Chemical Equilibrium - General Chemistry II | CHEM 1220, Study notes of Chemistry

Download Chemical Equilibrium - General Chemistry II | CHEM 1220 and more Study notes Chemistry in PDF only on Docsity! CHEMISTRY 1220 NOTES CHAPTER 14 — CHEMICAL EQUILIBRIUM In many chemical reactions, one side or the other is highly favored, and one may see no evidence of the species on the disfavored side, e.g., E E E (AH) ° (AH) E, Exothermic Endothermic (product favored)«——(generally)- (reactant favored) In other cases, there can be an observable balance, e.g., 2NO2(g) === N204(g) Observe: (1) Start with pure NO2 (brown), one forms some N2O4 (colorless). or (2) Start with pure N2O,, one forms some NOz. ==> Each proceeds until a balance is reached. So, if one puts 41 g of NOz in a1 L flask, or 1 g of N2Oz in a 1 L flask, eventually both flasks contain the same mass of NO2 and the same mass of N20, (if T is constant). When this happens, one says that a dynamic (reversible) equilibrium has been established. Chapter 14 - 1 In such a situation: (1) The position of equilibrium is independent of the direction of approach. (Useful check in practice.) (2) Both the forward and reverse reactions continue to occur (i.e., dynamic vs. static), but we see no change. (3) The rates of these reactions may be slow or fast — could take 10° sec or less to 10° years or more to reach equilibrium. (Ea, [ ], T control these rates.) Critical Question: | How do we define the equilibrium state? = > Equilibrium Constant For: 2 NO2(g) === Ne2Oa(g) Pn2o« one observes that (Pnoz)? is constant (for a given T). Thus, if we take 3 flasks and add NO2 and/or N2O,4 we might find the following: (1) | (2) | @) Pro, (atm.) | 0.22 | 0.07 | 0.42 Pro, 1.56 0.88 | 2.16 | | K, = 2:22 = 9.090: K, = 9:97 = 0.090; K, = 9:42 = 0.090 (1.56)? (0.88)? (2.16)? Note: Units ignored, assume atm. Chapter 14 - 2 Psos -Pno _— —— D thi "Ko? Peo, Pno. oes this equal K;:K2 Psos _Pxo-P"o2 _ Psos-Pyo KyKz = -— um Pso2-P%02, PNoz Pso2-PNoz => Yes (Note: K = KK = 2.24.0 = 8.8) Question: What if we are dealing with a heterogeneous reaction (often the case in industry — 20% GDP)? First consider vapor pressures as an equilibrium: 1L flask water(?) == water(g) 100 mL wat 0 MEST "at RT, Pio) = 17.5 mm Hg (This is K) What if we add more H2O(£)? 1 L flask 7200 mL water H20() H20(g) KK = Pro => We still observe Py,og) = 17.5 mm Hg! (Note: relative humidity) Therefore we see that the amount of a heterogeneous substance does not affect the equilibrium pressures of any other species, as long as some of that substance is present. Chapter 14-5 The same would be true of a more complicated reaction, e.g., COa(g) + Hog) == CO(g) + H20(2) (reverse of “water gas shift” reaction) H20 In such a case, we do know that since H20 is volatile, there will be some H20 in the gas phase. However, as long as there is liquid H2O present, Px,o@ = vapor P(H2O) = a constant at that temperature. In essence we can simplify K for the following: CO2(g) + H2(g) === CO(g) + H20@g) Poo:PH0 into a new K, K 2 Pos K' whose K = _ Pco2-Pre Puzo Pco2: Pre (If the reaction occurs in the liquid phase the volume of the solution is important — not for k or rate, but for conversion (larger volume, more gas dissolved to react). What would K then be?) What about equilibria in other phases? => Gases still appear as partial pressures (usually atmosphere units), Henry's Law, p. 533. == Pure (heterogeneous) solids/liquids do not appear, nor do solvents for dilute solutions, in K ([H20] = ~ 55.5 M). == Reactants in solution appear as molar concentrations. Chapter 14 -6 So, for: Zn(s) + 2H"(aq) == Zn**(aq) + Haig) K = Pre [Zn?] [HP Or, for: == Zn**(aq) + Ho(g) + 2H20 aaa (same) DETERMINATION OF K Simplest situation — one is given all partial pressures (e.g., NOz/N2O4 equilibrium on p. 14-2). Then one simply plugs the values into the K expression. What about the following? 2HI(g) === Ha(g) + Ia(g) Suppose we fill a flask at 520° with HI until Py; = 1.00 atm. (assuming no equilibration yet to Hz and Iz). One then allows the system to reach equilibrium, and finds that Pi, = 0.10 atm. What is K? Pate so we need P;, and Pi. P7H K= Since Hp» and Ip are formed in a 1:1 ratio, Pi, = Pus = 0.10 atm. Chapter 14 - 7 Second Example Consider: N2O4(g) === 2NO,(g) K = 11.0@ 100°. Start with Puno, = 1.00 atm., and allow the system to reach equilibrium. Determine Pyo, and Pn,o,. PNo04 Pnoy initial 1.00 0.00 2 K = Pino adjusted 1.00-x 2x Pnzo. However, since K > 1, we might expect the equilibrium to favor the products. So, let's shift the starting point to the other side. Sometimes this simplifies the math. This would give us: Pnoo4 Pno, Note: A == 2X,K=2 initial 0.00 —-2.00 Pa Px 0.5 1.0 adjusted x 2.00 - 2x 1.0 1.41 2.0 2.0 _ P*no, _ (2-2x)?? _ 8.0 4.0 Ba 10= Puor OX” ” 50.0 10.0 (Could use method of successive approximation — see text.) (2 - 2x)? = (11.0)x —~ 4x? - 8x + 4 = (11.0)x a b c @x’ C19x + @ = 0 > quadratic: ax’ + bx +c = 0 Chapter 14 - 10  - -—— -bivb*-4ac _ 19+y(-19)'-4-4-4 2a 8 (19? = 361) _ 19+ 297 _ 19417.23 8 g = 4.53 or 0.22 Since x = 4.53 would lead to a negative Pnos, x = 0.22 must be the solution. “. Po, = 2 - 2x = 1.56 atm. (not 2-x!) (1.56)? 0.22 (0.78 change from other conditions.) Puro, = X = 0.22 atm. [ confirm: Altered Equilibria There are a number of ways in which a system may be perturbed from an equilibrium state: (1) Adding or removing a reactant or product in the Q expression. (2) Compressing or expanding the system (P or V change). (Not including P of nonparticipants.) (3) Changing the temperature. Only this changes K. To determine what would happen after such a perturbation, one can consider Le Chatelier's Principle: When the conditions of a system are altered, the system responds in such a way as to oppose the change. Let's look at these three possible perturbations. Chapter 14 - 11 First: Adding or Removing a Reactant or Product Consider: aA + bB cC + dD K (org) = Pek Poy" (Pa)? -(Pa)? At equilibrium, K = Q. If we now add B (too much reactant), Q < Kand we are out of equilibrium. The system can be restored to equilibrium by then shifting reactants (including B) to products. (Note: by increasing reactant concentrations, reactants collide more with each other, so the rate of product formation increases). Note: Agreement with Le Chatelier's Principle: We added B, and the system responded by reducing B. Second: Pressure (or V) Effects Very important reaction: No(g) + 3 Hag) === 2 NHs(g) at 25°C, K = 610° (Nature can do). Commercially, the reaction rate is too slow at 25°, due to high Ez. Going to higher T increases rate, but at 400°, K = 1.510%. What can be done? (Catalyst helps some.) Observe: P?nts ==? PNe«P#He bc® Qe such thata = Pyu,,b = Pno,¢ = Pu, gives Q = K under some equilibrium condition (e.g.,a = 3, b = 60,andc = 10 at 400°). Let's now compress the system by 10 fold, thereby increasing each P by 10. Chapter 14 - 12