1. Unit 11 Equilibrium (Chapter 17) And you

3. 11-1 Equilibrium ~ A Conceptual Introduction (Sections 17.3, 17.4) Two half-filled beakers of water are allowed to evaporate. The only difference between the two is that the beaker on the right is covered, thereby allowing a greater rate of condensation than the one on the left. Note the difference in water levels after some time. Once the rate of condensation equals the rate of evaporation in the covered beaker, the level of the water will remain constant and a physical equilibrium has been established. Molecules continue to move between the two phases, gas and liquid; it is still a dynamic system.

4. Why does the beaker on the left never reach equilibrium? Additional Examples: 1) World of Chem Ski Area 2) Highway 36 during rush hour

5. Chemical Equilibrium A chemist is interested in producing hydrogen iodide; she mixes hydrogen and iodine in the following ratios:H 2 + I 2 ↔ 2 HIInitially, the reactants produce HI and its concentration increases. However, after a period of time, the reaction appears to stop and the concentration of HI remains constant. Strangely, there are still some reactants remaining. What is happening?

6. H 2 + I 2 2HI

7. Some reactions don’t “run to completion”; that is, the reactants can produce products and the products can make reactants. Essentially, a reverse reaction can occur. In our example, as the reactant diatomics produce HI their concentrations drop and the reaction rate of the forward reaction slows. As the concentration of HI increases, the reverse reaction (a decomposition) begins to occur at a faster rate. Eventually, the forward reaction rate will equal the reverse reaction rate and chemical equilibrium is established. The concentrations of reactants and products will remain unchanged once equilibrium is established.

8. Again, as in the case of a physical equilibrium, the reaction is still a dynamic process, reactants are turning into products and vice versa. It appears to have stopped, though, from a macroscopic perspective: no more color change, no more gas produced, etc.

9. Once equilibrium is established, the concentrations of the reactants and products will remain constant; however; they do NOT have to be equal. If the [products] > [reactants], chemists say the “products are favored.” If the [reactants] > [products], then the “reactants are favored.” Which situation do you think a chemist would prefer?

11. Once a reaction has reached equilibrium, the set of concentrations of reactants and products is called a “position.” And that position remains constant even though the reaction is still a dynamic process Unless conditions are changed!!!!!!!!!!! We will discuss this later.

12. https://www.youtube.com/watch?v=g5wNg_d KsYY#aid=P-KaF1ewdEM https://www.youtube.com/watch?v=g5wNg_d KsYY#aid=P-KaF1ewdEM

13. 11-2 The Equilibrium Constant (K) (Sections 17.5, 17.6) For any chemical reaction:aA + bB ↔ cC + dD where a, b, c, and d are the coefficients of the balanced chemical equation and A, B, C, and D are the chemical species (atoms or molecules). In the (g) or (aq) phases, the following relationship will hold true at equilibrium: K eq = C c D d so if d=2 then D 2... A a B b Raised to the power of coefficient in bal. rxn.

14. Yes, include: aqueous and gasses in K eq NO, do NOT put liquids and solids in K eq The above ratio is called the Law of Mass Action is a constant at a given temperature! In other words, K is independent of [ ], but dependent on temperature. The concentrations of a pure solid (s) or a pure liquid (l) are constant (value depends only upon the density) and are therefore NOT included in the K eq. Enter a 1 for solids and/or liquids.

15. So what is this equilibrium expression? a ratio of product/reagent at a given temperature Can also be found by a ratio of the percentages Forward / backward

16. Example 1: N 2 (g) + 3 H 2 (g) ↔ 2 NH 3 (g) K eq = NH 3 2 H 2 3 N 2 Example 2: 2 HgO(s) ↔ 2 Hg(l) + O 2 (g) K eq = O 2 Example 3: 4 Fe(s) + 3 O 2 (g) ↔ 2 Fe 2 O 3 (s) K eq = 1 O 2 3

17. The K eq value indicates which is favored at equilibrium, reactants or products. If K eq > 1then[products] > [reactants]and products are “favored” If K eq [products]and reactants are “favored” If K eq = 1then[products] = [reactants]and neither is “favored” If K is REALLY large, then the reaction is thought to go to completion.

18. Example 1: 2 H 2 O(g) ↔ 2 H 2 (g) + O 2 (g) K eq = 0.0024 or 2.4 x 10 -3 Which is greater, [products] or [reactants]? [reactants] Which is “favored” at equilibrium: H 2 O(g) or H 2 (g) + O 2 (g) ? H 2 O(g)

19. The value of K eq can be used to calculate equilibrium concentrations: 3 A(g) + 4 C(s) ↔ D(g) + 2 B(aq) with K eq = 10 If, at equilibrium, [A] = 2.0 M and [B] = 4.0 M, calculate the [D] at equilibrium: K eq = 10 = [D] [B] 2 = [D] [4] 2 [A] 3 [2] 3 So [D] = 5.0 M

20. Another way to express the equilibrium constant is in terms of pressure units when the chemicals are gases. In this case, a K p is written, where p is pressure. A flask is filled with dinitrogen tetroxide gas and nitrogen dioxide gas at 25 o C and allowed to reach equilibrium. It was found that the equilibrium pressures of the two gases are: P N2O4 = 3.476 atm and P NO2 = 1.048. Calculate the K p for the reaction: N 2 O 4 (g) ↔ 2 NO 2 (g) K p = 1.048 2 / 3.476 =.3160 = reagent favored reaction at this temperature