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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