Chemical Equilibrium

1
Chapter 6

• Chemical Equilibrium

2
Chemical Equilibria in the Environment
pH7.2
pH4.5
3
Equilibria

• Equilibria governs all chemical reactions
• In Chapter 6, we look at
• Solubility of ionic compounds
• Complex formation
• Acid-base reactions

4
The Equilibrium Constant

• The equilibrium constant, K for the reaction,
• when K gt1, the reaction is favored
• Each species is expressed as the ratio to its
  concentration in its standard state
• What numbers do we use?
• Solutes in M
• Gases in bars
• Solids, liquids, solvents are omitted because 1

5
Manipulating Equilibrium Constants

• For a reaction,
• For the reverse reaction,

6
Manipulating Equilibrium Constants

• For the addition of two reactions,
• The equilibrium constant is the product

7
Example

• For the reaction,
• If for
• What is the equilibrium constant for

8
Thermodynamics

• The equilibrium constant derives from
  thermodynamics
• Whether a reaction is favored or not depends upon
  two parameters
• Enthalpy, the heat absorbed or released during a
  reaction
• Entropy, the degree of disorder among the
  products and reactants

9
Enthalpy

• The enthalpy change, ?H, for a reaction is the
  heat absorbed or released under constant pressure
• Standard enthalpy change, ?H, is heat absorbed
  when all species are in standard states
• A negative sign for ?H indicates that heat is
  released during the reaction, or is exothermic.
• A positive sign for ?H indicates that heat is
  absorbed, or is endothermic. This means that the
  solution gets colder.

10
Entropy

• Entropy, S, of a species is a measure of its
  disorder. A gas is more disordered than a
  liquid, a liquid more than a solid, and ions in
  solution are more disordered than their salt.
• ?S is the change in entropy for a reaction when
  all species are in their standard states. A
  positive value means products are more disordered
  than reactants.

11
Free Energy

• Since enthalpy, ?H, and entropy, ?S, both exert
  an influence on whether a reaction proceeds, one
  must account for both to determine whether the
  reaction is favored. We use the change in Gibbs
  free energy, ?G for this
• If ?Glt0, then the reaction is favored.

12
Free Energy Example

• What is the Gibbs free energy change for the
  dissociation of HCl when all species are in their
  standard states?
• Since ?G lt0, then the reaction is favored.

13
How Does Free Energy Relate to Equilibrium?

• The equilibrium constant of a reaction, K, can be
  related to free energy by
• where R is the gas constant (8.314 J/molK) and T
  is temperature in K.

14
Example

• What is the equilibrium constant for the
  dissociation of HCl?
• The equilibrium constant is large, so HCl(g) is
  very soluble in water and nearly completely
  ionized. If ?G lt0 (or Kgt1) we say that the
  reaction is spontaneous.

15
Le Châteliers Principle

• If a system at equilibrium is changed, Le
  Chatliers principle states that the system will
  proceed in the direction to offset the change.
• One set of concentrations that exists in
  equilibrium is
• H5.0 M Cr2O72-0.10 M Cr30.0030 M
• Br-1.0 M BrO3-0.043 M

16
Le Châteliers Principle

• If we increase Cr2O72- to 0.20 M, what
  direction will the reaction proceed?
• Le Châtelier says to the left. Lets confirm
  this
• Q is the reaction quotient--the same as K except
  when the system is not at equilibrium
• Since Q gt K, the reaction must indeed go to the
  left to reduce the numerator and decrease the
  denominator

17
Effects on Equilibrium

• If products are added or reactants are removed,
  the reaction goes left
• If reactants are added or products are removed,
  the reaction goes right
• K is also dependent on temperature
• For an endothermic reaction (?Hgt0), K increases
  if the temperature increases
• For an exothermic reaction (?Hlt0), K decreases
  if the temperature increases

18
Thermodynamics

• Remember that equilibrium equations make
  thermodynamic predictions about how a chemical
  system will proceed.
• We calculate what must happen for the system to
  reach equilibrium
• Equilibrium equations cannot say anything about
  how fast the reaction proceeds. The rate or
  kinetics of the move to equilibrium may be fast
  or extremely slow.

19
Solubility Product

• The solubility product is the equilibrium
  constant for a reaction in which a solid
  dissolves and gives ions in solution.
• When a solution has excess, undissolved solid,
  the solution is saturated.

20
Solubility Product Example

• What is the concentration of Hg22 in a solution
  saturated with Hg2Cl2?
• Every molecule of Hg2Cl2 that dissolves produces
  1 Hg22 ion in solution and 2 Cl-, so if xHg2
  than Cl- must equal 2x.

21
Example with a 2nd Source of Ions

• What will the concentration of Hg22 in a
  solution containing 0.030 M NaCl saturated with
  Hg2Cl2?
• So what do you do here? Do you want to solve
  this equation?

22
Example 2 (Cont.)

• We learned in the first example that Cl- from
  saturated Hg2Cl2(aq) was 6.7x10-7 M. Weve added
  Cl-, and based on Le Châtliers principle, we
  would expect even less Cl- to be soluble. Lets
  assume that 2x is small compared to 0.030 M.
• Our assumption was valid 2x 2.6 x10-15 M which
  is much, much smaller than 0.030 M

23
When to Approximate?

• Approximations are a legitimate way to solve
  equilibria problems that would difficult
  otherwise. But, you must always confirm your
  assumptions with the final result.
• If you have to solve the equation and cant
  figure out how to do it, trial-and-error guessing
  can often be used

24
Trial-and-Error Guessing

• For the original question we had
• A series of guesstimates gives the following
• x0.01 2.5x10-5 (way too big)
• x0.00001 9.0x10-9 (still too big)
• x1x10-13 9.0x10-17 (too big)
• x1x10-15 9.0x10-19 (too small close!)
• Note if a specific level of accuracy is
  specified (e.g. to 1) you will have to make a
  lot of guesses!

25
Example 3

• What is the maximum concentration of Cl- at
  equilibrium in a solution where Hg22 is fixed
  at 1.0x10-9 M?

26
Separations by Precipitation
Precipitation reactions can be used to separate
and identify ions in solution. This has
historically been a very important analytical
chemistry technique. This is an important and
quick way to identify metal ions in particular.
Pb(NO3)2 soln.
PbI precipitate
KI soln.
27
A Metals Analysis Scheme
K, NH4, Ca2, Ba2, Zn2, Hg2, Al3, Pb2,
Cr3, Fe3, Ag
K, NH4, Ca2, Ba2, Zn2, Al3, Cr3
Hg2, Pb2, Fe3, Ag
NH3(aq)
Pb2, Ag
Hg2, Fe3
K, NH4, Ca2, Ba2
Zn2, Al3, Cr3
SO42-
SO42-
Excess NH3(aq)
CO32-
No Rxn
No Rxn
ppt Rxn
ppt Rxn
ppt Rxn
Green ppt.
dissolves
No Rxn
White ppt.
Zn2
Pb2
Ag
Fe3
Hg2
Cr3
K, NH4
Ca2, Ba2
Al3
OH-
No Rxn
ppt Rxn
Ca2
Ba2