Equilibrium Expression (Keq)

Equilibrium Expression (Keq)
Also called “Mass Action Expression”
Relates the concentration of products to reactants once
equilibrium has been reached.
For this general reaction:
aA + bB ↔ cC + dD
Keq =
[C]c x [D]d
[A]a x [B]b
[ ] the brackets mean
“the concentration of”
So basically concentration of products over
concentration reactants raised to the power
of their coefficient in balanced equation
Keq =
[A]a x [B]b
Exclude solids and pure liquids as they do
not have concentration values.
Write Keq expression for:
N2(g) + 3H2(g) ↔ 2NH3(g)
All gases (nothing excluded)
Keq =
[N2] x [H2]3
Ex: Write Keq expression for:
2NO(g) + 2H2(g) ↔ N2(g) + 2H2O(l)
Keq =
[NO]2 x [H2]2
Take out pure liquid
Ex: Write Keq expression for:
NaCl(s) + H2SO4(l) ↔ HCl(g) + NaHSO4(s)
Keq =
Take out solids and pure liquid
Value of Equilibrium Constant (Keq)
At equilibrium if you put the concentration
values (Molarity) into the Keq expression you
will get a specific number
(The is a unitless number and is unique to that reaction.)
The only thing that can change the value of Keq
is a change in temperature.
Value of Equilibrium Constant (Keq)
If Keq = 1
If Keq > 1
Conc. products = reactants at equilibrium
Favors Products
Large Keq = large quantities of product at equilibrium
If Keq < 1
Keq = [Products]x
Favors Reactants
Small Keq = large quantities of reactant at equilibrium
Plugging in Values
2A(g) + 3B(aq) ↔ 2AB(g)
At equilibrium [A] = .3M, [B] = .1M, [AB] = .8M
find the Keq.
Keq =
= 7111
[.3]2 x [.1]3
Favors Products
Plugging in Values
Find concentration of Cl2 at equilibrium if,
[PCl5] = .015M, [PCl3] = .78M and Keq = 35
PCl5 (g) ↔ PCl3(g) + Cl2(g)
35 = [.78] x [“X”]
[Cl2] = .67M
 The Keq value for the “reverse” reaction
will be the inverse of the “forward” reaction
 Products become reactants
Keq interactive
 http://glencoe.com/sites/common_assets/a
ICE Problems (Honors)
Keq problems where you are
given INITIAL concentrations.
Use stoich ratios to find the
CHANGE in concentration
Subtract this from initial
concentration find the
EQUILIBRIUM concentration
that can then go into Keq
Crash Course: Equilibrium Equations (mostly Honors)
Solubility Equilibrium for Ionics
Ionic Solids:
Dissociate when placed in solution.
 Positive
and negative ions are pulled apart.
 Polyatomic Ions stay together!!
If an ionic solid dissolves in a polar liquid, this
process is called dissolution.
Dissolution Equation:
AgCl(s) ↔ Ag+1(aq) + Cl-1(aq)
Try to write a dissolution equation for CaCl2(s)
CaCl2(s) ↔ Ca+2(aq) + 2Cl-1(aq)
Ksp Expression
Equilibrium expressions for ionic solutions are called
Ksp (sp = “solubility product”).
Set up “K” expression as before
include (g) and (aq), cross out (s) and (l)
AgCl(s) ↔ Ag+1(aq) + Cl-1(aq)
AgCl(s) ↔ Ag+1(aq) + Cl-1(aq)
Cross out solid
Answer is the “product” of the
Ksp = [Ag+1] x [Cl-1]
concentrations of the ions at
equilibrium or “ion product”
Try Writing the Ksp Expression
AlPO4 (s) ↔ Al+3 (aq) + PO4-3 (aq)
Ksp = [Al+3 ] x [PO4-3 ]
Ca3(PO4)2 (s) ↔ 3Ca+2 (aq) + 2PO4-3 (aq)
Ksp = [Ca+2]3 x [PO4-3 ]2
Value of Ksp
Higher Ksp = more soluble
Lower Ksp = less soluble
Large Ksp = more solid is dissolved at equilibrium
Ksp = 5 x 10-33
Ksp = 2 x 10-9
It would also indicate a higher level of conductivity since
ionics are electrolytes!
Value is temperature dependant, (usually given for 25 °C)
Just Read
Ionic compounds have different degrees of
solubility (none are truly insoluble as it
may indicate on your ref. tables), Ksp
allows us to compare solubility.
This is useful when looking at how much
relatively insoluble compounds will
dissolve in things like drinking water or
blood plasma.
Ksp Problems (Honors)
Find Ksp from Solubility:
A sat. solution of BaSO4 has a conc. of 3.9 x 10-5M
of Ba+2 ions, find Ksp.
BaSO4 (s) ↔ Ba+2 (aq) + SO4-2 (aq)
Ksp = [Ba+2] x [SO4-2]
Concentration of ions
is the same. (1:1 ratio)
Ksp = [3.9 x 10-5M] x [3.9 x 10-5M]
Ksp = 1.5 x 10-9
If [Pb+2] = 1.9 x 10-3 in a saturated solution
of PbF2 find Ksp.
PbF2 ↔ Pb+2 + 2F-1
Ksp = [Pb+2] x [F-1]2
Don’t know either but
one is double the other
Ksp = [X] x [2X]2
= [1.9 x 10-3] x [2(1.9 x 10-3)]2
= 2.7 x 10-8
Ksp Problems (Honors)
Find Solubility from Ksp
If the Ksp of RaSO4 = 4 x 10-11 calculate its
solubility in pure water.
RaSO4 (s) ↔ Ra+2 (aq) + SO4-2 (aq)
Ksp = [Ra+2] x [SO4-2]
We don’t know either
concentration but they
are the same
4 x 10-11= [Ra+2] x [SO4-2]
4 x 10-11= [X] x [X]
4 x 10-11= X2
X = the square root of 4 x 10-11 = 6 x 10-6 M
If Ksp of PbCl2 = 1.6 x 10-5, calculate solubility.
PbCl2 ↔ Pb+2 + 2Cl-1
Ksp = [Pb+2] x [Cl-1]2
1.6 x 10-5 = [X] x [2X]2
1.6 x 10-5 = 4X3
“X” the cube root of 1.6 x 10-5 = .016
[Pb+2] = .016M, [Cl-1] = .032M
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