Chapter 18 – Equilibrium Notes
Equilibrium and Reversible Reactions: (p. 559-561): 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 products are
turning into reactants at the same RATE 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).
Earth is in equilibrium. Water can be in the form of icebergs, lakes, clouds, rivers, oceans... But it is a closed environment no water can come or go from the earth. The water that is here can transform from one phase into another - and it does so at the
same RATE! If it's raining somewhere, then somewhere else water is evaporating off of the oceans or lakes.
Do all reactions stay at equilibrium?
Not all reactions go to equilibrium. By looking at a reaction sometimes there are equal amounts of product and reactant.
This can stay constant in the reaction. This is said to be at equilibrium.
At equilibrium, reactants are forming products at the same rate as products are forming reactants.
N2 + 3H2  2NH3
The above equation shows how nitrogen and hydrogen gases together form ammonia but at the same time, ammonia
decomposes to form nitrogen and hydrogen gas.
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 rate of 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) -- 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) O2(aq). The equilibrium of dissolved oxygen in water depends
upon the temperature of the water. There is more O2(aq) when it's 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)-> 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) --- 2NH3 (g)
How to Shift Equilibrium (Le Chatelier's Principle) p. 569-574: If you disturb a system in equilibrium, it will respond to
create equilibrium again - this is Le Chatelier's Principle. You can change Temp, Pressure, Volume, Amount of Reactants or the
Amount of Products. (see fig 18-8 on p. 571 and fig 18-9 on p. 572).
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 volume or pressure change will not make a difference.
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) - 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)
If you increase the temperature, the reaction will move from the endothermic side to the exothermic side (p. 572).
Reactants + energy ----- Products (Endothermic reaction – requires heat)
Reactants ------ energy + products (Exothermic reaction – gives off heat)
Quantitative analysis of Equilibrium (read p. 563-564 about the Keq). : This next part starts to use math, oh yes!
Though Equilibrium means that there are both products and reactants. It doesn’t necessarily mean that there is equal amounts on
both sides of the equation. As long as both sides are making and breaking bonds, there is equilibrium. Equilibrium usually
favors one side or another. You will find this out very soon.
Equilibrium Concentration : [HCl] = the concentration of hydrochloric acid in Molarity. [ ] means
concentration. Molarity is # of moles per 1 liter of water. Keq is a constant, the equilibrium constant. It
tells you if the reaction's equilibrium favors the products or the reactants. The K eq is found based on the ratio of molarity
concentrations. We use brackets [ ] to indicate concentration in units of Molarity. For example: [NH 3] = The concentration of
ammonia in Molarity (moles/liter)
This constant is found by finding the [products]/[reactants]. Multiply all the products concentrations and the reactants
concentrations and divide them both, just like the equation. The Keq is different for every reaction. By the way, just to make it
more mathematical, if you have a coefficient in the reaction say like:
H2(g) + Cl2(g)  2HCl(g) Then you raise the [HCl] to [HCl]2 You just make it a power number. So finding the K eq
of this reaction you would write it like this:
Keq = [HCl]2/ [H2][Cl2]
*** Pure liquids and solids do not really have any concentration values as they are pure. In these cases, you leave them out of
the equation for finding the Keq. In all these cases, you should see a (l) or a (s) after a compound showing you whether it is a
solid or a liquid, you leave the out. See below:
CaCO3 (s)  Ca +2 (aq) + CO3-2
(aq)
Keq = [Ca][CO3]
CaCO3 is a solid so it is not plugged into the equation for finding Keq. Make sure you look at this to understand it.
By knowing the Equilibrium Constant, you can determine which side of the reaction will be favored. Look at the number. Is it
greater or less than 1? What side of the reaction needs to have more concentration for the number to be greater than 1? Which
side of the reaction needs to have more concentration for the number to be less than 1? Look at the equation.
[products]/[reactants] It’s ALGEBRA and it’s easy!
NOTE: If Keq > 1, then the reaction favors the products
If Keq < 1, then the reaction favors the reactants.
Note: Keq means the equilibrium constant. Ksp is the equilibrium constant specifically for ionic salts. K a is the equilibrium
constant for Acids. Kb is the equilibrium constant for bases…etc. Keq is just the "generic" expression for the equilibrium
constant.
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 MgCO 3 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 [Ca 2+] [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. 577-580)
Homework:
Solving for Keq:
Practice Problems pg. 565-568 # 1-4
Section Review pg. 568 # 5-9
Practice Problems Pg. 576 # 16
Pg. 591 # 52 & 55
Le Chatelier’s Principle:
Section Review Pg 754 #11-14
Pg. 591 #59-62