Equilibrium Expression (Keq)

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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 =
[C]c
[D]d
x
[A]a x [B]b
Products
Reactants
IMPORTANT
Exclude solids and pure liquids as they do
not have concentration values.
Ex:
Write Keq expression for:
N2(g) + 3H2(g) ↔ 2NH3(g)
All gases (nothing excluded)
Keq =
[NH3]2
[N2] x [H2]3
Ex: Write Keq expression for:
2NO(g) + 2H2(g) ↔ N2(g) + 2H2O(l)
Keq =
[N2]
[NO]2 x [H2]2
Take out pure liquid
http://www.kentchemistry.com/links/Kinetics/EquilibriumExpression.htm
Ex: Write Keq expression for:
NaCl(s) + H2SO4(l) ↔ HCl(g) + NaHSO4(s)
Keq =
[HCl]
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
[Reactants]y
Favors Reactants
Small Keq = large quantities of reactant at equilibrium
http://employees.oneonta.edu/viningwj/sims/equilibrium_constant_s.html
Plugging in Values
2A(g) + 3B(aq) ↔ 2AB(g)
At equilibrium [A] = .3M, [B] = .1M, [AB] = .8M
find the Keq.
Keq =
[.8]2
= 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”]
[.015]
[Cl2] = .67M
Note:
 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
dvanced_placement/chemistry_chang10e/
animations/kim2s2_5.swf

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
expression
http://www.youtube.com/watch?v=rog8ou-ZepE&safe=active
Crash Course: Equilibrium Equations (mostly Honors)
http://www.youtube.com/watch?v=DP-vWN1yXrY&safe=active
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)
http://programs.northlandcollege.edu/biology/Biology1111/animations/dissolve.html
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
Ca3(PO4)2
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

Ex:

Large Ksp = more solid is dissolved at equilibrium



Al(OH)3
BaCO3
Much
more
soluble!
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
4
[Pb+2] = .016M, [Cl-1] = .032M
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