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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, HClO3
Perchloric, HClO4
Nitric, HNO3
Sulfuric, 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 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 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 SO4
2–(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 car’s 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 ½ 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.
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
First Law of Thermodynamics First Law of Thermodynamics Conservation of EnergyConservation of Energy
ΔE = Efinal – Einitial
ΔE is a state function
System Surroundingsenergy ΔE is –
Surroundings Systemenergy ΔE is +
ΔE = 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
ΔE = qp + wexpansion
this is negligible
ΔH = qp
ΔH is the enthalpy changeΔH is the quantity of thermal energy
transferred into a system at constantpressure
Like energy, enthalpy is a state function
ENTHALPY OF REACTIONENTHALPY OF REACTION
ΔH = H(products) – H(reactants)
Endothermic ΔH > 0Exothermic 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)
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 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.ΔHo delta H standard
standard P (1 bar) & T (usually 25 °C)
2 HI(g) products
H2(g) + I2(g) reactants
HΔH = 53 kJ
ENDOTHERMIC REACTION
CH4(g) and 2 O2(g) reactants
CO2(g) + 2 H2O(l) products
HΔH = –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 evolvedExothermicΔH 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)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 ionsand gases in solution
HEAT CAPACITYHEAT CAPACITY
Experimental measurement of heat flow
q = C m ΔT
temp changemass
specific heat
For H2O: 4.184 Jg °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 J°C g H2O
(200 g H2O)(8.0 °C) = 6694.4 J
6.7 kJ
q = C m ΔT
or
HESSHESS’’S LAWS LAW
ΔH for a sum of steps is the sameas for the overall process
Reason: ΔH is a state function
Analogy to location and distance
HESSHESS’’S LAW EXAMPLES LAW EXAMPLE
N2(g) + 2 O2(g) → 2 NO2(g) ΔH = ?
Two steps
N2(g) + O2(g) → 2 NO(g) ΔH = 180 kJ2 NO(g) + O2(g) → 2 NO2(g) ΔH = – 112 kJ
N2(g) + 2 O2(g) → 2 NO2 (g) ΔH = 68 kJ
sum of ΔH values gives ΔH of net reaction
SMOG REACTION
ΔΔH OF FORMATIONH OF FORMATION
Hess’s law applied to combination reactions
Elementscombine
Compound ΔHfheat of
formation
When all substances in standardstates (T, P, state) then
ΔHfo standard heat of formation
ΔHfo for stable form of element = 0
ener
gy
reactants
products
-(ΔHFo)R
(ΔHFo)P
ΔHRXNo = (Δ HF)P – (Δ HF)R
elements
ΔΔHHFFoo to calculate to calculate ΔΔHHRXNRXN
oo
C6H12O6 + 6 O2
C(s) + H2(g) + O2(g)
6 CO2 + 6 H2O
ΔHRXNo
ΔΔHHrxnrxnoo
ΔHrxno = Σn ΔHf
o (products) – Σm ΔHfo (reactants)
an application of Hess’s law
EXAMPLE: What is ΔHrxno for the reaction ?
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)ΔHf
o
valueskJ/mol
–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 allin standardstates 1 bar25 °C
ΔHfo = – 277.7 kJ/mol
produce 1 mol of C2H5OH
ΔΔHHrxnrxnoo
ΔHrxno = Σn ΔHf
o (products) – Σm ΔHfo (reactants)
an application of Hess’s law
EXAMPLE: What is ΔHrxno for the reaction ?
C6H6(l) + 15/2 O2(g) → 6 CO2(g) + 3 H2O(l)ΔHf
o
valueskJ/mol
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 allin standardstates 1 bar25 °C
ΔHfo = 49.0 kJ/mol
produce 1 mol of C6H6(l)
ΔΔHHrxnrxnoo ExampleExample
Calculate ΔHRXNo for
2 C(s) + H2(g) C2H2(g)
From the following information
C2H2(g) + 5/2 O2(g) 2 CO2(g) + H2O(l)ΔHRXN
o = -1299.6 kJ
C(s) + O2(g) CO2(g)ΔHRXN
o = -393.5 kJ
H2(g) + ½ O2(g) H2O(l)ΔHRXN
o = -285.8 kJ
WHERE DOES THE WHERE DOES THE ENERGY COME FROM?ENERGY COME FROM?
Bond breaking and formation
BOND ENTHALPYenergy 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 thebreaking and forming of bonds -- thenumbers of bonds and their strengths
(more in Chap. 8)
COVALENT BONDSCOVALENT BONDSLENGTH, ENERGYLENGTH, ENERGY
BOND LENGTH • •
bond length
Larger atoms have longer bondsMultiple bonds are shorter:
C–O 143 pm but C=O 122 pm
BOND ENERGY
As electron density (and number of bonds)between 2 atoms increases, the bond getsshorter and stronger.
Bond length & bond energy closely related
COVALENT BONDSCOVALENT BONDSLENGTH, ENERGYLENGTH, ENERGY
Covalent bonds are strongOverall stability of a molecule is related
to the covalent bonds it containsStrength of bond = energy needed to break it
BOND ENERGY
Value depends on identity of atoms involvedand number of shared electron pairs
Value is always positive -- it takes energyto break a covalent bond
C
H
HH
H
C 4 H.... + .
ΔH = 1660 kJ/mol
D(C–H) = 1660/4 = 415 kJ/mol
COVALENT BONDCOVALENT BONDENERGIESENERGIES
C–C 356 kJ/molC=C 598 kJ/molC≡C 813 kJ/mol
moreelectronsshared
shorterbond length
C–H 416 kJ/molC–Cl 327 kJ/molC–Br 285 kJ/mol
shorterbondlength
Table 8.4
BOND ENERGIESBOND ENERGIESAND AND ΔΔHH
The ΔH for a reaction can be estimatedfrom bond energies
ΔHrxn = ΔH(bonds broken) – ΔH(bonds formed)
EXAMPLE
Estimate ΔHrxn for
C C
HH
=R R′
C C
HH
R R′
+ Br2
BrBr
BOND ENERGIESBOND ENERGIESAND AND ΔΔHH
C C
HH
=R R′ C C
HH
R R′+ Br2
BrBr
ΔHrxn = 1[C=C] + 1[Br–Br] – {1[C–C] + 2[C–Br]} break break form form
598 193 356 276x 2
791taken up
908given off
ΔHrxn = 791 – 908 = – 117 kJ/mol
exothermic reaction
COMBUSION OF METHANECOMBUSION OF METHANE
CH4 + 2 O2 → CO2 + 2 H2O
C – H 413 kJ/molO = O 495 kJ/mol
C = O 799 kJ/molO – H 463 kJ/mol
CONSUMECONSUME RELEASERELEASE
4 x 413 = 16522 x 495 = 990
2642
2 x 799 = 15984 x 463 = 1852
3450
– 808 kJ/mol
expt value = – 802 kJ/mol
The thermite reaction below is used for welding.What is the ΔHrxn
o for the reaction involving1 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. + 847.6 kJB. – 847.6 kJC. +1895 kJD. – 423.8 kJ E. – 2,492 kJ
OLD EXAM QUESTIONOLD EXAM QUESTION
EXAM QUESTIONEXAM QUESTION
What is the ΔHrxn for this reaction?
2 HCl(g) + F2(g) → 2 HF(g) + Cl2(g)
Bond Bond Dissociation EnergyH−Cl 431 kJ/molF−F 155 kJ/molH−F 567 kJ/molCl−Cl 242 kJ/mol