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  • LIMITING REACTANTLIMITING REACTANT

    The limiting reactant is used up first

    The limiting reactant determines the

    amount of product

    Need balanced equation to proceed

  • Limiting ReagentsLimiting Reagents

    2H2 + O2 2H2O

    10 H2 and 7 O2 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 YIELDPERCENT YIELD

    PercentYield =

    actual yieldtheoretical yield

    x 100%

    Calculation is just one more stepbeyond a standard stoichiometrycalculation

  • PERCENT YIELDPERCENT YIELD

    EXAMPLESilicon carbide (SiC) is made from sand (silicon dioxide, SiO2) and carbon at high T. CO is alsoformed. 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:

    100.0 kg x 103 g/kg x 1 mol60.09 g

    = 1664 mol SiO2

    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 kg66.73 kg

    x 100% = 82.4%

  • TITRATIONTITRATION

    Goal: find conc. of unknown React solution of known conc.

    (std. soln.) with soln. of unknown conc. Find equivalence point (vol.)

    soln. of unknown conc.

    add std. soln.

    Find volume at equivalence point

    (need way to signal the equiv. pointsuch as indicator)

  • ACIDSACIDS--BASEBASENEUTRALIZATIONNEUTRALIZATION

    ACID + BASE SALT + WATER

    HNO3 + KOH KNO3 + H2O

    H+(aq) + NO3(aq) + K+(aq) + OH(aq)

    H2O(l) + K+(aq) + NO3(aq)

    completeionic

    equation

    H+(aq) + OH(aq) H2O(l) net ionicequation

    eliminate spectator ions

  • Strong Acids and BasesStrong Acids and BasesTable 4.2

    Strong Acids

    Hydrochloric, HClHydrobromic, HBrHydroiodic, HIChloric, HClO3Perchloric, HClO4Nitric, HNO3Sulfuric, 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 EXAMPLETITRATION EXAMPLEA 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 thereaction. How many grams of HCl were there?

    HCl + NaOH NaCl + H2O

    (1) Find moles of NaOH used

    (23.35 x 103 L NaOH)(0.101 mol/L) = 2.36 x 103 mol

    (2) Find moles of HClsame as NaOH

    (3) Find g of HCl

    (2.36 x 103 mol)(36.5 g/mol) = 86.1 x 103 g

    86.1 mg HCl

    If we knew the volume of HCl soln., we couldcalc. the M of the HCl soln.

  • TITRATING HTITRATING H22SOSO44 with with NaOHNaOH

    2 H+(aq) + 2 OH (aq) 2 H2O(l) net ionicequation

    H2SO4(aq) + 2 NaOH(aq) Na2SO4(aq) + 2 H2O(l)

    Begin with 20 mL of 0.1 M H2SO4(aq), titrate with0.1 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)

  • STOICHIOMETRYSTOICHIOMETRY

    When chemical reactions involve gases,the balanced equation provides the numberof moles of reactants and products.

    The ideal gas equation provides the linkbetween number of moles and P, V, Tof gases.

    How much gas is produced or consumedby a chemical reaction?

    find n and then V or P

  • Air BagAir Bag2 NaN3(s) 2 Na + 3 N2(g)

    sodium azide gas in air bag

    How many L of N2 at 735 mm Hg and26 C are produced from 125 g NaN3?

    Moles NaN3125 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

    (40 msec)

  • IDEAL GAS LAW EXAMPLEIDEAL GAS LAW EXAMPLECalculate pressure change in cylinder of a cars engine when 0.250 g of octane [C8H18(l)] is burned with a stoichiometric amount of O2assuming complete combustion.

    Cylinder volume = 0.100 L

    Initial Temp = 80 oC; final Temp = 700 oC

  • ENERGYENERGY

    KINETICENERGY

    POTENTIALENERGY

    mechanical(moving mass mv2)

    [ joule = kg m2/s2]

    mechanical(mass in a placewhere force can act)

    chemical (bonds)

    nuclear(binding energy)

    electrical(moving charge)

    light (photons)

    sound (moleculesmoving uniformly)

    heat (moleculesmoving randomly)

  • ENERGY CONVERSIONENERGY CONVERSION

    Energy can be converted from oneform to another

    When it is converted, the total energyremains constant

    Law of Conservation of EnergyFirst Law of Thermodynamics

    All energy lost by a system underobservation is gained by the surroundings

    During energy conversion, some heatis always produced

  • The energy of the universe is constant.

    Euniverse = Esystem + Esurroundings = 0

    Energy is neither created nor destroyed,

    only converted from one form to another.

    Esystem = q + w

    q is heat gained or lost by the system

    w is work done by or on the system

    First Law of Thermodynamics First Law of Thermodynamics Conservation of EnergyConservation of Energy

  • E = Efinal Einitial

    E is a state function

    System Surroundingsenergy E is

    Surroundings Systemenergy E is +

    E = q + wwork (work done to system +)

    heat (heat added to system +)

    State function: a function whose value doesnot depend on pathway used to get topresent state.

    ENERGY CHANGESENERGY CHANGES

  • ENTHALPYENTHALPYWhen changes occur at constant pressure

    E = qp + wexpansionthis is negligible

    H = qp

    H is the enthalpy changeH is the quantity of thermal energy

    transferred into a system at constantpressure

    Like energy, enthalpy is a state function

  • ENTHALPY OF REACTIONENTHALPY OF REACTION

    H = H(products) H(reactants)

    Endothermic H > 0Exothermic H < 0

    2 H2(g)+ O2(g) 2 H2O(g) + heat

    H = 483.6 kJ

    Characteristics of Enthalpy(1) Enthalpy is an extensive property(2) H for a reaction is equal in magnitude but

    opposite in sign to H for reverse reaction(3) H for a reaction depends on states of

    reactants and products (gas, liquid)

  • 2 H2 + O2 2 H2O H = 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 H for 2 H2O O2 + 2 H2 ? _______

    How many kJ of heat are needed to convert9.0 g of H2O into H2 and O2 ?

    2 H2O O2 + 2 H2

    THERMOCHEMICAL THERMOCHEMICAL EQUATIONSEQUATIONS

    A balanced chemical equation that alsoincludes the enthalpy change.Ho delta H standard

    standard P (1 bar) & T (usually 25 C)

  • 2 HI(g) products

    H2(g) + I2(g) reactants

    HH = 53 kJ

    ENDOTHERMIC REACTION

    CH4(g) and 2 O2(g) reactants

    CO2(g) + 2 H2O(l) products

    HH = 890 kJ

    EXOTHERMIC REACTION

  • EXOTHERMIC REACTIONEXOTHERMIC REACTIONDEMONSTRATIONDEMONSTRATION

    SO32(aq) + OCl(aq) SO42(aq) + Cl(aq) + heat

    sodium sulfite and bleach

    Laundry bleach is 5% NaOCl

    This is a redox reaction

    S4+ S6+ + 2 e

    Cl1+ + 2 e Cl

    Heat is evolvedExothermicH is negative

  • FLAMING COTTONFLAMING COTTON

    Na2O2(s) + 2H2O(l) 2Na+ + 2OH + H2O2(aq)

    sodiumperoxide

    2H2O2(aq) 2 H2O + O2

    reaction produces heathave high conc. of O2cotton has low ignition temperature

    .so..

    flame

  • ENDOTHERMIC REACTIONENDOTHERMIC REACTIONDEMONSTRATIONDEMONSTRATION

    Ba(OH)28H2O(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 ionsand gases in solution

  • HEAT CAPACITYHEAT CAPACITY

    Experimental measurement of heat flow

    q = C m T

    temp changemass

    specific heat

    For H2O: 4.184 J

    g C

    heat gained or lost

  • QUANTITATIVEQUANTITATIVECALORIMETRYCALORIMETRY

    EXAMPLEA calorimeter with 200 g H2O is used for areaction. 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

    4.184 JC g H2O

    (200 g H2O)(8.0 C) = 6694.4 J

    6.7 kJ

    q = C m T

    or

  • HESSHESSS LAWS LAW

    H for a sum of steps is the sameas for the overall process

    Reason: H is a state function

    Analogy to location and distance

  • HESSHESSS LAW EXAMPLES LAW EXAMPLE

    N2(g) + 2 O2(g) 2 NO2(g) H = ?

    Two steps

    N2(g) + O2(g) 2 NO(g) H = 180 kJ2 NO(g) + O2(g) 2 NO2(g) H = 112 kJ

    N2(g) + 2 O2(g) 2 NO2 (g) H = 68 kJ

    sum of H values gives H of net reaction

    SMOG REACTION

  • H OF FORMATIONH OF FORMATION

    Hesss law applied to combination reactions

    Elementscombine

    Compound Hfheat of

    formation

    When all substances in standardstates (T, P, state) then

    Hfo standard heat of formation

    Hfo for stable form of element = 0

  • ener

    gy

    reactants

    products

    -(HFo)R(HFo)P

    HRXNo = ( HF)P ( HF)R

    elements

    HHFFoo to calculate to calculate HHRXNRXNoo

    C6H12O6 + 6 O2

    C(s) + H2(g) + O2(g)

    6 CO2 + 6 H2O

    HRXNo

  • HHrxnrxnoo

    Hrxno = n Hfo (products) m Hfo (reactants)an application of Hesss law

    EXAMPLE: What is Hrxno for the

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