6 kJ +1895 kJ – 423.8 kJ – 2,492 kJ EXAM QUESTION What is the ΔHrxn for this reaction? 2 HCl(g) + F2(g) → 2 HF(g) + Cl2(g) Bond H−Cl F−F H−F Cl−Cl Bond Dissociation Energy 431 kJ/mol 155 kJ/mol 567 kJ/mol 242 kJ/mol