3 H 2 (g) + N 2 (g) 2 NH 3 (g)

Transcription

1 TPIC 7: EQUILIBRIUM Dynamic equilibrium chapter 16.1 Chemical equilibrium chapters , 16.5 pen system: 2 Na (s) + Cl 2 (g) 2 NaCl goes virtually to completion. 3 2 (g) + N 2 (g) 2 N 3 (g) 2 (l) 2 (g) T 373 K Water boiling Water evaporating Some molecules have sufficient Ek to escape surface Eventually: all water from clothes evaporates does not and is said to be reversible. Closed/isolated system (matter/energy not lost/gained): In a closed system the position of equilibrium is reached At 273 K: 2 (l) Air saturated with water vapor: DYNAMIC EQUILIBRIUM 2 (l) 2 (g) A pressure is produced = vapor pressure of water (dependent on T) 2 (s) 1) when rate of the forward reaction = rate of reverse reaction 2) when the relative concentrations of the reactants and products in the equilibrium mixture are the same 3) by starting w/ either reactants or products Dynamic equilibrium: At equilibrium of a process the rate of the forward process is equal to the rate of the reverse process. The process AS NT STPPED. c (mol/l) c (mol/l) Static equilibrium: ccurs when all movement ceases once equilibrium is established. 0.4 t t N 2 4 (g) 2 N 2 (g) 2 N 2 (g) N 2 4 (g) colorless brown brown colorless 0.4 1

2 An equilibrium expression can be written for a homogenous reaction, for which all the reactants & products are in the same phase. A phase is similar to a state (s, l, g) but there is a physically distinct boundary between 2 phases. a A + b B c C + d D At equilibrium the concentrations are written: [A] eqm [B] eqm [C] eqm [D] eqm The equilibrium expression: [products]coefficient [reactants] coefficient = esterification: equilibrium constant at stated T C 3 C (l) + C 2 5 (l) + [C] c [D] d [A] a [B] b C 3 CC 2 5 (l) + 2 (l) dissociation of I: 2 I (g) 2 2 (g) + I 2 (g) K c is essentially a measure of the amount of products in an equilibrium compared with the amount of reactants. K c >> 1: position of equilibrium lies far toward the products reaction proceeds far toward completion Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) at 298 K K c << 1: position of equilibrium lies far toward the reactants extremely small amounts of products are formed C 3 C (aq) C 3 C - (aq) + + (aq) at 298 K K c 1: position of equilibrium lies approximately midway between reactants and products concentration of reactants and products nearly the same at equilibrium C 3 C (l) + C 2 5 (l) + C 3 CC 2 5 (l) + 2 (l) 4 at 298 K 2

3 Le Châtelier's principle chapter 16.8 Le Châtelier's principle: A dynamic equilibrium that is disturbed by changing the conditions (c / T / p / V) moves the position of equilibrium to counteract the change. Note! This does not explain WY the position of equilibrium changes! It merely enables the prediction of what will happen. C 3 C (l) + C 2 5 (l) + C 3 CC 2 5 (l) + 2 (l) 1. Removal of water concentration of product decreases Remember, that constant colorless N24 (g) 2 N2 (g) N24 (g) 2 N2 (g) brown 3. pressure increase volume decrease Equilibrium shifts to the side that counteracts this, i.e. to the side with lower number of molecules of gas. This can be seen as the color of the equilibrium mixture becomes... 2(g) + I2(g) 2 I (g) 2(g) + I2(g) 2 I (g) 2. Addition of ethanol concentration of reactant increases In this case has pressure increase no effect on the position of equilibrium! This because: concentrations of both the reactants & products are affected in the same way no change in the total number of molecules 3

4 4. The effect of T on the position of equilibrium depends on whether the reaction is exo or endothermic. In exothermic reactions, the heat released can be considered as a product. In endothermic reactions the heat consumed is "one of the reactants". The application of the concepts of kinetics and equilibrium 1) the aber process to manufacture ammonia from natural gas from air 3 2 (g) + N 2 (g) 2 N 3 (g) Δ θ = 92 kj/mol N 2 4 (g) 2 N 2 (g) Δ θ = 24 kj/mol 80 % N 3 20 % exothermic Lowering the temperature " = removal of heat" : more "product" will be produced position of equilibrium shifts to the right value of K c increases 140 million tons What happens when we a) increase the pressure: N 3 polymers 2 (g) + C 2 (g) 2 (g) + C (g) Δ θ = +41 kj/mol endothermic Increasing the temperature " = addition of heat" : the added heat is consumed by forward reaction the position of equilibrium shifts to the left value of K c decreases Remember! When T changes, K c changes! b) increase the temperature: c) add a catalyst: " [products] " [reactants] 350 o C < t < 450 o C 150 atm < p < 250 atm 15 % yield process repeated: total yield 98 % N 3 separated by cooling, unreacted gases recycled Catalyst has no effect on the position of equilibrium. Fe (s), 450 o C, 200 atm 3 2 (g) + N 2 (g) 2 N 3 (g) Δ θ = 92 kj/mol the aber process 4

5 2) the Contact process to manufacture sulfuric acid 2 S 2 (g) + 2 (g) 2 S 3 (g) 2 S 3 (g) + 2 (l) 2 S 4 (l) Exothermic process high yield when: Δ θ = -192 kj/mol Liquid vapor equilibrium more particles escape the system due to higher E k And fast rate when: only a few particles have sufficient E K to overcome the attractive forces that hold them in liquid state V25 (s), 450 o C, 2 atm (This gives a yield of over 98%. A higher pressure would give a greater yield but would not be costeffective.) The S 3 is not directly added to water (greatly exothermic). It is instead absorbed by conc. 2 S 4 (aq) to which small quantities of water are added. the chemicals needed for reducing pressure ABVE the liquid reduces the boiling point increasing pressure ABVE the liquid increases the boiling point Practical applications: 1)? fertilizers 2S4 paints & pigments detergents & soaps Manufacture of sulfuric acid electrolyte in car batteries 2) Lactic acid (2 hydroxypropanoic acid) decomposes below its normal boing point. SL ENDS ERE. 5

6 Relationship between Δ vap, boiling point and intermolecular forces Boiling points are normally measured under a pressure of 1 atm. 1) the liquid phase Equilibrium calculations chapter 16.9 C 3 C (l) + C 2 5 (l) C 3 CC 2 5 (l) + 2 (l) a mol b mol x mol x mol + A liquid boils when the intermolecular forces of attraction between particles in the liquid state are completely broken. initial n (mol) Total volume of mixture V dm 3 C 3C (l) + C 2 5 (l) C 3CC 2 5 (l) + 2 (l) + liquid (l) vapor (g) Δn (mol) equil. n (mol) 2 (l) 2 (g) Δ θ = 44 kj/mol equil. c (mol/l) SiCl 4 (l) SiCl 4 (g) Δ θ = 30 kj/mol The difference between the enthalpies of vaporization is explained by: 1) 2) 3) 4) ow do the boiling points compare? (Notice that volume does not appear in this equilibrium expression!) When 46 g of ethanol was reacted with 30,0 g of ethanoic acid at 373 K the equilibrium mixture contained 37,0 g of ethyl ethanoate. Calculate the value of K c at 373 K to the nearest integer. initial n (mol) Δn (mol) equil. n (mol) C 3C (l) + C 2 5 (l) + C 3CC 2 5 (l) + 2 (l) K c (373 K) = compound M r bp ( o C) Δ vap θ kj/mol SiCl 4 170,09 57,6 30,0 Br 2 159,80 58,8 30,7 6

7 2) the gaseous phase: no overall volume change 2 (g) + I 2 (g) 2 I (g) 3) the gaseous phase: overall volume decrease 3 2 (g) + N 2 (g) 2 N 3 (g) 2 (g) + I 2 (g) 2 I (g) initial c (mol/l) initial n (mol) Δn (mol) equil. n (mol) Δc (mol/l) equil. c (mol/l) (Notice AGAIN that V does not appear in this equilibrium expression! This agrees with Le Châtelier's principle & Boyle's law.) 3) the gaseous phase: overall volume increase initial c (mol/l) Δc (mol/l) equil. c (mol/l) PCl 5 (g) PCl 3 (g) + Cl 2 (g) (V appears in this equilibrium expression! Increase in V decrease in p increase in the value of x) (Decrease in V increase in p increase in the value of x) 7