Equilibrium (p. 553-555): Not all chemical rxns go to “completion”. Sometimes, reactants are left over even
though they haven’t been used up (as we have seen in a limiting reagent rxn). When the same amount of products are turning
into reactants as reactants are turning into products, the rxn is in
equilibrium. Equilibrium must happen in
a closed environment (eg. nothing can
be added or subtracted from the reaction for it to be considered to be in
equilibrium).
How
to know if a rxn is at equilibrium(macroscopic level): (1)If the concentration of the products and
reactants are constant. (2) If the color, temp, volume, pressure and mass are
constant. There may be a higher
concentration of products or of reactants, but so long as it is constant, then
you are at equilibrium.
How to
know if a rxn is at equilibrium(sub-microscopic level): Reactants are forming products at the same
rate products are forming reactants.
Even though you cannot see a change, on the sub-microscopic level
there is a constant change between product and reactants.
Physical Example of
Equilibrium – Vaporization of a liquid: Have you
ever noticed water in an airtight container has some “vapor” at the top? The water in the container is evaporating
and the vapor it created begins to condense back into a liquid: H2O(l) ß--à H2O(g). The water cycle! SEWER TOUR RULES!
Another Physical Example of
Equilibrium – Solutions: Water will dissolve a certain
amount of oxygen from the air. The
colder the water, the more oxygen will be absorbed: O2(g) ßàO2(aq). The equilibrium of dissolved oxygen in water depends upon the
temperature of the water. There is more
O2(aq) when its cold.
For solids which dissolve completely in
water, there is no equilibrium – this physical rxn goes to completion. But, if you have a solid which is not
completely dissolved (eg. an insoluble salt will partially dissociate in water,
and partially not), then you have an equilibrium situation (PbI2(s)ß-> Pb2+ (aq)+ 2I-(aq))
Chemical Example of
Equilibrium – Gases: Nitrogen and Hydrogen gases
together will combine to form ammonia gas in equilibrium: N2(g)
+ 3H2(g) ß---à 2NH3
(g)
Read
p. 562-565 aloud in class to understand how Equilibrium can be shifted.
Le Chatelier’s Principle: If you disturb a system in equilibrium, it will respond to create
equilibrium again. You can change Temp,
Pressure, Volume, Amount of Reactants
or the Amount of Products. (see photo on p. 563). If you increase pressure, the equation will go to the side with
the least number of gas molecules. If
you increase the volume, the equation will go to the side with the most number
of gas molecules. If there is equal
numbers of gases on each side of the equation, a change in volume or pressure
will not make a difference. If you
increase the temperature, the reaction will from the endothermic side to the
exothermic side (p. 565).
Reactants + energy -----à Products (Endothermic reaction – requires
heat)
Reactants ------à energy + products (Exothermic reaction –
gives off heat)
Practical Application
(making Ammonia in the Haber Process): Ammonia is used in
fertilizers and it is made in the following process which can go in both
directions: N2(g) + 3H2(g)
ß-à 2NH3(g) + 46 kJ
(exothermic). You can increase the
amount of Ammonia (NH3) made by 1) increasing the amount of
reactants. 2) decreasing the product as
it is made 3) increasing the pressure or decreasing the volume 4) decreasing
the temperature of the rxn to favor the exothermic product. (if the product were endothermic, you would
heat the system to favor the product)
Quantitative analysis (the
math stuff) of Equilib:
[NH3] = The concentration of
ammonia in Molarity
Keq
= the equilibrium constant (read p. 556 Aloud in class). This is determined by
[products]/[reactants]. Multiply the
concentration of all of the products together and divide that by the
concentration of all the reactants.
Each type of equilibrium rxn will have its own Keq.
NOTE: If Keq > 1, then the reaction
favors the products
If Keq < 1, then the reaction favors the reactants.
If there is a co-efficient in front of
one of the products or the reactants, raise that up to a power of the
concentrations (example: H2(g) +
Cl2(g) ß-à 2HCl
(g) will be [HCl]2
/ [H2] [Cl2] ) (see p. 558 for problem)
As you might have guessed, Ke q
is dependent upon temp and pressure.
Pure liquids and pure solids
are not included in the equilibrium constant expression(p. 577-581): There is no concentration of a pure solid or of a pure liquid in
moles/Liter. Their concentrations are
considered to be constant
(example: 10 mL of water has the same concentration of
water molecules as does 20 mL of water)
(example
#2: 10 g of BaSO4 has the same
concentration per gram as does 20 g of BaSO4). They are, however, inherently a part of the
Keq, but they are left out of the expression itself.
Therefore, the Keq for CaCO3(s)
ßà Ca2+(aq) +CO32- (aq) is: Keq =
[Ca2+] [CO32-]. See Example on p. 580 (whenever you see the (s) or (l),
you can assume that those compounds are NOT
going to be in the problem. But an (aq) or (g) will be in the
problem.)
The
ppt of CaCO3 will begin when [Ca2+] [CO32-]
is greater than the Keq for this equation.
When
[Ca2+] [CO32-] equal the Keq for this
equation, the ppt of CaCO3 will stop.
Water
Softening(a practical application of all of this): Tap water contains Ca+2 and Mg+2
ions which will bind with soap and make a precipitate (ppt) - in other words, a
“ring” in your bathtub. If you add Na2CO3
to your water, the Ca and the Mg will single displace the Na and will ppt out
as MgCO3 and CaCO3 (realize
that Na2CO3 is soluble in water while CaCO3 and MgCO3 are not). You will be left with Na+1 ions
in your water which will not ppt with soap.
This is basically an exchange of one ion (Ca or Mg) for another (Na),
but Na is the lesser of the evils since it will not make a “ring” in the tub
with the soap.
What this means is you need to add a lot
of Na2CO3 to the water to make the product of the
concentrations of [Ca2+] [CO32-] exceed the Keq. The more Na2CO3 you
add, the more Ca+2 ions you can bind and create a ppt of CaCO3.
(same is true for any Mg+2 ions in the water). This is known as solubility equilibrium because it is an equilibrium between the
solubility of a solid and its ions in solution.
Big
hint: Solubility equilibrium
problems will never have a denominator!
They will always be a product of the concentrations of the anions (-)
and cations (+) raised to their respective powers. (p. 579)
HW #1 Ch. 18: #1-11 on worksheet. P. 559
PRACTICE PROPBLEMS#1-4, P. 559
SECTION REVIEW #5-9 (general and mathematical questions about Keq)
HW#2 Ch. 18:
#13-26 on worksheet (Le
Chatelier’s Principle general questions
HW #3: #5-#6
section review p. 568, P. 586 #11 (Le Chatelier’s)