Download 2012 Coaches Institute Presentation

Presenter:
Gina Barrier
Director, NW Outreach Office
The Science House, NCSU
State Supervisor, NCSO Chemistry Lab Event: John Kiser
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Topics for 2012-2013: Equilibrium & Periodicity
Regional vs State Topics
Must bring chemical splash goggles & apron or lab
coat
Must bring a non-programmable, non-graphing
calculator!
May bring 1 (8.5 x 11 ) 2-sided page of notes
containing info in any form from any source
Need to know topics
}  Formula Writing/Nomenclature
}  Mole & Stoichiometry Calculations (incuding
solution stoichiometry, molarity calculations, and
dilution calculations)
Closed-toed shoes
}  Pants or skirts that cover the leg to the ankle
}  Chemical splash goggles (ANSI Z87)
}  A Long Sleeve Shirt under lab apron or a
long-sleeved Lab Coat
}  Gloves (optional)
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Dynamic equilibrium - forward and reverse
processes occur at same rate
}  Results in the formation of a mixture of
reactants and products whose
concentrations are in dynamic equilibrium
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For a reaction…
aA(aq) + bB(aq) ⇔ cC(aq) + dD(aq)
Keq = Kc when working with Molar concentrations
Kc - equilibrium constant for a system at
equilibrium at a given temperature
Depends on stoichiometry, not reaction mechanism
Catalyst will not affect equilibrium constant
No units
Concentrations of pure solids and pure liquids are not
included
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Keq = Kp when working with partial pressures
(atm)
Kp – equilibrium constant in terms of gas
pressures
aA(g) + bB(g)
⇔ cC(g) + dD(g)
Kp = (PC)c(PD)d
______________
(PA)a(PB)b
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Kp is related to Kc
Kp = Kc(RT)∆n (where ∆n = #moles product - #moles
reactant)
N2(g) + 3H2 (g) ⇔ 2NH3(g)
K = [NH3] 2
[N2][H2]3
K >> 1 favors products
K << 1 favors reactants
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The value of the equilibrium constant
changes with Temperature
2 NO2 (g) <=> N2O4 (g)
0oC Kp= 58
100oC Kp= 0.066
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K expression for a reaction written in one
direction is the reciprocal for the expression
written in the other direction
2NO(g) ⇔ N2(g) + O2 (g) at 25oC
Kc = [N2][O2] = 1 x 1030
[NO]2
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N2 (g) + O2(g) ⇔ 2NO(g) at 25oC
Kc = [NO] 2 = 1 x 10-30
[N2][O2]
Keq of a reaction that has been multiplied by a
number = Keq raised to a power equal to that
number
}  HF(aq) ⇔ H+(aq) + F-(aq)
Kc= 6.8x10-4
}  2HF(aq) ⇔ 2H+(aq) + 2F-(aq)
Kc= (6.8x10-4)2 = 4.6x10-7
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Keq for a net reaction made up of 2 or
more steps is the product of the Keq for
the individual steps
2HF(aq) ⇔ 2H+(aq) + 2F-(aq)
Kc=
4.6x10-7
2H+(aq) + C2O4- 2(aq) ⇔ H2C2O4 (aq) Kc=2.5 x
105
Net equation:
2HF(aq) + C2O4- 2(aq) ⇔ 2F- (aq) + H2C2O4(aq)
Kc = (4.6x10-7)(2.5x105) = 0.12
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If you know initial concentrations and 1
equilibrium concentration, use stoichiometry to
find the other concentrations.
2 A(aq) + B(aq) ⇔ 4 C(aq)
[A]
[B]
Initial Molarity
1.00 M
1.00 M
Change in
Concentration
-1/2(0.50)
-1/4(0.50)
Equilibrium
Molarity
0.75 M
0.88 M
[C]
0
+0.50
0.50 M
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Kc =
[C]4
=
(0.50) 4
[A]2[B]
(0.75)2(0.88)
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Kc = 0.13
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Reactants favored
If you know Keq and initial concentrations/
pressures
}  H2(g) + I2(g) ⇔ 2HI(g)
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A 1.000L flask is filled with 1.000 mol H2
and 2.000 mol I2 at 448oC. Kc for the
reaction at 448oC is 50.5. What are the
equilibrium concentrations?
H2
I2
HI
Initial Molarity
1.000 M
2.000 M
0M
Change in
Concentration
-x
Equilibrium
Molarity
(1.000 – x) M
-x
(2.000 – x) M
+ 2x
2x M
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Kc = [HI]2 =
(2x) 2
= 50.5
[H2][I2] (1.000-x)(2.000-x)
4x2 = 50.5(x2 - 3.000x +2.000)
* Note: We will not include quadratic
equations in the event, but students will
need to show how to set up the K
expression
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Reaction Quotient, Q, obtained by substituting
initial concentrations into the Keq expression
o Q = K at equilibrium
o Q > K reaction will move from right to left (reverse)
o Q < K reaction will move from left to right (forward)
If a system at equilibrium is disturbed, the
equilibrium position will shift to counteract
the effect of the disturbance
}  Change may occur in…
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◦  Temperature
◦  Pressure
◦  Concentration of one of the components
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N2(g) + 3H2 (g) ⇔ 2NH3(g)
Action
Shift
Add N2 or H2
⇒
Add NH3
⇐
Remove N2 or H2
⇐
Remove NH3
⇒
Yellow
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Colorless
Red-Brown
Fe+3(aq) + SCN-(aq) ⇔ FeSCN+2(aq) + heat
Fe+3(aq) + H2PO4-(aq) à Fe(HPO4)+2(aq) + H+
(aq)
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Decreasing the volume of a gaseous
equilibrium mixture (increasing pressure)
shifts the system in the direction that reduces
the number of moles of gas
N2O4(g) ⇔ 2NO2(g)
If pressure is increased, the reaction will shift
in the reverse direction
Sinc
mole
side
the p
the p
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Endothermic:
o Reactants + heat ⇔ products
o Increasing T, shift to right, K increases
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Exothermic:
o Reactants ⇔ products + heat
o Increasing T, shifts to left, K decreases
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Co(H20)6+2(aq) + 4Cl-(aq)
Pink
⇔
CoCl4-2(aq) + 6H2O(l)
blue
pH = - log [H+]
}  pOH = - log [OH-]
}  pH + pOH = 14
}  Kw = [H+][OH-] = 1 x 10-14
}  Kw = Ka x Kb
}  pKw = pKa + pKb = 14 (at 25oC)
}  pKa = - log Ka
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Completely ionize in solution
}  [H3O+] = [HA]
}  The stronger the acid the weaker the
conjugate base
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HA(aq) + H2O(l) ⇔ H3O+(aq) + A-(aq)
Ka = [H3O+][A-]
[HA]
}  Ka = acid dissociation constant
}  Large Ka = stronger acid
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Calculate the pH of a 0.20 M solution of HCN.
Ka = 4.9 x 10-10
HCN(aq) ⇔ H+(aq) + CN-(aq)
Ka = [H+][CN-] = 4.9 x 10-10
[HCN]
Initial Molarity
Change in
Concentration
Equilibrium
Molarity
[HCN]
[H+]
[CN-]
0.20 M
0M
0M
-x M
+x M
+x M
(0.20 – x) M
x M
x M
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Ka = [H+][CN-] = 4.9 x 10-10
[HCN]
Ka = (x)(x) = 4.9 x 10-10
(0.20 – x)
Approximate 0.20 –x ≅ 0.20 (weak acid)
x2 = 4.9 x 10-10
x = [H+] = 9.9 x 10-6 M
0.20
Since [H+] is much smaller than 5% of 0.20 M,
the simplifying approximation is appropriate.
pH = - log [H+] = - log(9.9 x 10-6 ) = 5.00
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The percentage of acid molecules
that ionize in water is another
measure of the strength of an acid
% Ionization = M(ionized acid) x
100
M(initial)
Percent ionization of a weak acid
increases as its concentration
decreases
NH3 + H2O ⇔ NH4+ + OH-
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B(aq) + H2O(l) ⇔ HB+(aq) + OH-(aq)
Kb = [HB+][OH-]
[B]
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Kb = base dissociation constant
Polyprotic acids ionize in steps, each H has
a separate Ka
}  Ka1 > Ka2 > Ka3
}  Ka1 is the largest
}  Usually, the second ionization is not large
enough to affect the pH
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Ksp - equilibrium constant for the dissociation of
a solid salt into its aqueous ions
For ionic solid MnXm, the dissociation reaction is:
MnXm(s) ⇔ nMm+(aq) + mXn−(aq)
Solubility product would be
Ksp = [Mm+]n[Xn−]m
Example, the dissociation reaction for PbCl2 is
PbCl2(s) ⇔ Pb2+(aq) + 2 Cl−(aq)
and its equilibrium constant is
Ksp = [Pb2+][Cl−]2
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Assume AgCrO4 dissociates completely in water at
25oC. [Ag+] = 1.3 x 10-4
AgCrO4(s) ⇔ 2Ag+(aq) + CrO4-2(aq)
Ksp = [Ag+]2[CrO4-2]
[CrO4-2] = 1.3 x 10-4 mol Ag+ x 1 mol CrO4-2
1L
2 mol Ag+
[CrO4-2] = 1.3 x 10-4M
Ksp = [Ag+]2[CrO4-2] = (1.3 x 10-4)(1.3 x 10-4) =
1.1 x 10-12
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How (and why) do many chemical and
physical properties of elements relate to
position on periodic table?
Building Up Principle – Start at lowest
energy level and sublevel, work your way
up. An atom builds on the previous
atom before it.
}  Pauli Exclusion Principle – No more than
2 electrons in an orbital, one is spin up,
the other spin down.
}  Hund s Rule – If 2 or more orbitals have
the same energy, one electron goes in
each until all are half full. Then,
electrons go back and pair up.
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Valence Electrons = In main group or
representative elements, valence
electrons are in the outmost energy level
}  In examples below, both Na and Mg have
10 core electrons. Na has 1 valence
electron, Mg has 2 valence electrons.
}  For representative element, column #
before A gives number of valence
electrons
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What positive charge do valence electrons
feel from nucleus?
}  Zeff increases from left to right across a
row – number of protons in nucleus
increase
}  Zeff slightly increases from top to bottom
down a column – more diffuse inner
electrons slightly reduce shielding.
}  Zeff ≈ Column number before A
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Electronegativity - Ability of an atom in a
molecule to attract shared electrons in a
covalent bond.
1. 
2. 
3. 
Low ΔEN (< 0.5) the bond is nonpolar covalent.
0.5 < ΔEN < 1.6 the bond is polar covalent
High ΔEN (> 2.0) then the bond is ionic.
Main group metals typically lose all valence
electrons when an ion forms. This allows metal
ion to be isoelectric to a noble gas and have an
octet (8 valence electrons)
Na atom: 1s22s22p63s1 - 1 valence electron
When ion forms, 1 valence electron is lost
Ion: 1s22s22p6 8 valence electrons, isoelectric to
Ne
Because 1 electron is lost, charge of ion is 1+ (Na
+)
Prediction: Column number before A is charge
on ion.
While electrons fill first in ns sublevel and
then
(n-1)d sublevel, a transition metal
usually loses the ns electrons first when
an ion forms, THEN the (n-1)d electrons.
Fe atom: [Ar]4s23d6
Fe2+ ion: [Ar]4s03d6 or [Ar]3d6
Fe3+ ion: [Ar]4s03d5 or [Ar]3d5
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When a nonmetal atom forms an ion, electrons
are gained to complete an octet in outer energy
level.
Chlorine Atom: [Ne]3s23p5 7 valence electrons
When ion forms, 1 more electron is gained to
complete octet
Ion: [Ne]3s23p6 – 8 valence electrons – octet!
Ion is isoelectric to Argon.
Since 1 electron is gained, charge is 1- (Cl-)
Rule of Thumb: 8-Column number before A =
charge on ion
OR, count number of columns from noble gases.
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The amount of energy needed to remove the
highest energy electron from an isolated neutral
atom in the gas state.
Energy change that occurs when an electron is
added to an isolated atom in the gaseous state.
X + e- → X- ∆E is < 0 kJ, energy released
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For metals, mp will decrease down a column in
the periodic table.
For nonmetals, mp will usually increase down a
column in the periodic table.
Chang, Chemistry, 9th edition, McGraw Hill
Metal + Nonmetal → Ionic compound
Be sure to use ionic charges to write
formula of product
Then, go back and balance equation.
Al (s) + O2 (g) → Al2O3 (s)
Al3+ and
O24 Al (s) + 3 O2 (g) → 2 Al2O3 (s)
1) 
2) Reaction of metals with water
Reactivity usually decreases from left to
right across a row, increases down a
column
3) Oxide reactivity
Most metallic oxides react with water to
produce a base (hydroxide)
Li2O (s) + H2O (l) → 2 LiOH (aq)
Most nonmetallic oxides react with water
to produce an acid.
SO3 (g) + H2O (l) → H2SO4 (aq)
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Aluminum oxide can produce an acidic or
basic solution!
Silver Salts Lab Demo
Please email me at [email protected]
Or John Kiser at [email protected]
For follow-up questions, concerns, etc..