Chem 17 Concepts and Equations 2nd LE
-
Upload
monique-ramirez -
Category
Documents
-
view
11 -
download
1
description
Transcript of Chem 17 Concepts and Equations 2nd LE
Chem 17 Concepts and Equations
2nd LE Topics
The Equilibrium Constant
• Similarly, for the general reaction:
we can define a constant
( ) ( )( ) ( )
tscoefficien
tscoefficien
products ofactivity
reactants ofactivity b
B
a
A
d
D
c
CeqK
←
←=
aa
aa
(aq)(aq)k
k
(aq)(aq) dD cC bB aA
r
f
+←→
+
NOTE: Each species is raised to their respective coefficients in the
balanced chemical equation when expressing the Keq value.
[ ]CγCC =a
[ ]DγDD =a
[ ]AγAA =a
[ ]BγBB =a
Activity of species i -
[ ]ia speciesγ ii = i
ia
i
i
species oft coefficienactivity γ
species ofactivity
−
−
At low concentrations
(dilute, ideal solutions):
( ) ( )( ) ( )bB
a
A
d
D
c
CeqK
aa
aa=
[ ]
1 γ i
iai →∴
→
[ ] [ ][ ] [ ]ba
dc
cBA
DCKK ==
• Kc - Equilibrium constant in terms of concentrations. It is the
product of the equilibrium concentrations (in M) of the
products, each raised to a power equal to its stoichiometric
coefficient in the balanced equation, divided by the product of
the equilibrium concentrations (in M) of the reactants, each
raised to a power equal to its stoichiometric coefficient in the
balanced equation.
Keq becomes Kc � ONLY at ideal dilute conditions
• Kc values are dimensionless
Relationship of K to the Balanced Chemical
Equation
• Reverse an equation, invert the value of K.
• Multiply the coefficients in a balanced equation by a common factor, equilibrium constant is raised to the corresponding power
• Divide the coefficients in a balanced equation by a common factor, take the corresponding root of the equilibrium constant (square root, cube root, )
N2(g) + 3H2(g) � 2NH3(g) K = 5.8 × 105
NH3(g)� 1/2N2(g) + 3/2H2(g) Knew = 1.313 × 10-3
Combining Equilibrium Constant
Expressions• When individual equations are combined (that is,
added), their equilibrium constants are multiplied to obtain the equilibrium constant for the overall reaction.
N2O(g) + 1/2O2(g) � 2NO(g) Koverall = 8.54 × 10-13
2N2(g) + O2(g) � 2N2O(g) K = 2.9 × 10-37 (inverse, take square root)
N2(g) + O2(g) � 2NO(g) K = 4.6 × 10-31(retain)
Solving Equilibrium Problems: How to use
ICE table
Example: At a given temperature 0.80 mole of N2 and 0.90 mole of H2 were placed in an evacuated 1.00-liter container. At equilibrium 0.20 mole of NH3 was present. Calculate Kc for the reaction.
[ ][ ][ ]
( )( )( )
26.060.070.0
20.0
HN
NHK
M 0.20 M 0.60 M 0.70 mEquilibriu
M 0.20+ M 0.30- M 0.10- Change
0 M 0.90 M 0.80 Initial
2NH 3H N
3
2
3
22
2
3c
3(g)2(g)2(g)
===
+ →
←
The Reaction
Quotient, Q
• The major difference between Q and Kc is that the concentrations used in Q are not equilibrium values.
• Q helps predict how the equilibrium will respond to an applied stress - compare Q with Kc.
When
Q = Kc : the system is in equilibrium
Q > Kc : the system goes to the left (), towards reactants
Q < Kc : the system goes to the right (����), towards products
[ ] [ ]
[ ] [ ]bnonequila
nonequil
d
nonequil
c
nonequil
BA
DCQ
)conditions mequilibriu-non(at
dD+cC bB+aA
=
→
←
Le Chatelier’s Principle
� If a change of conditions (stress) is applied to a system in equilibrium, the system responds in the way that best tends to reduce the stress in reaching a new state of equilibrium.
� Some possible stresses to a system at equilibrium are:
1. Changes in concentration of reactants or products.
2. Changes in pressure or volume (for gaseous reactions)
3. Changes in temperature (effect depends on sign of ∆H)
reactants) gaseous of moles of (#-products) gaseous of moles of (#=n∆
Relationship Between Kp and Kc
( ) ( ) n
pc
n
cp RTKKor RTKK∆−∆
==
For gas phase reactions the equilibrium constants can be expressed in partial pressures rather than concentrations.
Gibbs Free Energy and Equilibrium
General equation: THERMODYNAMIC
FORM
All gaseous reactants and products
All solutions of reactants and products
Used when both gaseous and solution
forms appear in the chemical equation
∆Gorxn K Spontaneity condition
< 0 > 1Forward reaction spontaneous,
More products than reactants at equilibrium
= 0 = 1 IDEAL system, very RARE
> 0 < 1Reverse reaction spontaneous,
More reactants than products at equilibrium
Gibbs Free Energy and Equilibrium
EVALUATION OF EQUILIBRIUM CONSTANTS AT
DIFFERENT TEMPERATURES
• From the value of ∆Ho and K1 at one temperature, T1, we can use the van’t Hoff equation to estimate the value of K2 at another temperature, T2.
OR
Gibbs Free Energy: Standard and Non-
standard forms
quotientreaction =Q
Kelvinin re temperatuabsolute = T
K-mol
J 8.314constant gas universal =R
Q log RT 303.2G=G
or lnQ RTG=G
o
o
+∆∆
+∆∆
� ∆Go - standard free energy change.
� ∆G - free energy change at nonstandard conditions
eq
o
eq
o
K log RT 2.303 -=G
or Kln RT -=G
∆
∆ � If concentrations and partial pressures of
species are in equilibrium, then ∆G = 0, and the
equations at left follows
ACID-BASE EQUILIBRIA
[HA]
]][AH[K
-
a
+
=
[B]
]][BHOH[Kb
+−
=
HA ⇄ H⇄ H⇄ H⇄ H++++ + A+ A+ A+ A----
B + H2O ⇄ OH⇄ OH⇄ OH⇄ OH---- + BH+ BH+ BH+ BH++++
• Relating Ka to Kb
HA + H2O ⇄ A- + H3O+ Ka (HA is acid)
A- + H2O ⇄ HA + OH- Kb (A- is base)
H2O + H2O ⇄ H3O+ + OH- Kw
baw K KK ×=
a
wb
K
KK =
b
wa
K
KK =
The stronger the
acid/base, the
weaker is its
conjugate
11-
a2(aq)3
-2
3(aq)(l)2
-
3(aq)
-7
a1(aq)3
-
3(aq)(l)23(aq)2
10 x 4.7 K OH COOH HCO
10 x 4.4K OH + HCOOHCOH
=+↔+
=↔+
+
+
8-
b2(aq)3(aq)2(l)2
-
3(aq)
-4
b1(aq)
-
3(aq)(l)2
-2
3(aq)
10 x 2.3 K HO COHOH HCO
10 x 1.2K HO + HCOOHCO
=+↔+
=↔+
−
−
b2a2
w
b1a1
KK
K
KK
12-
b3(aq)4(aq)3(l)2
-
4(aq)2
7-
b2(aq)
-
4(aq)2(l)2
-2
4(aq)
2
b1(aq)
-2
4(aq)(l)2
-3
4(aq)
101.33 K HO POHOH POH
101.63 K HO POHOH HPO
1078.2K HO + HPOOHPO
×=+→+
×=+→+
×=→+
−
−
−−
b3a3
b2wa2
b1a1
KK
KKK
KK
13-
a3(aq)3
-3
4(aq)(l)2
-2
4(aq)
8-
a2(aq)3
-2
4(aq)(l)2
-
4(aq)2
-3
a1(aq)3
-
4(aq)2(l)24(aq)3
10 3.60 K OH POOH HPO
10 6.20 K OH HPOOH POH
10 50.7K OH + POHOHPOH
×=+→+
×=+→+
×=→+
+
+
+
Strengths of Acids
• BINARY Acids HX - acid strength increases with
decreasing H-X bond strength.
• Down a group: ����size, ����energy to break H- bond (����electronegativity),
����acidity
• Across a period: ����electronegativity, ����acidity
• TERNARY ACIDS HXOn - hydroxides of nonmetals
that produce H3O+ in water.
• For acids with same central element, increase acidity with increase
oxidation state of central element, or increase in attached O atoms
• For acids with different central element but with same number of
O atoms, increase acidity with increase elctronegativity of central
element
Strength of Amine Bases
� The electronic
properties of the
substituents (alkyl
groups enhance the
basicity, aryl groups
diminish it).
� Steric hindrance
offered by the groups
on nitrogen.
Buffers and Henderson-Hasselbalch Equation
HHE in acidic form
HA + H2O ⇄ A⇄ A⇄ A⇄ A---- + H+ H+ H+ H3333OOOO++++
B + H2O ⇄ BH⇄ BH⇄ BH⇄ BH++++ + HO+ HO+ HO+ HO----
HHE in basic form
Preparation of Buffers
• Buffer components must be chosen (acid and conjugate-base, or base and conjugate-acid) based on how near the pKa (of pKb) is near the required buffer pH
• 2 equations needed when solving buffer systems
base conjugate of mole acid of mole componentsbuffer moles total +=
acid moles
base conjugate moleslog pK pH a +=
Other Acid-Base Equations
[ ]( ) HAlog pK2
1 pH initiala -=
pH of weak acid solution,
assumptions valid
[ ][ ]
HA
AlogpK pH
-
a += pH of weak acid solution
added with base (limiting
reactant); forms a buffer
2
pK pK pH a2a1 +=
pH of an amphiprotic weak
acid, amphiprotic form
between the 1st and 2nd Kas
(e.g. HCO32-)
Acid-Base Titration curve• Weak Acid (analyte) titrated with Strong base (titrant)
SOLUBILITY EQUILIBRIA
Example:
• Note: When comparing the solubility of
different salts, look at molar solubility
derived from the Ksp, instead of just the
Ksp value.
• Which of the following salts is more soluble in
water?
Ag3AsO4 27(molar solubility s)4 = Ksp (1.00 × 10-22)
FeAsO4 (molar solubility s)2 = Ksp (5.70 × 10-21)
Cu3(AsO4)2 108(molar solubility s)5 = Ksp (7.60 × 10-36)
The Common Ion Effect in Solubility Calculations
• Calculate the molar solubility of barium sulfate, BaSO4, in 0.010 M sodium sulfate, Na2SO4, solution at 25oC.
The Reaction Quotient in Precipitation
Reactions
• The reaction quotient, Qsp, and the Ksp of a compound are used to analyze whether or not a precipitate will form upon mixing of two ionic species
�Set-up the Ksp reaction
�Calculate the Qsp
�If Qsp > Ksp, then PRECIPITATION happens (← reaction toward ppt formation)
�If Qsp < Ksp, then DISSOLUTION happens (→ reaction toward dissociation into ions)
Fractional Precipitation
• The method of precipitating some ions from solution
while leaving others in solution is called fractional
precipitation.
If a solution contains 0.010 M (each) Cu+, Ag+, and Au+ at
100.0 mL, and a solution of NaCl is added (0.010 M), each ion
can be precipitated as chlorides.
[ ][ ][ ][ ][ ][ ] 13
sp
-
(aq)(aq)(s)
10
sp
-
(aq)(aq)(s)
7
sp
-
(aq)(aq)(s)
100.2Cl AuK Cl Au AuCl
108.1Cl AgK ClAgAgCl
109.1Cl CuK ClCu CuCl
−−++
−−++
−−++
×==+↔
×==+↔
×==+↔
• To determine which species will precipitate first upon addition of
Cl-, calculate the molar solubility
– ↓[molar solubility], will precipitate first (since it is least soluble)
– ↑[molar solubility], will precipitate last (since it is more soluble, it will
stay more in solution)
• For CuCl ⇌ Cu(�)� + Cl(�)
:
[molar solubility of CuCl] = x = [Cu+]eq = [Cl-]eq
∴ Ksp = 1.97 × 10-7 = x ∙ x
∴ x = Ksp = 4.44 × 10-4
• AgCl ⇌ Ag(�)� + Cl(�)
:
[molar solubility of AgCl] = x = [Ag+]eq = [Cl-]eq
∴ Ksp = 1.80 × 10-10 = x ∙ x
∴ x = Ksp = 1.34 × 10-5
• AuCl ⇌ Au(�)� + Cl(�)
:
[molar solubility of AuCl] = x = [Au+]eq = [Cl-]eq∴ Ksp = 2.00 × 10-13 = x ∙ x
∴ x = Ksp = 4.47 × 10-7
Determining [1st ion to be precipitated]
remaining when the 2nd ion starts to
precipitate
• For two ions in solution in which the Solubility Product expressions can be written as 1:1 cation:anion dissolution process
1stion(tobeprecipitated)remaininginsolution
=K!"#!$%&'
K!"(')%&'× 2ndionabouttobeprecipitated
NOTE: This expression changes when solubility
expression is not 1:1 cation:anion dissolution