CHEMICAL EQUILIBRIUM: Kc, Kp, and Ksp
I.
Chemical equilibrium
A. Definition: the state in which the concentrations of all reactants and
products remain constant with time. (dynamic process)
B. Law of mass action (equilibrium expressions)
1. xA (aq) + yB (aq)  zAB (aq)
K = [AB]z / [A]x [B] y
* always products / reactants
2. In equilibrium expressions:
always include gases (g) and aqueous(aq)
do not include solids (s) and pure liquids (l)
3. If an equation is reversed, the new value for K is 1/K.
4. When the balanced equation for a reaction is multiplied by a factor
of n, the new equilibrium constant is K n . K new = (Korig)n
C. Relationship between K and equilibrium:
1. Large K values- equilibrium lies to the right (favors the products)
2. Small K values- equilibrium lies to the left (favors the reactants)
D. Calculations involving Kc, Kp, and Ksp
1. Kc= Keq= K: use when given concentrations (molarity) [ ]
2. Kp: use when given pressures (atm)
3. Ksp: use when dealing with solid salts breaking apart into ions
4. When given concentrations and looking for Kp, or when given pressures
and looking for Kc: Kp = Kc(RT) n
R = 0.0821 L atm/ mol K
T= Kelvin temp
n= moles of products (g) – moles of reactants (g)
II.
Equilibrium Calculations
A. If given equilibrium conditions- plug directly into the equilibrium
expression
B. If initial conditions are given, use a RICE chart
R= reaction
I= initial concentrations or pressures
C= change (stoichiometry)
E= equilibrium concentrations or pressures (use these to plug into the
equilibrium expression)
C. Reaction quotient (Q)
1. Used to determine if a system is at equilibrium, and if not, which
direction the reaction will shift in order to establish equilibrium
2. Calculate Q and compare to K
- Q > K, too much product, equilibrium shifts to the left
- Q < K, too much reactant, equilibrium shifts to the right
- Q = K, system is at equilibrium
III.
Le Chatelier’s principle
A. Definition: If a stress is placed on a system at equilibrium, the equilibrium
will shift in a direction to alleviate the stress
STRESS
DIRECTION OF SHIFT
Addition of reactant
Right (away from excess)
Addition of product
Left
Removal of reactant
Removal of product
Addition of an inert gas
Addition of a catalyst
Increase temp – exothermic rxn
Increase temp- endothermic
Decrease temp- exothermic
Decrease temp- endothermic
Increase pressure by decreasing
the volume
Decrease pressure by
increasing the volume
IV.
Left (towards reactants)
Right
No change
No change (speeds up rxn in both directions)
Heat is in the products, increasing the temp is adding
heat- shifts left
Heat is in the reactants, increase the temp is adding heat
to the reactants- shifts right
Right
Left
Shifts from largest # of moles of a gas to smalles
Shifts from smallest to largest gaseous moles
Solubility Product (Ksp)
A. Concept:
1. Ksp is used to determine the solubility in mol/L of “insoluble” salts
2. The higher the Ksp value, the more soluble the compound
3. Dissociation equations: solid ions
Example: PbCl2 (s) Pb 2+ (aq) + 2 Cl – (aq)
B. Effect of a common ion on solubility
1. Dissolving a solid in a solution that contains one of the ions
present in the salt decreases the solubility of the solid- pushes
equilibrium to the left (LeChatelier’s principle)
2. Example: PbCl2(s)  Pb 2+ (aq) + 2Cl- (aq)
a. in water
-If the solubility of PbCl2 = x, [Pb 2+] = x and [ Cl-] = 2x
b. in a 0.20 M NaCl solutions
- [Pb 2+] = x , [Cl-] = 0.20 + 2x
3. You must also consider acid base properties when looking for
common ions
C. Determining whether a precipitate will form
1. Calculate the concentration of the ions in solution
2. Determine Q
3. Compare Q to Ksp
Q > Ksp a precipitate will form
Q < Ksp, no precipitate will form
D. Determining which solid will precipitate first:
1. Calculate the concentration of the ion needed to precipitate each solid
2. The lowest concentration will precipitate first
E. Determining the concentration of ions remaining when two solutions react to
form a precipitate
1. Find the concentrations of each of the ions in solution
2. the ion present in the smallest amount will react completely. The other
will be left over.
3. Equilibrium calculation (Ksp) with the leftover ion present as a common
ion in solution