Chapter 6 Chemical Equilibrium

Transcription

1 1 Chapter 6 Chemical Equilibrium Section 6 1: The Equilibrium constant For the generalized reaction aa + bb cc + dd where lower case letters are stoichiometric coefficients in the balanced reaction. Upper case letters are reactants and products in the reaction. [A], [B], [C], [D] in solution use molarity in gas phase reactions use pressure (bars) Pure solids, liquids, and solvents = 1, so they do not affect the equilibrium constant, and are therefore ignored. Why are these values 1? Because what goes into the K expressions are not really molarity. (Ignore gas phase reactions from now on, everything we do is in aqueous (i.e. water) solutions.) [A] = [A]/1 M. The standard state for anything dissolved in solution is 1 molar. So really the concentration put into the expressions is not concentration, but a unitless value.

2 2 To add one additional level of complexity that nobody told you about in gen. chem., you really put in activities (Chapter 8, which we must skip for lack of time) rather than concentration. These last 2 things are pretty much just FYI, they do not affect anyone here. So the practical result for us is the same as in gen. chem. 1. For solute concentrations use molarity (though remembering K is unitless can avoid confusion), and 2. Pure solids, liquids and gases simply leave out of the K expression. If K is large then the reaction is product favored, and the reverse reaction is reactant favored. (Kinetics?) If the forward reaction has an equilibrium constant = K, then the reverse reaction has an equilibrium constant = 1/K. Adding 2 reactions together, K of the sum of these 2 reactions is the product of the 2 equilibrium constants of the original reactions added. (Adding reactions = multiplying K values)

3 3 Section 6 2 Equilibrium and Thermodynamics Large K => reaction is product favored Small K => reaction is reactant favored Remember Negative ΔG => reaction is product favored Positive ΔG => reaction is reactant favored Both the equilibrium constant and Gibbs free energy contain the same information, how are they related? Where R = J/mol.K and T = Kelvin temperature Le Chatelier s Principle Manipulating chemical reactions to proceed in a desired direction. (Grav. Cl lab common ion effect, Section 6 3, pages ) More generally discussed on page 99. Add reactant(s) then reaction proceeds to products Add product(s) then reaction proceeds to reactants. But why? Because nature is always striving to achieve equilibrium.

4 4 Section 6 3 is solubility product. This was covered in grav. Cl prelab, you are responsible for material in this section for lecture, including for salts which dissociate not only 1:1 like AgCl but also salts which dissociate 1:2, etc. such as Hg 2 Cl 2 discussed on page 100 of the text. Example: Determine the [Cl ] in a saturated solution of Hg 2 Cl 2.

5 5 Section 6 4 is complex formation. This section will be covered near the end of the semester when this type of chemistry is discussed. Section 6 5 is protic acids and bases. From now on the material will build on itself for pretty much the rest of the semester. You know what that means? Acid = substance that increases [H + ] Base = substance that decrease [H + ] Protic = H 1e = H + In aqueous solution H + + H 2 O H 3 O + (Hydronium ion) In this class use H + /H 3 O + interchangeably depending on convenience. The Bronsted Lowry theory of acids and bases will be used until complex ion chemistry is discussed much later. In B R theory: Acid = proton donor Base = proton acceptor HCl is an acid in H 2 O because it donates a proton to water, increasing [H 3 O + ]. In all Bronsted Lowry acid base reactions there is a proton transfer from an acid to a base. Conjugate acid/base pairs are recognized as such by the gain or loss of a H +.

6 6 In the above reaction H 2 O is a base, accepting a H + from HCl. Depending on circumstances H 2 O can also act as an acid donating a H + to a base: Since H 2 O can act as either an acid or a base depending on the circumstances (not unique), the following reaction may not be surprising: H 2 O + H 2 O H 3 O + + OH Since H 3 O + = H + (use interchangeably), the above reaction normally written this way: H 2 O H + + OH This reaction is so fundamental for aqueous solution chemistry that K is given its own subscript: W (for water) Section 6 6 ph In pure H 2 O [H + ] = [OH ] So If [H + ] = 1 x 10 5, then what is [OH ]?

7 7 ph log [H + ] (Activity) In pure H 2 O: [H + ] = 1 x 10 7 ph = log (1 x 10 7 ) = 7 Unless one makes a great effort, H 2 O ph 5.5 because If ph = 5.5 then 5.5 = log [H + ] = [H + ] 3.16 x 10 6 = [H + ] panything = log Anything. Can and will define: poh = pk w = pk a = Now to derive a useful relationship In Pure H 2 O: Acidic solution: Basic (alkaline) solution:

8 Section 6 7 Acid/Base Strength (strong vs. weak) 1. Strong a. Acids react completely in water to form H 3 O + b. Bases react completely in water to form OH You should know a strong acid or a strong base when you see one. Table 6 2 page 108 gives a list of strong acids/bases. Strong acids = HX where X = halide (except F) + H 2 SO 4, HNO 3, HClO 4. 8

9 9 Strong bases = Groups I and II metal hydroxides. 2. Weak acids and bases. Acids: Partially react to donate a proton to water to form H 3 O +. For a weak acid K a << 1 because the reaction proceeds to a very limited extent (the definition of a weak acid). Weak bases react partially with water to abstract (i.e. take) a proton. For a weak base K b << 1 because the reaction proceeds to a very limited extent (the definition of a weak base).

10 10 Note: For a weak acid reaction H 2 O does not need to appear explicitly in the equation. For a weak base reaction H 2 O must appear explicitly in the written reaction. It is an essential skill to be able to write K a (not hard) and K b (harder) reactions and expressions from those reactions. Required skills discussed thus far: Recognize conjugate acid/base pairs Write a K a reaction and K a expression Write a K b reaction and K b expression Recognize the difference between strong/weak acids/bases For those of you who have had organic chemistry: The most common class of weak acids is carboxylic acids. And the most common class of weak bases is amines. Amino acids contain both acidic and basic groups in the same molecule amphiprotic. See this a little later here, and in biochemistry. To treat well gets very complicated.

11 11 Why is recognizing conjugate acids and bases so important? Here is derived a fundamental relationship between conjugate acid/base pairs. What are the implications of this? If you know K a for a weak acid, then you know K b for its conjugate base. If you know K b for a weak base, then you know K a for its conjugate acid. The stronger the acid, the weaker its conjugate base. The stronger the base, the weaker its conjugate acid. To illustrate: K a for ethanoic (acetic) acid = 1.8 x K a for methanoic (formic) acid =

12 12 In the book Appendix G gives K a for weak acids and weak bases. There are no K b values given. Ex. If you want K b for methylamine (CH 3 NH 2 ), look up K a for its conjugate acid and compute it. Salts The salt of a weak acid is a weak base. Recognizing solutions like this as weak base solutions is hard but necessary. The salt of a weak base is a weak acid. Recognizing solutions like this as weak acid solutions is hard but necessary. The salt of a strong acid is neutral. (Remember the stronger the acid the weaker the conjugate base.) In this section (6 7) we are for now skipping all material with polyprotic acids.

13 13 Also skipping entirely section 6 8. Recall we also skipped section 6 4. Section 6 3 was covered in Grav Cl pre lab, as well as Ch. 27. Chapter 6 Exercises C, E, H Problems 1 4, 6, 8, 14, 16, 17, 31, 34, 36, 37, 40, 42, 43 46, 48, 49