Chemical Equilibrium

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Chemical Equilibrium
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Chemical Equilibrium

• When a reaction is not totally converted from
  reactants to products, a condition can be set up
  known as chemical equilibrium.
• In this state, the concentrations of all
  reactants and products remain constant with time,
  as both forward and reverse reactions are
  occurring at the same rate.
• On the molecular level, there is frantic
  activity. Equilibrium is not static, but is a
  highly dynamic situation.

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The Haber Process

• The most common reversible reaction known is the
  Haber process of making ammonia from nitrogen and
  hydrogen
• N2 (g) 3H2 (g) 2NH3 (g)
• This reaction is very slow in either direction
  (implying a large Ea for both)
• The reaction can be influenced to increase in
  either direction if the conditions are right

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The Law of Mass Action

• Gundberg and Waage developed the equilibrium
  expression from the law of mass action.
• For the reaction
• jA kB ? lC mD
• The law of mass action is represented by the
  equilibrium expression

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Equilibrium Expression

• 4NH3(g) 7O2(g) ? 4NO2(g) 6H2O(g)

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Notes on Equilibrium Expressions (EE)

• The Equilibrium Expression for a reaction is the
  reciprocal of that for the reaction written in
  reverse.
• When the equation for a reaction is multiplied by
  n, EEnew (EEoriginal)n
• The units for K depend on the reaction being
  considered.

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Examples 13.2

• Calculations can be made if the equlibrium
  concentations of reactants and products are
  known.
• If NH3 3.1 x 10-2 mol/L, N28.5 x 10-1
  mol/L, and
• H2 3.1 x 10-3 mol/L,
• N2 (g) 3H2 (g) 2NH3 (g)
• a. What is the value of K?
• b. What is the value of K (for the reverse
  reaction)?
• c. What would be the K for this reaction?
• ½ N2 (g) 3/2 H2 (g) NH3 (g)
• (3.1 x 10-2 )2
• (8.5 x 10-1 )(3.1 x 10-3 )3 3.8 x 104
  L2/mol2
• K 1/K 1/ 3.8 x 104 2.6 x 10-5 mol2/L2
• For the reaction in (c), Kc K ½ (3.8 x 104
  L2/mol2) ½ 1.9 x 102 L/mol

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K v. Kp

• Equilibrium constants can be calculated using
  partial pressures of gases in the reaction
• For
• jA kB ? lC mD
• Kp K(RT)?n
• ?n sum of coefficients of gaseous products
  minus sum of coefficients of gaseous reactants.

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Example 13.4 5

• 13.4 The reaction forming nitrosyl chloride is
• 2 NO (g) Cl2 (g) 2 NOCl (g)
• If the pressures at equilibrium are PNOCl 1.2
  atm,
• PNO5.0 x 10-2 atm, and PCl2 3.0 x 10-1
  atm, what is the value of Kp?
• Kp PNOCl2 (1.2)2
  1.9 x 103 atm-1
• (PNO )2(PCl2) (5.0 x 10-2)2(3.0 x 10-1)
• 13.5What is the value of K from the above value
  of Kp at 25º C?
• In the reaction, there are 3 mol of gas reactants
  and 2 of gas products, so Dn -1.
• Since Kp K(RT)n K(RT)-1 and therefore KKp
  (RT)
• 1.9 x 103(.0821)(298) 4.6 x 104 L/mol

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Heterogeneous Equilibria

• . . . are equilibria that involve more than one
  phase.
• CaCO3(s) ? CaO(s) CO2(g)
• K CO2
• The position of a heterogeneous equilibrium does
  not depend on the amounts of pure solids or
  liquids present, so only gases and/or aqueous
  species are included in the expression.

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Example 13.6

• For the reaction PCl5 (s) PCl3 (l) Cl2
  (g), what are the expressions?
• K Cl2 and Kp PCl2
• For the reaction
• CuSO4 5 H2O (s) CuSO4 (s) 5 H2O (g)
• What are the equilibrium expressions?
• KH2O5 Kp PH2O5

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Reaction Quotient

• If the law of mass action is applied with initial
  concentrations, the expression becomes a value
  called the reaction quotient (Q). This helps to
  determine the direction of the move toward
  equilibrium.
• If Q gt K then the reaction will favor reactants
• If Q lt K then the reaction will favor products
• If QK the reaction is at equilibrium already

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Reaction Quotient (continued)

• H2(g) F2(g) ? 2HF(g)

Compare Q with K to determine direction Q gt K
to left Q lt K to right Q K at Equilibrium
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Example 13.7

• If the value of K for the Haber process reaction
  at 500º C is 6.0 x 10-2 L2/mol2 , predict the
  direction each of these will proceed from the
  initial conditions
• NH30 1.0 x 10-3 M N201.0 x 10-5 M H20
  2.0 x 10-3 M
• b. NH30 2.00 x 10-4 M N201.50 x 10-5 M
  H20 3.54 x 10-1 M
• c. NH30 1.0 x 10-4 M N205.0 M H20
  1.0 x 10-2 M
• Q NH302 / N20 H203 (1.0 x 10-3 M)2 /(1.0
  x 10-5 M)(2.0 x 10-3 M)3
• 1.3 x 107 This is much greater than K, so
  the equilibrium will shift toward reactants N2
  and H2
• Q (2.00 x 10-4 M)2 / (1.50 x 10-5 M)(3.54 x 10-1
  M)3 6.01 x 10-2
• This is equal to K, so the reaction is already
  at equilibrium
• Q (1.0 x 10-4 M)2 / (5.0 M)(1.0 x 10-2 M)3
  2.0 x 10-3 L2/mol2
• This is less than the value of K, so the
  reaction will shift toward the formation of
  products

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Application of Equilibria

• The extent of a reaction can be expected from the
  value of K If Kgt 1 products predominate, if Klt
  1 reactants predominate.
• The value of K is also related to DG, since DG -
  RT ln K . If Kgt1, ln K gt0 and DGlt0, thus
  spontaneous. If Klt1 , ln K lt0 and DG gt0 thus
  non-spontaneous

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Homework 13-a

• p. 630ff 9, 13, 14, 15, 23, 25

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Solving Equilibrium Problems From Initial
Conditions

• 1. Balance the equation.
• 2. Write the equilibrium expression.
• 3. List the Initial concentrations.
• 4. Calculate Q and determine the shift (Change)
  to equilibrium.

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Solving Equilibrium Problems(continued)

• 5. Define Equilibrium concentrations.
• 6. Substitute equilibrium concentrations into
  equilibrium expression and solve.
• 7. Check calculated concentrations by calculating
  K.
• Take note of ICE calculations

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Example 13.9
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Example 13.10

• Carbon monoxide reacts with steam to produce
  carbon dioxide and hydrogen. At 700K the value of
  K is 5.10. Find equilibrium concentrations if
  1.00 mol of each component is mixed in a 1.00 L
  flask.

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Example 13.12
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Example 13.13For Small Values of K
We will use this approximation later in
Acids-Bases
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Le Châteliers Principle

• . . . 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.

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Effects of Changes on the System

• Concentration The system will shift away from
  the added component or toward a removed
  component.
• Temperature K will change depending upon the
  type of reaction. Since exothermic produces heat,
  it will shift away from added, endo the opposite.

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Example 13.14

• Arsenic can be extracted from its ores by first
  reacting the ore with oxygen to form solid As4O6,
  which is then reduced with carbon
• As4O6 (s) 6 C (s) As4 (g) 6 CO (g)
• Predict the effect of these changes upon the
  position of equilibrium
• a. Adding carbon monoxide
• b. Adding or removing C or As4O6
• c. Removing gaseous As4
• LeChateliers principle states that addition
  shifts away from what is added, so (a) shifts
  toward the reactants, that is left. Solids being
  added have no effect, so (b) remains the same.
  Removing any component shifts toward what is
  removed, so (c) shifts right.

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Effects of Changes on the System (continued)

• 3. Pressure
• a. Addition of inert gas does not affect the
  equilibrium position.
• b. Decreasing the volume shifts the
  equilibrium toward the side with fewer moles.

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Example 13.15 16

• 15Predict the shift in equilibrium when volume
  is reduced in each of these(causing increased
  pressure)
• P4 (s) 6 Cl2 (g) 4 PCl3 (l)
• Since there are no moles of gas on the right,
  volume reduction will favor that side.
• b. PCl3 (g) Cl2 (g) PCl5 (g)
• Since there are fewer moles of gas on the right,
  the equilibrium will shift that direction.
• c. PCl3 (g) 3 NH3 (g) P(NH2)3 (g) 3 HCl (g)
• Since there are 4 moles on either side of the
  equation, there is no shift due to pressure
  change.
• 16For each of these reactions, predict the shift
  of equilibrium with an increase in temperature
• N2 (g) O2 (g) 2NO (g) DHº 181 kJ
• This reaction is endothermic, so heat acts as
  an added reactant, favoring the products to the
  right
• 2 SO2 (g) O2 (g) 2 SO3 (g)
  DHº -198 kJ
• This reaction is exothermic, heat a product, so
  it shifts to the left.

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Summary of LeChatelier
Concentration Equilibruim shifts away from what
is added, toward what is removed. Temperature
Equilibrium shifts away from added temperature,
depending on type of reaction, Exo-toward
reactants Endo-toward products Pressure
Equilibrium shifts toward less molecules when
pressure is added
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Homework 13-b

• P.631ff 22, 28 adf, 31, 32, 34, 35, 36, 38

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Common Ion Effect

• The shift in equilibrium that occurs because of
  the addition of an ion already involved in the
  equilibrium reaction.
• AgCl(s) ? Ag(aq) Cl?(aq)

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Solubility Product

• For solids dissolving to form aqueous solutions.
• Bi2S3(s) ? 2Bi3(aq) 3S2?(aq)
• Ksp solubility product constant
• and
• Ksp Bi32S2?3

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Solubility Product

• Solubility s concentration of Bi2S3 that
  dissolves, which equals 1/2Bi3 and 1/3S2?.
• Bi2S3 (s) ltgt 2 Bi3 (aq) 3 S2?(aq)
• 2 s 3 s
• Note Ksp is constant (at a given temperature)
• s is variable (especially with a common
• ion present)

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Relation of Ksp to s

• Ionic compounds will dissociate according to set
  equations, which will give set Ksp to s relation.
  Use ICE to find it.
• A2B ltgt 2A B
• 2s s Ksp (2s)2(s)4s3
• AB3 ltgt A 3B
• s 3s Ksp (s)(3s)327s4

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Ksp from s
If 4.8 x 10-5 mol of CaC2O4 dissolve in 1.0 L of
solution, what is its Ksp? CaC2O4 mol / L
4.8 x 10-5 / 1.0 L 4.8 x 10-5 M CaC2O4 (s)
ltgt Ca2 (aq) C2O4-2 (aq) s s Ksp
(s)(s) s2 (4.8 x 10-5 )2 2.3 x 10-9 If the
molar solubility (s) of BiI3 is 1.32 x 10-5, find
its Ksp. BiI3 (s) ltgt Bi3 3I- s
3s Ksp (s)(3s)3 27 s4 Ksp 27(1.32 x
10-5)4 8.20 x 10-19
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s from Ksp
If Ksp for Ag2CO3 is 8.1 x 10-12, find its molar
solubility. Ag2CO3 (s) ltgt 2 Ag (aq) CO3-2
(aq) 2 s s Ksp (2s)2(s) 4 s3 8.1 x
10-12 s 1.27 x 10-4 If Ksp for Al(OH)3 is 2 x
10-32, find its molar solubility. Al(OH)3 (s)
ltgt Al3 (aq) 3 OH- (aq) s 3 s
Ksp (s)(3s)3 27s4 2 x 10-32 s 5.22 x
10-9
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Equilibria Involving Complex Ions

• Complex Ion A charged species consisting of a
  metal ion surrounded by ligands (Lewis bases).
• Coordination Number Number of ligands attached
  to a metal ion. (Most common are 6 and 4.)
• Formation (Stability) Constants The equilibrium
  constants characterizing the stepwise addition of
  ligands to metal ions.

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Complex Ion Examples

• Coordination number2
• Ag(NH3)2
• Coordination number4
• Al(OH)4-1 Cu(NH3)42
• Coordination number6
• Fe(CN)6-4 Fe(SCN)6-3

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Homework and XCR

• p. 762ff 50, 51, 59, 60
• XCR 72, 76, 86
• Super XCR 90