Equilibrium
• All reactions are reversible
• In labs, many products of chem.reactions
can be directly reacted to form the reactants
• Sometimes the conditions for the reaction to
occur are similar for the forward and
reverse reactions, other times they are
different
– Ex. Fe3O4 + 4H2  3Fe + 4H2O(l)
3Fe + 4H2O(g)  Fe3O4 + 4H2
If steam is not allowed to escape
• Both reactions are taking place
simultaneously at different rates
• Eventually the two rates are equal
• Chemical Equilibrium
• Write one equation with a double arrow
Characteristics
• State of balance between opposing changes can be
physical
• Vapor-liquid
• Changes can be chemical
• The original changes before an equilibrium exists
are observable
• Changes in
• concentration
• Temperature
• State
• Definite relationship between
concentrations of reactants and products
when reversible reaction is in Equilibrium
• Example:
H2 (g) + I2(g) D 2HI(g) at 4900C
[H2]= 1mol/dm3 [I2]= 1mol/dm3
 When H and I come into contact, reaction starts
 Equil.is achieved when reaction rate = product rate
• The relationship between Reactants and
Products is expressed as Keq and can be
expressed as:
•
Kc(eq) = [Products]
[Reactants]
• Coefficients in the balanced equation are
used as superscripts in the equation
• The Equilibrium Constant (Kc) will not
change in a closed system
• It is affected by changes in temperature
• What does the Kc value determine?
• Size of value determines the extent a
reaction proceeds to the right hand side of a
reversible reaction before an equilibrium is
achieved
• A constant value <1 indicates little product
forming
• Constant >1 shows product formation
favored
• The higher the value, the greater the amt.of
product
• Reactions are sensitive to changes in conc.,
temp. and press.
• A change in one of these places stress on the
system
• Stress causes the equil. to Shift
• A new equil. is eventually achieved with the
new factors
• LeChatlier’s Principle
• LeChatelier’s Principle
– System in equilibrium subjected to stress(
change in con., temp. or press.) the equilibrium
will shift in the direction that tends to
counteract the effect of the stress
Concentration changes
A+BD C+D
Add A more AB collisions
Forward reaction faster
Equil. No longer exists
B consumed and its conc. Decreases the
reaction slows
• C + D conc. Increases and its collisions
increase and its reaction begins to speed up
• A new equilibrium is established
•
•
•
•
•
Temperature changes
•
endo
N2(g) + 3H2(g) D
2NH3(g) + 92kJ
exo
Haber Process
• Every equilibrium has an endo and an exo component
 temp. stress countered by favoring the
endothermic reaction since it absorbs heat
$temp. stress countered by favoring the
exothermic reaction as heat is given off
Pressure changes
• Applies mostly to reactions with gases
endo
N2(g) + 3H2(g) D
2NH3(g) + 92kJ
exo
Haber Process
• Increase pressure- favors side with fewer moles ( # of
particles) of substance
• Decrease pressure favors side with more moles of
substance
• Calculate an equilibrium constant K
• Don’t always know conc. of all substances
• Knowing conc. Of one, can stoichiometrically
determine others
• 1. Tabulate known initial and equil.conc. for all
species in the equilibrium
• 2. For species with known conc. of initial and
equilibrium conc., calculate change in conc. as
reaction approaches equilibrium
• 3. Use stoich to determine conc. of all others
• 4. From init. and changes in conc. Calculate the
equilibrium concs. Evaluate Kc
• Example: a mixture of 5.000x10-3 mol H2
and 1.000x10-2 mol I2 is placed in a 5.000L
container at 448oC and allowed to come to
equilibrium. Analysis of the equil. mixture
shows the conc. of HI is 1.87x10-3 M.
Calculate Kc at 448oC for the reaction.
• First: Calculate concentrations of all
reactants and products
• [H2]init = 5.000x10-3 mol = 1.000x10-3 M
5.000L
• [I2]init = 1.000x10-2 mol = 2.00x10-3 M
5.000L
•
•
•
•
Place above values into a ICE table
I initial conc.
C change in conc.
E equilibrium conc.
•
H2(g) +
Initial
1.000X10-3 M
Change
Equilibrium
I2(g) D
2.00x10-3 M
2HI(g)
0M
1.87x10-3M
• Second: calculate change conc. of HI using
initial and equil. values
• Change is +1.87x10-3 M
• Third: use stoichiometry to calculate
changes in other substances
• (1.87x10-3 mol HI)(1 mol H2) =
L
2 mol HI
• 0.935x10-3 mol H2/L
• The value for I2 is the same as H2
•
H2(g) +
Initial
1.000x10-3 M
Change
-0.935x10-3 M
Equilibrium
I2(g) D
2HI(g)
2.000x10-3 M
0M
-0.935x10-3 M +1.87x10-3 M
1.87x10-3 M
• Fourth: calculate equil. conc. using initial
conc. and changes
• Equilibrium conc. H2 is init. minus
consumed
• [H2] = 1.000x10-3 M – 0.935x10-3 M =
0.065x10-3 M
• [I2] = 2.00x10-3 M – 0.935x10-3 M =
1.065x10-3 M
•
H2(g) +
I2(g) D
2HI(g)
Initial
1.000x10-3 M
2.000x10-3 M
0M
Change
-0.935x10-3 M -0.935x10-3 M +1.87x10-3 M
Equilibrium 0.065x10-3 M 1.065x10-3 M 1.87x10-3 M
• Finally: now know equilibrium concentrations of reactants
and products
• Use equilibrium expression to calculate Kc
• Kc = [HI]2 =
(1.87x10-3)2
[H2][I2]
(0.065x10-3)(1.065x10-3)
= 51
• For solutions that have solutes dissolved in
water, the same principle can be applied
• Ksp is the solubility constant
• Ksp = [Products]
• Example:
• MgCl2  Mg+2(aq) + 2Cl-1(aq)
• Ksp = [Mg+2][Cl-1]2
K of acids and bases
•
•
•
•
Ka Kb
Ka = [H+][cation] = [H+]2
Kb = [anion][OH-] = [OH-]2
Strong acids and bases dissociate completely into
H+ or OH- and use above equations
• Weak acids and bases do not and are similar to
equilibrium equations
• Ka or Kb = [products]
[reactants] use ICE chart