Chemical Equilibrium

1
Chemical Equilibrium

• What is equilibrium
• Expressions for equilibrium constants, Kc
• Calculating Kc using equilibrium concentrations
• Calculating equilibrium concentrations using
  initial concentration and Kc value
• Relationship between Kc and Kp
• Factors that affect equilibrium
• Le Chateliers Principle

2
What is Chemical Equilibrium?

• Consider the following reactions
• CaCO3(s) CO2(aq) H2O(l) ? Ca2(aq)
  2HCO3-(aq) ..(1)
• and
• Ca2(aq) 2HCO3-(aq) ? CaCO3(s) CO2(aq)
  H2O(l) ..(2)
• Reaction (2) is the reverse of reaction (1).
• If the two reaction occur simultaneously and at
  the same rate, there will be no net change in the
  concentrations of all chemical species in the
  reaction mixture.
• We say that this reaction has attained equilibrium

3
Expression for Equilibrium Constant

• Consider the following reaction at equilibrium
• wA xB ? yC zD
• Kc
• This is the law of mass action
• The numerical value of Kc is calculated using the
  concentrations at equilibrium

4
Expressions for Equilibrium Constants

• Examples
• N2(g) 3H2(g) ? 2NH3(g) Kc
• PCl5(g) ? PCl3(g) Cl2(g) Kc
• CH4(g) H2(g) ? CO(g) 3H2(g)
• Kc

5
Calculating Equilibrium Constantfor reaction
H2(g) I2(g) ? 2HI(g)

• H2(g) I2(g) ? 2 HI(g)
• Initial , M 0.1000 0.1000
  0.0000
• Change in , M -0.0790 -0.0790
  (2 x 0.0790)
• Equilibrium , M 0.0210 0.0210
  0.1580
• Kc 56.6

6
Equilibrium Constant for a Reaction

• Depends on how the equation is written
• Does not depend on how the reaction actually
  occurs.
• Consider the reaction H2(g) I2(g) ?
  2HI(g), in which the actual reaction proceeds
  from right to left,
• H2(g) I2(g) ? 2 HI(g)
• Initial , M 0.00000 0.00000
  0.05000
• Change in , M 0.00525 0.00525
  -(2 x 0.00525)
• Equilm , M 0.00525 0.00525
  0.03950
• Kc (0.03950)2/(0.00525)2 56.6

7
Equilibrium Constant Using Partial Pressures

• Consider the following reaction in gaseous phase
• 2 SO2(g) O2(g) ? 2 SO3(g)
• Kp (PSO3)2/(PSO2)2(PO2)

8
Relationships Between Kc and Kp

• Consider the following reaction
• 2 SO2(g) O2(g) ? 2 SO3(g)
• Kc
• Kp (PSO3)2/(PSO2)2(PO2)
• PSO2 SO2RT PO2 O2RT PSO3
  SO3RT
• This yields Kp Kc(RT)-1

9
Relationship Between Kc and Kp

• Consider a general reaction
• wA xB ? yC zD
• Kp
• PA ART PB BRT PC CRT PD
  DRT
• Kp
• Kp Kc(RT)Dn (where Dn (yz)-(wx))

10
Homogeneous Heterogeneous Equilibria

• Homogeneous equilibria
• CH4(g) H2O(g) ? CO(g) 3 H2(g)
• CO(g) H2O(g) ? CO2(g) H2(g)
• Heterogeneous equilibria
• CaCO3(s) ? CaO(s) CO2(g)
• HF(aq) H2O(l) ? H3O(aq) F-(aq)
• PbCl2(s) ? Pb2(aq) 2 Cl-(aq)

11
Use of Equilibrium Constant

• Knowledge of the equilibrium constant allows us
  to predict
• whether a given reaction mixture is at
  equilibrium
• Qc Kc ? equilibrium (no net reaction)
• direction of net reaction if mixture is not at
  equilibrium
• Qc lt Kc ? a net forward reaction
• Qc gt Kc ? a net reverse reaction
• the extent of the reaction when equilibrium is
  established.
• Knowing Kc and initial concentrations, we can
  calculate concentrations at equilibrium.

12
Calculating Concentrations at Equilibrium

• Consider the following reaction
• H2(g) I2(g) ? 2HI(g), with Kc 55.6
  at 425oC.
• Suppose that the initial concentrations of H2 and
  I2 are 1.000 M and that of HI is 0.000 M. What
  would be the concentrations of H2, I2, and HI at
  equilibrium?
• Solution Assume that when the reaction reaches
  equilibrium, x molar of H2 and I2 were reacted,
  and 2x molar of HI were formed. Then, the
  equilibrium concentrations of H2, I2, and HI are
  (1.000 x), (1.000 x), and 2x molar,
  respectively.
• (see the ICE table for this reaction in the next
  slide)

13
Calculating Concentrations at Equilibrium

• H2(g) I2(g) ? 2 HI(g)
• Initial , M 1.000 1.000
  0.000
• Change in , M -x -x
  2x
• Equilm , M (1.000 x) (1.000 x)
  2x
• Kc HI2/(H2I2) (2x)2/(1.000 x)2
  55.6
• Taking the square-roots of both sides yields
• 2x/(1.000 x) ?(55.6) 7.46
• 2x 7.46 7.46x 9.46x 7.46 ? x
  0.789
• At equilibrium, H2 I2 0.211 M, and HI
  1.58 M

14
What Factors Affect Equilibrium?

• Le Châtelier's principle states that,
• when a system at equilibrium is subjected to a
  change, the system will response by shifting in
  the direction that tends to minimize the change
  while reaching a new equilibrium position.
• Factors that affect equilibrium
• Change in concentrations
• Change in pressure by expansion or compression of
  volume
• Change in temperature.

15
Shift due to Concentration Change

• Consider the reaction N2(g) 3H2(g) ? 2NH3(g)
  
• Equilibrium will shift to the right if either
  N2 or H2 is increased, or NH3 is decreased
• Equilibrium will shift left if either N2 or
  H2 is reduced, or NH3 is increased.

16
Shift due to Pressure Change1

• For the reaction N2(g) 3H2(g) ? 2NH3(g),
• Compression causes pressure build up system
  responses by shifting to the right, decreasing
  the number of molecules and pressure.
• Expansion causes pressure to decrease
  equilibrium shifts left, increasing the number of
  molecules and pressure.
• This type of reactions favor high pressure for
  product formation.

17
Shift due to Pressure Change2

• For the reaction CH4(g) H2O(g) ? CO(g)
  3H2(g),
• Compression will increase pressure equilibrium
  shifts left, reducing the number of molecules and
  pressure.
• Expansion causes pressure to decrease
  equilibrium shifts right, increasing the number
  of molecules pressure.
• This type of reactions favor low pressure for
  product formation.

18
Shift due to Pressure Change3

• For the reaction CO(g) H2O(g) ? CO2(g)
  H2(g),
• There are equal number of gaseous molecules on
  both sides
• Compression or expansion will not cause a shift
  in the equilibrium position
• Stress caused by pressure change cannot be
  relieved by shifting one way or the other.

19
Shift due to Temperature Change

• Consider the following reactions
• N2(g) 3H2(g) ? 2NH3(g) DHo -92 kJ
• CH4(g) H2O(g) ? CO(g) 3H2(g) DHo 206 kJ
• For exothermic reactions, like reaction (1),
  raising the temperature causes a shift to the
  left
• For endothermic reactions, like reaction (2),
  increasing the temperature causes a shift to the
  right
• Exothermic reactions favor low temperature, while
  endothermic reactions favor high temperature for
  products formation.

20
High Pressure of Low Pressure?

• Predict whether the following reactions favor
  high pressures or low pressures (or neither)?
• (a) 2SO2(g) O2(g) ? 2SO2(g)
• (b) CO(g) 2H2(g) ? CH3OH(g)
• (c) PCl5(g) ? PCl3(g) Cl2(g)
• (d) COCl2(g) ? CO(g) Cl2(g)
• (e) H2(g) I2(g) ? 2HI(g)
• (f) 2NOBr(g) ? 2NO(g) Br2(g)

21
High Temperature or Low Temperature?

• Predict whether the following reactions favor
  high temperatures or low temperatures?
• (a) 2SO2(g) O2(g) ? 2SO3(g) DHo -180 kJ
• (b) COCl2(g) ? CO(g) Cl2(g) DHo 108 kJ
• (c) N2O4(g) ? 2NO2(g) DHo 57 kJ
• (d) CO(g) H2O(g) ? CO2(g) H2(g) DHo -41
  kJ
• (e) CO(g) 2H2(g) ? CH3OH(g) DHo -90.5 kJ