Equilibrium
I. Reversible Reactions
A.
Definition and Analogies:
Pigs….Fish…. Fish (stressed)…..
B.
Chemical Equilibrium Example
Chickens…
CO2(g) ó CO2(aq)
CO2(aq) ó H2CO3(aq)
H2CO3(aq) ó H+ + HCO3-(aq)
HCO3-(aq) ó H+(aq) + CO32Ca 2+(aq) + CO32- à CaCO3(s)
C. Equilibrium Simulation:
1. Forward rate potential = Reverse rate potential
[reactants] = [products]
2. Forward rate potential > Reverse rate potential
[reactants] < [products]
3. Forward rate potential < Reverse rate potential
[reactants] > [products]
(very rare)
II. Conditions of Equilibrium
A. Takes place in a
closed system
B. Is dynamic
C. Rate of opposing
reactions are
equal.
III. Le Chatelier’s Principle
Le Chatelier’s
Principle – If a stress
is applied to a system
at Equilibrium, the
Equilibrium is shifted
in a way that tends to
reduce or relieve the
effects of the stress.
A. Changes in concentration
N2(g) + 3H2(g) ßà 2NH3(g)
1. Adding or removing reactants or products
Ex: if you add more N2, there will be more
collisions between N2 and H2, therefore more
NH3 is produced
2. Now relate this to Le Chatelier’s Principle
What if we remove N2? Shifts left
Shifts right
What if we remove NH3?
Shifts right
What if we add H2
B. Effect of Pressure
1. Only relates to gases
2. Increased pressure favors smaller
volumes (moles)
3. N2 + 3H2 ßà 2NH3
1
3
2
=4
Shift to the right, only 2 moles
C. Effect of Temperature
Exo
N2 + 3H2 ßà 2NH3 + Energy
Endo
1. In general, increasing temperature
increases both rxn rates, and decreasing
temperature decreases both rxn rates.
2. Increase temp – favors endothermic rxn
(absorbs heat)
3. Decrease temp – favors exothermic rxn
D. Catalyst
• – speeds up rxn to reach Equilibrium, but
does not favor either forward or reverse
rxn.
IV. General Equilibrium Expression
aA + bB ßà cC + dD
1. Place products in numerator
2. Place reactants in denominator
Keq = C x D – products
A x B – reactants
3. Brackets indicate concentration in moles/liter
Keq = [C] x [D]
[A] x [B]
4. Concentration in raised to the power of
the coefficient used for balancing
Keq = [C]c x [D]d
[A]a x [B]b
Do not include in equilibrium expression:
Liquids (usually water)
Solids
Example: When 1.00 mole of HI is heated to
510 C in a sealed 1-liter flask until
equilibrium is reached, it decomposes to form
0.14 mole of each of the products Hydrogen
gas and Iodide gas. All reactants are in the
gaseous phase. Calculate the equilibrium
constant.
2 HI
H 2 + I2
Initial (I)
1.0 M
0M
0M
Change (C)
-.28 mol +.14 mol +.14 mol
Equilibrium (E)
0.72 mol 0.14 mol 0.14 mol
2 HI
H 2 + I2
Keq = [H2 ] x [I2 ]
[HI]2
Keq = [.14 ] x [.14 ]
[0.72]2
Keq = 0.038
What is the meaning of the size of
the Keq?
Keq is large (>1) = numerator is larger,
large [products] so forward reaction
favored.
Keq is small (<1) = products are less
than reactants, so the reverse reaction
is favored.
Special Types of Equilibrium
• Ka – Equilibrium in Weak Acids
i.e. HC2H3O2(aq) + H2Oßà
ßà H3O+(aq) + C2H3O2(aq)
• Kb – Equilibrium in Weak Bases
i.e. NH3 + H2O ßà NH4+ + OH-
• Ksp – Equilibrium in precipitation
i.e. AgCl(s) ßà Ag+(aq) + Cl-(aq)
Warm Up!
#1. Hydrogen and carbon disulfide react to
form methane and hydrogen sulfide
according to this equation.
Calculate Keq if the equilibrium concentrations
are [H2] = 0.205 M, [CS2] = 0.0664 M, [CH4] =
0.0196 M, and [H2S] = 0.0392 M.
Keq = 0.257
#2 - How would decreasing the
volume of the reaction vessel affect
these equilibria?
shift to the right
shift to the left
Warm Up!
#3. At 350°C, Keq = 66.9 for the formation of
hydrogen iodide from its elements.
What is the concentration of HI if [H2] = 0.0295
mol/L and [I2] = 0.0174 mol/L?
The equilibrium concentration of HI is 0.185 mol/L.
#4. The Ksp for lead(II) fluoride (PbF2) is 3.3
x 10–8 at 25ºC. Use this Ksp value to calculate
the following.
The lead and fluoride ion concentrations in a
saturated solution of PbF2.
2.0 x 10-3 M = [Pb2+]
4.0 x 10-3 M = [F-]