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CHAPTER 13
Chemical Equilibrium
Equilibrium

Some reactions go to completion.


All the reactants are converted into products.
There are many reactions that do not go to
completion:


Example: 2 NO2 (g)

N2O4 (g)
Dark Brown
Colorless
If the reaction went to completion the container
would become colorless – it doesn’t
EXAMPLE




Should:
2 NO2 (g)

Dark Brown
N2O4 (g)
Colorless
Actual:
2 NO2 (g) 
N2O4 (g)
Light Tan
CHEMICAL EQUILIBRIUM


Some Reactions go both way.
Sometimes very few products are created.


2 CaO (s)  2 Ca (s) + O2 (g)
Very Little appears to happen.
Sometimes a lot of products are created.
Cu (s) + 2 AgNO3 (aq)  2 Ag (s) + Cu(NO3)2 (aq)

Very Little of the reactant remains.
CHEMICAL EQUILIBRIUM


CuCl2 (aq)  Cu+2 (aq) + 2 Cl- (aq) + Heat
Green
Blue
Colorless
Observe what happens when the following occurs:




Add Heat
Remove Heat
Add NaCl
Add AgNO3
CHEMICAL EQUILIBRIUM

Reaction Graph – Concentration of CuCl2
nevers reaches zero.
EQUILIBRIUM


Is not static.
It is Highly Dynamic.


At the Macro Level everything appears to have
stopped.
At the Micro Level, reactions are continuing.




The reaction is travels both directions.
As much Reactant is being created as Products.
The Reaction Rates in both directions are in Equilibrium.
Not the concentrations of Reactants and Products.
CHEMICAL EQUILIBRIUM
EQUATION

jA + kB  mC + nD
j is coef for Reactant A
k is coef for Reactant B
m is coef for Product C
n is coef for Product D
Keq = [C]m[D]n
[A]j[B]k
EXAMPLE PROBLEM
The following reaction is allowed to go to equilibrium.
CuCl2 (aq)  Cu+2 (aq) + 2 Cl- (aq) + Heat
The final concentrations of all the reactants and products
CuCl2 = [0.0250]
Cu+2 = [1.25]
Cl- = [0.625]
What is the Equilibrium constant?
SOLUTION
Keq = [C]m[D]n
[A]j[B]k
Keq =
[1.25][0.625]2 = 19.5
[0.0250]
What does it mean?




If the Keq < 1, then Reactants are favored
If the Keq = 1, then Products and Reactants
are equal
If the Keq > 1 then Products are favored
Since the Keq is 19.5 and 19.5 is greater
then 1, more Products will be present then
reactants.
Another Example

The Keq for the following reaction is 130, If
the concentration of Nitrogen is 0.100 M and
the concentration of Hydrogen is 0.200 M,
what is the concentration of Ammonia?
N2
+
3 H2  2 NH3
CONTINUED

130 = [NH3]2
[N2][H2]3

130 = [NH3]2
[0.100 ][3 0.200 ]3

130 = [NH3]2
[0.0008 ]

130 * 0.0008 = NH32 = 0.322 M/2 = 0.161 M
HETEROGENOUS EQUILIBRIUM

Reactants or Products that are solids, and/or
water are not included in the expression.
H2SO4 (aq) + 2 NaOH (aq)  Na2SO4 (aq) + 2 H2O (l)




H2SO4 = 0.100 M
NaOH = 0.200 M
Na2SO4 = 0.150 M
What is the Keq?
Continued
Keq =
[0.150]
= 37.5
[0.100][0.200]2
In this reaction, are the products or reactants favored?
How do you know?
A BIT HARDER
3 NaOH (aq) + H3PO4 (aq)  Na3PO4 (aq) + 3 H2O (l)
Na3PO4 = 0.200 M
The Keq is 130
What is the concentration of H3PO4 and NaOH?
We have two unknowns so we must have two equations:
First Equation is: Keq = [0.200]
= 130
[NaOH]3[H3PO4]
What is the second equation?
A BIT HARDER CONTINUED
3 NaOH (aq) + H3PO4 (aq)  Na3PO4 (aq) + 3 H2O (l)
Let’s replace NaOH with X and H3PO4 with Y
Let’s look at the equation now:
Keq = [0.200]
= 130
[X]3[Y]
If we look at the coefficents, we can see that:
3 NaOH = 1 H3PO4
We replace NaOH with X and H3PO4 with Y we get:
3X=Y
Let’s look at the equation now:
Keq = [0.200]
= 130
[X]3[3X]
A BIT HARDER CONTINUED #2
Keq = [0.200]
= [X]3[3X]
130
0.00154 = 3X4
0.000513 = X4
0.1505 = X = [NaOH]
0.1505 * 3 = [.4515] = [H3PO4]
Ksp vs Keq



PbCl2, AgCl, and HgCl2 are considered
insoluble in water. That mean they do not
dissolve in water, right?
Actually, a very, very, very tiny amount will
dissolve in water.
We use the Ksp (Constant for Solid Products)
to determine the amount that will dissolve.
Ksp vs Keq


jAB (s)  mA (aq) + nB (aq)
Ksp = [A]m[B]n
Notice the Ksp equation uses only the
concentration of the PRODUCTS,
reactants are not included.
EXAMPLE

Lead (II) Chloride is considered insoluble in
water, but experiments show that a very tiny
amount will dissolve in water, if the Ksp for
PbCl2 is 3.40 x 10-15 what is the concentration
of Lead and Chlorine ions?

Ksp = [A]m[B]n

PbCl2 (s)  Pb2+ (aq) + 2 Cl- (aq)

3.40 x 10-15 = [Pb2+] [Cl-]2
Ksp Continued








Pb2+ = 2 ClX = 2Y
3.40 x 10-15 = XY2 = 2Y*Y2
3.40 x 10-15 = 2Y3
3.40 x 10-15 = 2Y3
2
2
1.70 x 10-15 = [Y]3
1.19 x 10-5 = [Y] = [Cl-]
2 (1.19 x 10-5) = [2.38 x 10-5] = [Pb2+]