Chapter 6 General Concepts of Chemical Equilibrium · 2013. 11. 6. · Chapter 6 General Concepts...
Transcript of Chapter 6 General Concepts of Chemical Equilibrium · 2013. 11. 6. · Chapter 6 General Concepts...
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Chapter 6
General Concepts of Chemical Equilibrium
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Introduction Chemical Equilibrium
Ø The state where the concentrations of all reactants and products remain constant with time.
Ø Equilibrium is not static, but is a highly dynamic situation.
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Chemical Reactions: The rate Concept
Ø A + B C + D ( forward) Ø C + D A + B (reverse) Ø Initially there is only A and B so only
the forward reaction is possible Ø As C and D build up, the reverse
reaction speeds up while the forward reaction slows down.
Ø Eventually the rates are equal
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Rea
ctio
n R
ate
Time
Forward Reaction
Reverse reaction
Equilibrium Dynamic?
Rate forwards = Rate reverse
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The equilibrium constant Law of Mass Action
Ø For any reaction Ø aA + bB cC + dD Ø [C]c[D]d
[A]a[B]b Ø K is called the equilibrium constant. Ø is how a reversible reaction is
identified
K =
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v Comments on Law of mass action v K is constant regardless of the amounts
of materials mixed initially v Equilibrium concentrations will not always be the same but K is the same v Each set of equilibrium concentrations
in an equilibrium system is called equilibrium position
v There is only one K value for a given system but infinite number of equilibrium positions
v The law of mass action applies to solution and gaseous equilibria
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Changing the chemical equation of an equilibrium system: Reciprocal rule
Ø If we write the reaction in reverse. Ø cC + dD aA + bB Ø Then the new equilibrium constant is
Ø K’ reversible= forwardK1
[D][C] [B][A]
dc
ba
= Reciprocal rule
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Multiplying the equation by a coefficient: Coefficient Rule
Ø If we multiply the equation by a constant
Ø naA + nbB ncC + ndD Ø Then the equilibrium constant is
Ø K’ = [C]nc[D]nd = ([C] c[D]d)n = Kn [A]na[B]nb ([A]a[B]b)n
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Rules of Multiple Equlibria Ø Reaction 3 = Reaction 1 + Reaction 2
A2 B A + AB K1 = 2.2 AB A + B; K2 = 4.0
A2B 2A + B
K3=2.2 X4.0 = 8.8
][][][
2
2
BABA
K3 = K1XK2
K3 = = K1XK2
Ø K (Reaction 3) =K (reaction 1) X K (reaction 2)
][]][[
21 BA
ABAK =
][]][[
2 ABBAK =
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Notes on Equilibrium Expressions v The Equilibrium Expression for a reaction is
the reciprocal of that for the reaction written in reverse.
v When the equation for a reaction is multiplied by n, the equilib expression changes as follows:
v (Equilib Expression) final = (Equilib Expression initial)n v Usually K is written without units
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Calculation of K
Ø N2 + 3H2 2NH3 Ø Initial At Equilibrium Ø [N2]0 =1.000 M [N2] = 0.921M Ø [H2]0 =1.000 M [H2] = 0.763M Ø [NH3]0 =0 M [NH3] = 0.157M = 9.47X10-3
322
23
]][[][HN
NHK =
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Ø N2 + 3H2 2NH3 Ø Initial At Equilibrium Ø [N2]0 = 0 M [N2] = 0.399 M Ø [H2]0 = 0 M [H2] = 1.197 M Ø [NH3]0 = 1.000 M [NH3] = 0.157M Ø K is the same no matter what the
amount of starting materials
Calculation of K
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Homogeneous Equilibria Ø All reactants and products are in
one phase, gases for example Ø K can be used in terms of either
concentration or pressure.
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Heterogeneous Equilibria Ø If the reaction involves pure solids or pure
liquids as well as gases, the concentration of the solid or the liquid doesn’t change.
Ø As long as they are not used up they are left out of the equilibrium expression.
Ø Thus, there is no term for L or S in “K” expression.
Ø However, the presence of L or S is a must for equilibrium to occur.
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Example: Equilibrium expression for heterogeneous equilibria
Ø H2(g) + I2(s) 2HI(g) Ø
Ø But the concentration of I2 does not change.
Ø
]][[]['22
2
IHHIK =
][][]['2
2
2 HHIIK = K=
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Ø The magnitude of K helps prediction of the feasibility (extent or direction but not the speed) of the reaction
Ø K> 1; the reaction system consists mostly products (equilibrium mostly lies to the right) Ø Systems with very large K go mostly to
completion Ø Systems with very small values of K do not
occur to any significant extent Ø There is no relation between the value of K and
the time to reach equilibrium (the rate of reaction) Ø Time to reach equilibrium depends on Ea for reactants and products
Applications of the equilibrium Constant
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The Reaction Quotient, Q (Quantitative prediction of direction of reaction)
Ø Q Tells how the direction of a reaction will go to reach equilibrium
Ø Q’s are calculated the same as K’s, but for a system not at equilibrium
Ø [Products]coefficient
[Reactants] coefficient
aA(g) + bB(g) cC(g) + dD(g)
Q =
bB
aA
dD
cC
PXPPXPQ)()()()(
=
Ø Compare value of Q to that of K
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What Q tells us? Ø If Q<K
Ø Not enough products Ø Equilibrium shifts to right; forward
reaction is predominant Ø If Q>K
Ø Too many products Ø Equilibrium shifts to left; reverse
reaction is predominant Ø If Q=K system is at equilibrium; there
is no further change
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Le Chatelier’s Principle Ø if a change is imposed on a system
at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.
Ø If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress.
Ø There are 3 Types of stress
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External conditions that cause a disturbance to a chemical equilibrium
Ø Adding or removing reactants or products
Ø Changing the volume (or pressure) of the system
Ø Changing the temperature
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The effect of a change in concentration of reactants and/or products
Ø Adding product makes Q>K Ø Removing reactant makes Q>K Ø Adding reactant makes Q<K Ø Removing product makes Q<K Ø knowing the effect on Q, will tell you
the direction of the shift Ø Adding or removing liquids or solids
does not affect the equilibrium
Ø The system will shift away from the added component
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Ø The pressure changes as a result of: Ø Adding or removing gaseous reactant or product Ø Changing the volume of the container
Ø By reducing the volume of the container, the system will move in the direction that reduces its volume.
The effect of a Change in Pressure
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Changes in volume
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Changes in Temperature Ø Affects the rates of both the
forward and reverse reactions. Ø changes the equilibrium constant. Ø The direction of the shift depends
on whether it is exo- or endothermic
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Exothermic Ø DH<0 Ø Releases heat Ø heat is a product Ø Raising temperature shifts the
reaction direction toward reactants (to the left)
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Endothermic Ø DH>0 Ø Heat is added to the system Ø Heat as a reactant Ø Raising temperature shifts the
direction of reaction toward products (to the right)
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Example Ø Consider: A + B C + D At 25 oC, K is 0.30. Calculate the equilibrium
concentrations of all species if 0.20 mol of A is reacted with 0.5 mol of B and dissolved in 1.00 L flask
A + B C + D Initial(mol) 0.2 0.5 0 0 [ ] (mol/L) 0.2 0.5 0 0
Change(mol/L-1) -x -x +x +x Equilib 0.2-x 0.5-x +x +x
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3.0)1()(
)5.0)(20.0())((
2
2
=−
=−−
=xx
xxxxK
3.0)1()(
)5.0)(20.0())((
2
2
=−
=−−
=xx
xxxxK
0.70 x2
+ 0.21 X - 0.030 = 0
aacbbx
242 −±−
−=
X = 0.11 M
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Example
Ø Consider: A + B C + D At 25 oC, K is 2.0x1016. Calculate the equilibrium
concentrations of all species if 0.20 mol of A is reacted with 0.5 mol of B and dissolved in 1.00 L flask
• Since K is very large, the reaction of A with B will be virtually complete to the right
• Thus, only traces of A will be left at equilibrium. • Let x represent the equilibrium concentration of A.
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Example Assume 0.10 mol of A is reacted with 0.20 mol of B in a
volume of 1000 mL; K = 1.0 X 1010. What are the equilibrium concentrations of A, B, and C
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Dissociation Equilibria Calculate the equilibrium concentrations of A and B in a 0.10 M solution of a weak electrolyte AB with an equilibrium constant of 3.0 X 10-6.
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The Common Ion Effect-Shifting the Equilibrium
Calculate the equilibrium concentrations of A and B in a 0.10 M solution of a weak electrolyte AB with an equilibrium constant of 3.0 X 10-6 and 0.20 molar B
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Systematic approach to equilibrium calcualations This approach involves writing expressions for mass balance of species and one for charge balance of Species Mass Balance
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Example
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Charge balance equations principle of electroneutrality
• All solutions arc electrically neutral that is, there is no solution containing a detectable excess of positive or negative charge
• The sum of the positive charges equals the sum or negative charges
• A single charge balance equation lor a given set of equilibria.
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Example
The total charge concentrations from all sources is always equal to the net equilibrium concentration of the species multiplied by its charge
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Example
Write a charge balance expression for a solution containing KCI, AI2(SO4)3 and KNO3. Neglect the dissociation of water.
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Example
Write the charge balance equation for a solution of [Ag(NH3)2]Cl
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Example Calculate the equilibrium concentrations of A and B In a 0.10 M solution of a weak electrolyte AB with an equilibrium constant of 3.0 X 10-6. Use the mass charge balance approach
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Example
Calculate the equilibrium concentrations of A and B In a 0.10 M solution of a weak electrolyte AB with an equilibrium constant of 3.0 X 10-6. Use the mass charge balance approach. Assume the charge A is + I, the charge on B is - I, and that the extra B (0.20M) comes from MB; MB is completely dissociated.
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