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Chapter 6 General Concepts of Chemical Equilibrium
Chapter 6
General Concepts of Chemical Equilibrium
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.
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
Rea
ctio
n R
ate
Time
Forward Reaction
Reverse reaction
Equilibrium Dynamic?
Rate forwards = Rate reverse
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 =
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
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
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
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 =
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
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 =
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
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.
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.
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=
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
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
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
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
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
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
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
Changes in volume
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
Exothermic Ø DH<0 Ø Releases heat Ø heat is a product Ø Raising temperature shifts the
reaction direction toward reactants (to the left)
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)
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
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
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.
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
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.
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
Systematic approach to equilibrium calcualations This approach involves writing expressions for mass balance of species and one for charge balance of Species Mass Balance
Example
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.
Example
The total charge concentrations from all sources is always equal to the net equilibrium concentration of the species multiplied by its charge
Example
Write a charge balance expression for a solution containing KCI, AI2(SO4)3 and KNO3. Neglect the dissociation of water.
Example
Write the charge balance equation for a solution of [Ag(NH3)2]Cl
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
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.
