Chemical Equilibrium In a chemical reaction, the reactions never go in only one direction In this...
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Transcript of Chemical Equilibrium In a chemical reaction, the reactions never go in only one direction In this...
Chemical Chemical EquilibriumEquilibrium
In a chemical reaction, the reactions never go in only one direction
In this chapter, we will study:
- the equilibrium concept
- equilibrium constant and its calculation
- activity of ionic species – ion effect
complex formation
- acid-base reactions
a A + b B c C + d D
The Rate ConceptThe Rate Concept
Similarly, in the backward reaction
Rate of backward reaction (Rb ) [C]c[D]d
= kb [C]c[D]d
If A is a solid or liquid molar concentration of A
If A is a gas pressure of A in atmosphere
Solvents are omitted from the equilibrium constant
rate constant: dependent on temperature, pressure and presence of catalyst
The Rate Concept
Rate of a chemical reaction is proportional to the “active masses” of the reacting substances present at any time
Rate of forward reaction (Rf ) [A]a[B]b
= kf [A]a[B]b
a A + b B c C + d D
[C]c[D]d
[A]a[B]b
kf
kb
At equilibrium:
rate of forward reaction = rate of backward reaction
kf [A]a[B]b = kb [C]c[D]d
Rearranging:
= = K
K can be evaluated by measuring the concentrations of A, B, C and D at equilibrium.
The larger the K value the farther to the right is the reaction at equilibrium the more favorable the rate constant for the forward reaction relative to the backward reaction
Manipulating Equilibrium Manipulating Equilibrium ConstantsConstants
Consider the reaction:
HA H+ + A-
K1 =
If the direction of a reaction is reversed, the new value of K is simple the reciprocal of the original value of K
H+ + A- HA
K’1 = = 1/ K1
[H+][A-]
[HA]
[H+][A-]
[HA]
Equilibrium constant for sum of reactions:
K3 = K1K2 = x
=
[H+][A-]
[HA]
[CH+]
[H+][C]
[A-][CH+][HA][C]
[A-][CH+]
If two reactions are added, the new K value is the product of the two individual values:
HA H+ + A- K1
H+ + C CH+ K2
HA + C A- + CH+ K3
Example :
The equilibrium constant at 25oC for the reaction:
H2O H+ + OH- Kw = 1.0 x 10-14
NH3(aq) + H2O NH4+ + OH-
K = 1.8 x 10-5
Find the equilibrium constant for the reaction:
NH4+ NH3 + H+
NH3
The 3rd reaction can be obtained by:
H2O H+ + OH- Kw
NH4+ + OH- NH3 + H2O 1/K
NH4+ H+ + NH3 K3 = Kw /K
= 5.6 x 10-10
NH3
NH3
Equilibrium and Equilibrium and ThermodynamicsThermodynamics
Enthalpy of a reaction : heat absorbed or released by the reaction
Entropy of a reaction : degree of disorder of reactants and products
contribute to the degree to which the reaction is favoured or disfavoured
entropy and enthalpy are related to equilibrium constant
Enthalpy change for a reaction (H) : Hproduct -
Hreactanct
H = + heat is absorbed endothermic
H = - heart is liberated exothermic
K value :
- does not tell us how fast a reaction will proceed toward equilibrium
- tells us the tendency of a reaction to occur and in what direction
EntropyEntropy
S = Sproduct - Sreactanct
S = + products are more disordered than reactants
S = - products are less disordered than reactants
Entropy of a substance, S : is the degree of disorder.
Greater the disorder greater the entropy eg gas is more disorder (has higher entropy) than a liquid which is turn is more disordered than a solid
A system will always tend toward lower energy and increased randomness ie lower enthalpy and higher entropy ie a chemical system is driven towards the formation of products by a negative H or a positive value of S, or both. The combined effects of these is given by the Gibbs free energy, G:
G = H – TS where T = temperature in Kelvins
G is a measure of the energy of the system, and a system spontaneously tends toward lower energy state.
Gibbs Free EnergyGibbs Free Energy
The change in energy of a system at a constant temperature is
G = H -TS
The equation combines the effects of H and S. Hence :G = + reaction is
disfavored
G = - reaction is favored
Go is related to the equilibrium constant of a reaction by:
K = e-G /RT or Go = -RTlnK
where R = gas constant = 8.314 JdegK-
1mol-1
From the equation: a large equilibrium constant results from a large negative free energy
o
Standard enthalpy, Ho, standard entropy, So, and standard free energy, Go represent the thermodynamic quantities at standard state (ie 1 atm, 298K and unit concentration)
Summary:
A chemical reaction is favored by (i) the liberation of heat (H negative), (ii) an increase in disorder (S positive), (iii) Go is negative or, equivalently, if K > 1.
Le Châtelier’s PrincipleLe Châtelier’s Principle
Suppose a system at equilibrium is subjected to a change that disturbs the system (eg the equilibrium concentrations of reactants and products are altered by changing the temperature, the pressure or the concentration of one of the reactants), the effects of such changes can be predicted from Le Châtelier’s Principle which states:
when a change is applied to a system at equilibrium, the equilibrium will shift in a direction that tends to relieve or counteract that change.
Example :
BrO3- + Cr3+ + 4H2O Br - + Cr2O7
2- + 8H+
for which the equilibrium constant is given by
K = =1 x 1011 at 25oC
In a particular equilibrium state of this system, the following concentrations exist:
[Br -][Cr2O72-][H+]8
[BrO3-][Cr3+]2
[H+] = 5.0 M; [Cr2O72-] = 0.10M;
[Cr3+] = 0.003M; [Br -] = 1.0M;
[BrO3-] = 0.043M
Dichromate is added to the solution to increase the concentration of [Cr2O7
2-] to
0.20M. In what direction would the reaction proceed to reach equilibrium?
According to Le Châtelier’s Principle, the reaction should move to the left to partially offset the increase in dichromate. This can be verified by setting up a reaction quotient, Q:
Q = = 2 x 1011 > K
(1.0)(0.20)(5.0)8
(0.043)(0.0030)2
Because Q > K reaction must move to the left to decrease the numerator and increase the denominator until Q = K
Hence to achieve equilibrium and:
If Q < K reaction must proceed to the right
If Q > K reaction must proceed to the left
What if the temperature is changed ?
temperature dependent temperature independent
From:
K = e-G /RT = e-(H - TS )/RT = e-H /RT .e-S /Ro o o o o
If Ho is negative, e-H /RT decreases with increasing temperature in an exothermic reaction, K decreases with increasing temperature
If Ho is positive, e-H /RT increases with increasing temperature in an endothermic reaction, K increases with increasing temperature
o
o
Solubility ProductSolubility Product
When substances have limited solubility and their solubility is exceeded, the ions of the dissolved portion exist in equilibrium with the solid material
AgCl(s) AgCl(aq) Ag+ + Cl-
- the substance will have a definite solubility at a given temperature
- a small very amount of undissociated compound usually exists in equilibrium in the aqueous phase and its concentration is constant
- the overall equilibrium constant for the solubility can be written for the stepwise equilibrium:
Ksp =
=
Since AgCl(s) is the pure solid [AgCl(s)] = 1
[AgCl(aq)]
[AgCl(s)] [AgCl(aq)]
[Ag+] [Cl-]
[Ag+] [Cl-][AgCl(s)]
Hence :
Ksp = [Ag+] [Cl-]
This relationship measures the compound’s solubility. It holds under all equilibrium conditions at the specified temperatures
Example:
What is the solubility of Hg2Cl2, in g/l, if the
solubility product is 1.2 x 10-18 ?
Hg2Cl2(s) Hg22+ + 2Cl-
Ksp = [Hg22+ ][Cl-]2 = 1.2 x 10-18
Let s represent the molar solubility of Hg2Cl2.
Then
[Hg22+ ] = s and [Cl-] = 2s
Thus: (s)(2s)2 = 1.2 x 10-18 s = 6.7 x 10 –7 M
Solubility in g/l = 6.7 x 10 –7 x 472.08g/mol
= 3.162 x 10 –10 g/l
Common Ion EffectCommon Ion Effect
What will be the concentration of Hg22+ if a
2nd source of Cl- was added to the solution (eg 0.030M NaCl) ?
Hg2Cl2(s) Hg22+ + 2Cl-
Initial conc solid 0 0.030
Final conc solid x 2x+ 0.030
Ksp = [Hg22+ ][Cl-]2 = x (2x + 0.030)2 = 1.2 x 10-18
(x)(0.030)2 = 1.2 x 10-18
x = 1.3 x 10-15 M
Without the NaCl, [Hg22+ ] = 6.7 x 10 –7 M
- addition of a product displaces the reaction toward the left
This application of Le Châtelier’s Principle is called the common ion effect : a salt will be less soluble if one of its constituent ions is already present in the solution
Separation by PrecipitationSeparation by Precipitation
Example :
Consider a solution containing lead(II)(Pb2+) and mercury(I)(Hg2
2+) ions, each at
a concentration of 0.010M
PbI2(s) Pb2+ + 2I- Ksp = 7.9 x 10-9
Hg2I2(s) Hg22+ + 2I- Ksp = 1.1 x 10-28
Is it possible to completely separate the Pb2+ and Hg2
2+ by selectively precipitating
the latter with iodide?
Precipitation reactions can sometimes be used to separate ions from each other.
Q = [Pb2+][I-]2 = (0.010)(1.0 x 10-11)2
= 1.0 x 10-24 < Ksp for PbI2
Hg2I2(s) Hg22+ + 2I-
Initial Conc 0 0.010 0
Final Conc solid 1.0 x 10 –6 x
Ksp = [Hg22+ ][I-]2 =(1.0 x 10 –6 )(x)2
= 1.1 x 10-28
x = [I-] = 1.0 x 10-11 M
Will this amount of I - cause Pb2+ to precipitate?
Consider lowering the Hg22+ concentration to
0.010% (ie .010% of 0.010M = 1.0 x 10 –6 M) of its original value without precipitating Pb2+.
Let x be the concentration of I- at equilibrium with 1.0 x 10 –6 M [Hg2
2+]
Complex FormationComplex Formation
If anion X- precipitates metal M+, it is sometimes observed that a high concentration of X- causes solids MX to redissolve. This phenomenon can be attributed to the formation of complex ions, such as MX2
-
In complex ions such as MX2-, X- is known as
the ligand of M+. A ligand is defined as any atom of group of atoms attached to the species of interest. M+ accepts electrons Lewis acid
X- donates electrons Lewis base
Room to accept electrons
Room to donate electrons
Example :
Pb++ + I - [Pb I ]+ adduct
dative or coordinate covalent bond
Effect of Complex Ion Effect of Complex Ion Formation on SolubilityFormation on Solubility
At low I- concentrations, the solubility of lead is governed by precipitation of PbI2 :
Pb2+ + 2I- PbI2(s) Ksp = 7.9 x 10-9
However at high I- concentrations, complex ion formation occurs as the reaction is driven to the right :
Pb2+ + I- PbI+ K1 = 1.0 x 102
Pb2+ + 2I- PbI2(aq) 2 = 1.4 x 103
Pb2+ + 3I- PbI3 – 3 = 8.3 x 103
Pb2+ + 4I- PbI4 – 4 = 3.0 x 104
Example :
Find the concentration of PbI+, PbI2(aq), PbI3–
and PbI42- in a solution saturated with PbI2(s)
and containing dissolved I- with a concentration of 1.0M.
Ksp = 7.9 x 10-9 = [Pb2+][I-]2 = [Pb2+]
[PbI+] = K1 [Pb2+][I-] = (1.0 x 102)(7.9 x 10-9)1
= 7.9 x 10-7 M
[PbI2(aq)] = 2[Pb2 +][I-]2 = (1.4 x 103 )(7.9 x 10-9) = 1.1 x 10-5 M
[PbI3–] = 3[Pb2 +][I-]3 = (8.3 x 103 )(7.9 x 10-9)
= 6.6 x 10-5M
[PbI42-] = 4[Pb2 +][I-]4 = (3.0 x 104 )(7.9 x 10-9)
=2.4 x 10-4M
The total concentration of dissolved lead is:
[Pb]total = [Pb2 +] + [PbI+] +[PbI2(aq)] +
[PbI3–] + [PbI42-]
= 3.2 x 10-4M
% dissolved Pb in PbI42- = (2.4 x 10-4)/(3.2 x 10-
4) x 100
= 75%
Each of the reaction is governed by an equilibrium and an equilibrium constant. The total concentration of dissolved lead is considerably greater than that of
Pb2+ alone
The concentration of Pb2+ that satisfies any one of the equilibria must satisfy all the equilibria.
Protic Acids and BasesProtic Acids and Bases
H+ is known as proton because it is what remains when a hydrogen atom loses its electron
Protic refers to the chemistry that involves the transfer of H+ from one molecule to another
In the Brønsted and Lowry classification:
acid proton donor
eg HCl + H2O H3O+ + Cl-
bases proton acceptors
eg HCl + NH3 NH4+Cl-
Salts an ionic compound
product of an acid-base neutralization reaction
typically strong electrolyte
Conjugate Acids and BasesConjugate Acids and Bases
In the Brønsted and Lowry classification: the products of a reaction between an acid and a base are also classified as acids and bases. What does this mean?
Example :
acetic acid methylamine acetate ion methyl-
ammonium ion
CH3 C
O
O H
+ CH3 NH
H
CH3 C
O
O -+ CH3 N
HH
H
+
acid basebase acid
Acetic acid and the acetate ion are said to be a conjugate acid-base pair
Conjugate acids and bases are related to each other by the gain or loss of one H+
AutoprotolysisAutoprotolysis
Some compounds are able to undergo self-ionization, in which they act as both an acid and a base
Example :
H2O H+ + OH-
2 H2O H3O+ + OH-
2 NH3 NH4+ + NH2
-
What compounds undergo autoprotolysis?Compounds with a reactive H+ protic solvent
Examples of protic solvents:
- CH3 CH2 OH (ethanol)
- CH3 COOH (acetic acid)
Compounds without a reactive H+
aprotic solvent (eg CH3 CH2 O CH2 CH3 , CH3
CN)
pHpH
pH -log[H+]
- value can lie from –2 +16
- pH = -1 means -log[H+] = -1.00 or
[H+] = 10M
very concentrated strong acid
-2 0 2 4 6 8 10 12 14 16
neutral
acidic basic
For water : H2O H+ + OH-
Kw = [H+][OH-] = 1.0 x 10-14 at 25oC
log Kw = log [H+] + log[OH-]
- log Kw = -log [H+] - log[OH-]
14.00 = pH + pOH
Strengths of Acids and Strengths of Acids and BasesBases
Acids and bases are commonly classified as strong or weak, depending on whether they react nearly “completely” or only “partially” to produce H+ or OH-
Example of strong acid: HCl(aq) H+ + Cl-
Example of strong base: KOH(aq) K+ + OH-Weak acids and bases react “partially” to produce H+ or OH-
Weak acids (HA) react with water by donating a proton to H2O ie
HA + H2O H3O+ + A-
or
HA + H2O H+ + A- Ka =
[H+][A-]
[HA]
acid dissociation constant
– value is small for weak acids
Weak bases (B) react with water by abstracting a proton from H2O ie
B + H2O BH+ + OH- Kb =[BH+][OH-]
[B]
Base hydrolysis constant
– value is small for weak bases
Acids and BasesAcids and Bases
HOC-COH H+ + -OC-COH
O O O O
-OC-COH H+ + -OC-CO-
O O O O
Compounds that can donate more than one proton polyprotic acids
Compounds that can accept more than one proton polyprotic bases
P
O
O-
O-O- + 3H2O P
O
OHHO+ 3HO-
HO
P
O
O-
O-O- + H2O P
O
OH+ HO-
O-
O-
P
O
O-O- + H2O P
O
OHOH
HO
+ HO-O-
P
O
O- + H2O P
O
OHOH
HO
+ HO-
HO
HO
Kb1 = 1.4 x 10-2
Kb2 = 1.59 x 10-7
Kb3 = 1.42 x 10-12
P
O
O-
O-O- + 3H2O P
O
OHHO+ 3HO-
Kf = Kb1 .Kb2 .Kb3
HO