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--ŕ 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) ßŕO2(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---ŕ 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 -----ŕ Products (Endothermic reaction – requires heat)
Reactants ------ŕ 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-ŕ 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-ŕ 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) ßŕ 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)