Community Contributions - Articles by goIITians
|
|
|
|
|
Hi guyz, Contin. of my stuffon Le chatelier's principle What is the equilibrium quotient? We define equilibrium constant as a A + b B  c C + d D as  In the general case in which the concentrations can have any arbitrary values (including zero), this expression is called the equilibrium quotient and its value is denoted by Q (or Qc if we wish to emphasize that the terms represent molar concentrations.) If the terms correspond to equilibrium concentrations, then the above expression is called the equilibrium constant and its value is denoted by K (or Kc .) K is thus the special value that Q has when the reaction is at equilibrium The value of Q in relation to K serves as an index how the composition of the reaction system compares to that of the equilibrium state, and thus it indicates the direction in which any net reaction must proceed. For example, if we combine the two reactants A and B at concentrations of 1 mol L?1 each, the value of Q will be indeterminately large(1÷0). If instead our mixture consists only of the two products C and D, Q = 0. It is easy to see (by simple application of the Le Châtelier principle) that the ratio of Q/K immediately tells us whether, and in which direction, a net reaction will occur as the system moves toward its equilibrium state. The three possibilities are shown in the table below. | Q/K | | | > 1 | Product concentration too high for equilibrium;
net reaction proceeds to left. | | = 1 | System is at equilibrium; no net change will occur. | | < 1 | Product concentration too low for equilibrium;
net reaction proceeds to right. | It is very important that you be able to work out these relations for yourself, not by memorizing them, but from the definitions of Q and K. A visual way of thinking about Q and K The formal definitions of Q and K are quite simple, but they are of limited usefulness unless you are able to relate them to real chemical situations. The following diagrams illustrate the relation between Q and K from various standpoints. Take some time to study each one carefully, making sure that you are able to relate the description to the illustration.  Each tiny dot on the graph represents a possible combination of NO2 and N2O4 concentrations that produce a certain value of Q for the chemical reaction system N2O4  2 NO2. (There are of course an infinite number of possible Q's of this system within the concentration boundaries shown on the plot.) Only those dots that fall on the red line correspond to equilibrium states of this system (those for which Q = K ). The line itself is a plot of [NO2] = ([N2O4]K)0.5. If the system is initially in a non-equilibrium state, its composition will change in a direction that moves it to one on the line.
 One of the simplest equilibria we can write is that between a solid and its vapor. Using the sublimation of iodine I 2(s) I 2 (g) as an example, we see that the possible equilibrium states of the system (shaded area in the diagram) are limited to those in which at least some solid is present, but that within this region, the quantity of iodine vapor is constant (red line) as long as the temperature is unchanged. The arrow shows the succession of states the system passes through when 0.29 mole of solid iodine is placed in a 1-L sealed container; the unit slope of this line reflects the fact that each mole of I 2 removed from the solid ends up in the vapor.
 The decomposition of ammonium chloride
NH4Cl(s) NH3(g) + HCl(g) is another example of a solid-gas equilibrium. Arrow 1 traces the states the system passes through when solid NH4Cl is placed in a closed container. Arrow 2 represents the addition of ammonia to the equilibrium mixture; the system responds by following the path 3 back to a new equilibrium state .which, as the Le Châtelier principle predicts, contains a smaller quantity of ammonia than was added. Lines 1 and 3 have unit slopes because changes in the quantities of HCl and NH3 are identical when the system undergoes any spontaneous change as it returns to an equilibrium state. Does everything stop when equilibrium is reached? When a reaction system is not at equilibrium, the quantities of reactants or products will change until Q = K, at which point no further change will occur as long as the system remains at the same temperature and pressure. So all net change does come to an end when equilibrium is reached. But the absence of any net change does not mean that nothing is happening! Since all reactions are reversible at least in principal, we can regard an equilibrium A  B as the sum of two processes | A B | forward reaction | ratef = kf [A] | | B A | reverse reaction | rater= kr [B] | The expressions given in the rightmost column above simply reflect the fact that the rate at which a substance undergoes change should be proportional to its concentration; this is just another statement of the Law of Mass Action. The proportionality constants kf and kr are the forward and reverse rate constants. If we start with substance A alone, the absence of B means that the forward reaction alone is proceeding. Then, as the concentration of B begins to build up, the reverse reaction comes into operation, the rate of the forward reaction diminishes due to the reduction in the concentration of B. At some point these two processes will come into exact balance so that the forward and reverse rates are the same, at which point we can write and combine the k?s to obtain thus showing that the equilibrium constant can in a sense be regarded as the resultant of the two opposing rate constants. If the rate constant of the forward reaction exceeds that of the reverse step, then the equilibrium state will be one in which the product dominates. (Note carefully that although the two rate constants will generally be different, the forward and reverse rates themselves will always be identical at equilibrium.) The preceding paragraph shows how the concept of the equilibrium constant follows from the Law of Mass Action, but it is not a proper derivation of the formula for equilibrium constants in general, which must be done through the laws of thermodynamics. The single most important idea for you to carry along with you from this section is that equilibrium is a dynamic process in which the forward and reverse reactions are continually opposing each other in a dead heat. To see if you really understand this, try explaining to yourself how the Le Châtelier Principle as it applies to concentrations of reaction components follows directly from the idea of opposing reaction steps. Although it is the ratio of kf/kr that determines K, the magnitudes of these rate constants also make a difference; if they are small (or as is often the case, zero) then the reaction is kinetically inhibited and equilibrium will be achieved slowly or not at all. When their values are large, equilibrium is achieved quickly; the equilibrium is said to be facile and the reaction is fast. Sometimes a very slow equilibrium can be made more facile in the presence of a suitable catalyst.
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the highlighted terms in the context of this topic. - When arbitrary quantities of the different components of a chemical reaction system are combined, the overall system composition will not likely correspond to the equilibrium composition. As a result, a net change in composition ("a shift to the right or left") will tend to take place until the equilibrium state is attained.
- The equilibrium state is characterised by the value of the equilibrium expression, whose formulation is defined by the coefficients in the balanced reaction equation; it may be expressed in terms of concentrations, or in the case of gaseous components, as partial pressures.
- The various terms in the equilibrium expression can have any arbitrary value (including zero); the value of the equilibrium expression itself is called the equilibrium quotient Q.
- If the concentration or pressure terms in the equilibrium expression correspond to the equilibrium state of the system, then Q has the special value K, which we call the equilibrium constant.
- The ratio of Q/K thus serves as an index of how far the system is from its equilibrium composition, and its sign indicates the direction in which the net reaction must proceed in order to reach its equilibrium state.
- When this ratio is unity (Q = K), then the equilibrium state has been reached, and no further net change will take place as long as the system remains undisturbed.
Using concentration and pressure terms in equilibrium expressions Although we commonly write equilibrium quotients and equilibrium constants in terms of molar concentrations, any concentration-like term can be used, including mole fraction and molality. Sometimes the symbols Kc, Kx, and Km are used to denote these forms of the equilibrium constant. Bear in mind that the numerical values of K?s and Q?s expressed in these different ways will not generally be the same. Most of the equilibria we deal with in this course occur in liquid solutions and gaseous mixtures. We can express Kc values in terms of moles per liter for both, but when dealing with gases it is often more convenient to use partial pressures. These two measures of concentration are of course directly proportional: so for a reaction A(g) B(g) we can write the equilibrium constant as All of these forms of the equilibrium constant are only approximately correct, working best at low concentrations or pressures. The only equilibrium constant that is truly ?constant? (except that it still varies with the temperature!) is expressed in terms of activities, which you can think of as ?effective concentrations? that allow for interactions between molecules. In practice, this distinction only becomes important for equilibria involving gases at very high pressures (such as are often encountered in chemical engineering) and in ionic solutions more concentrated than about 0.001 M. We will not deal much with activities in this course. For a reaction such as CO2(g) + OH?(aq) ? HCO3?(aq) that involves both gaseous and dissolved components, a ?hybrid? equilibrium constant is commonly used: Clearly, it is essential to be sure of the units when you see an equilibrium constant represented simply by " K". Don't show unchanging concentrations! Substances whose concentrations undergo no significant change in a chemical reaction do not appear in equilibrium constant expressions. How can the concentration of a reactant or product not change when a reaction involving that substance takes place? There are two general cases to consider. The substance is also the solvent This happens all the time in acid-base chemistry. Thus for the hydrolysis of the cyanide ion
CN?+ H2O ? HCN + OH?, we write in which no [H2O] term appears. The justification for this omission is that water is both the solvent and reactant, but only the tiny portion that acts as a reactant would ordinarly go in the equilibrium expression. The amount of water consumed in the reaction is so minute (because K is very small) that any change in the concentration of H2O from that of pure water (55.6 mol L?1) will be negligible. Be careful about throwing away H2O whenever you see it. In the esterification reaction CH3COOH + C2H5OH ? CH3COOC2H5 + H2O that we discussed in a previous section, a [H2O] term must be present in the equilibrium expression if the reaction is assumed to be between the two liquids acetic acid and ethanol. If, on the other hand, the reaction takes place between a dilute aqueous solution of the acid and the alcohol, then the [H2O] term would not be included. The substance is a solid or a pure liquid phase. This is most frequently seen in solubility equilibria, but there are many other reactions in which solids are directly involved: CaF2(s) ? Ca2+(aq) + 2F?(aq) Fe3O4(s) + 4 H2(g) ? 4 H2O(g) + 3Fe(s) These are heterogeneous reactions (meaning reactions in which some components are in different phases), and the argument here is that concentration is only meaningful when applied to a substance within a single phase. Thus the term [CaF2] would refer to the ?concentration of calcium fluoride within the solid CaF2", which is a constant depending on the molar mass of CaF2 and the density of that solid. The concentrations of the two ions will be independent of the quantity of solid CaF2 in contact with the water; in other words, the system can be in equilibrium as long as any CaF2 at all is present. Throwing out the constant-concentration terms can lead to some rather sparse-looking equilibrium expressions. For example, the equilibrium expression for each of the processes shown in the following table consists solely of a single term involving the partial pressure of a gas: The last two processes represent changes of state or phase equilibria which can be treated exactly the same as chemical reactions. In each of the heterogeneous processes shown in the table, the reactants and products can be in equilibrium (that is, permanently coexist) only when the partial pressure of the gaseous product has the value consistent with the indicated Kp. Bear in mind also that these Kp s all increase with the temperature. Problem Example 1 What are the values of Kp for the equilibrium between liquid water and its vapor at 25°C, 100°C, and 120°C? The vapor pressure of water at these three temperatures is 23.8 torr, 760 torr (1 atm), and 1489 torr, respectively. Comment: These vapor pressures are the partial pressures of water vapor in equilibrium with the liquid, so they are identical with the Kp's when expressed in units of atmospheres. Solution: | | | |  | |  | | The partial pressure of H2O above the surface of liquid water in a closed container at 25°C will build up to this value. If the cover is removed so that this pressure cannot be maintained, the system will cease to be at equilibrium and the water will evaporate. | This temperature corresponds, of course, to the boiling point of water. The normal boiling point of a liquid is the temperature at which the partial pressure of its vapor is 1 atm. | The only way to heat water above its normal boiling point is to do so in a closed container that can withstand the increased vapor pressure. Thus a pressure cooker that operates at 120°C must be designed to withstand an internal pressure of at least 2 atm. | Significance of the numerical value of an equilibrium constant Your ability to interpret the numerical value of a quantity in terms of what it means in a practical sense is an essential part of developing a working understanding of Chemistry. This is particularly the case for equilibrium constants, whose values span the entire range of the positive numbers. Although there is no explicit rule, for most practical purposes you can say that equilibrium constants within the range of roughly 0.01 to 100 indicate that a chemically significant amount of all components of the reaction system will be present in an equilibrium mixture and that the reaction will be incomplete or ?reversible?. As an equilibrium constant approaches the limits of zero or infinity, the reaction can be increasingly characterized as a one-way process; we say it is ?complete? or ?irreversible?. The latter term must of course not be taken literally; the Le Châtelier principle still applies (especially insofar as temperature is concerned), but addition or removal of reactants or products will have less effect. Although it is by no means a general rule, it frequently happens that reactions having very large or very small equilibrium constants are kinetically hindered, often to the extent that the reaction essentially does not take place. The examples in the following table are intended to show that numbers (values of K), no matter how dull they may look, do have practical consequences! > | reaction | K | remarks | | N2(g) + O2(g) ? 2 NO(g) | 5 10?31 at 25°C,
0.0013 at 2100°C | These two very different values of K illustrate very nicely why reducing combustion-chamber temperatures in automobile engines is environmentally friendly. | | 3 H2(g) + N2(g) ? 2 NH3(g) | 7105 at 25°C,
56 at 1300°C | See the discussion of this reaction in the section on the Haber process. | | H2(g) ? 2 H(g) | 10?36 at 25°C,
610?5 at 5000° | Dissociation of any stable molecule into its atoms is endothermic. This means that all molecules will decompose at sufficiently high temperatures. | | H2O(g) ? H2(g) + ½ O2(g) | 810?41 at 25°C | You won?t find water a very good source of oxygen gas at ordinary temperatures! | | CH3COOH(l) ? 2 H2O(l) + 2 C(s) | Kc = 1013 at 25°C | This tells us that acetic acid has a great tendency to decompose to carbon, but nobody has ever found graphite (or diamonds!) forming in a bottle of vinegar. A good example of a super kinetically-hindered reaction! | Do equilibrium constants have units? The equilibrium expression for the synthesis of ammonia
3 H2(g) + N2(g) ? 2 NH3(g) is so Kp and Qp for this process would appear to have units of atm?1, and Kc and Qc would be expressed in mol?2 L2. And yet these quantities are often represented as being dimensionless. Which is correct? The answer is that both forms are acceptable. There are some situations (which you will encounter later) in which K?s must be considered dimensionless, but in simply quoting the value of an equilibrium constant it is permissible to include the units, and this may even be useful in order to remove any doubt about the units of the individual terms in equilibrium expressions containing both pressure and concentration terms. In carrying out your own calculations, however, there is rarely any real need to show the units. Strictly speaking, equilibrium expressions do not have units because the concentration or pressure terms that go into them are really ratios having the forms(n mol L?1)/(1 mol L?1) or (n atm)/(1 atm) in which the unit quantity in the denominator refers to the standard state of the substance; thus the units always cancel out. For substances that are liquids or solids, the standard state is just the concentration of the substance within the liquid or solid, so for something like CaF2(s), the term going into the equilibrium expression is [CaF2]/[CaF2] which cancels to unity; this is the reason we don?t need to include terms for solid or liquid phases in equilibrium expressions. The subject of standard states would take us somewhat beyond where we need to be at this point in the course, so we will simply say that the concept is made necessary by the fact that energy, which ultimately governs chemical change, is always relative to some arbitrarily defined zero value which, for chemical substances, is the standard state. Equilibrium expressions and the reaction equation It is important to remember that an equilibrium quotient or constant is always tied to a specific chemical equation, and if we write the equation in reverse or multiply its coefficients by a common factor, the value of Q or K will be different. The rules are very simple: ? Writing the equation in reverse will invert the equilibrium expression; ? Multiplying the coefficients by a common factor will raise Q or K to the corresponding power. Here are some of the possibilities for the reaction involving the equilibrium between gaseous water and its elements: Heterogeneous reactions: the vapor pressure of solid hydrates Heterogeneous reactions are those involving more than one phase. Some examples: | Fe(s) + O2(g) ? FeO2(s) | air-oxidation of metallic iron (formation of rust) | | CaF2(s) ? Ca(aq) + F+(aq) | dissolution of calcium fluoride in water | | H2O(s) ? H2O(g) | sublimation of ice (a phase change) | | NaHCO3(s) + H+(aq) ? CO2(g) + Na+(aq) + H2O(g) | formation of carbon dioxide gas from sodium bicarbonate when water is added to baking powder (the hydrogen ions come from tartaric acid, the other component of baking powder.) | A particularly interesting type of heterogeneous reaction is one in which a solid is in equilibrium with a gas. The sublimation of ice illustrated in the above table is a very common example. The equilibrium constant for this process is simply the partial pressure of water vapor in equilibrium with the solid? the vapor pressure of the ice.  Many common inorganic salts form solids which incorporate water molecules into their crystal structures. These water molecules are usually held rather loosely and can escape as water vapor. Copper(II) sulfate, for example forms a pentahydrate in which four of the water molecules are coordinated to the Cu 2+ ion while the fifth is hydrogen-bonded to SO 42?. This latter water is more tightly bound, so that the pentahydrate loses water in two stages on heating: These dehydration steps are carried out at the temperatures indicated above, but at any temperature, some moisture can escape from a hydrate. For the complete dehydration of the pentahydrate we can define an equilibrium constant CuSO 4·5H 2O (s) ? CuSO 4(s) + 5 H 2O (g) Kp = 1.14 10 ?10 The vapor pressure of the hydrate (for this reaction) is the partial pressure of water vapor at which the two solids can coexist indefinitely; its value is Kp1/5 atm. If a hydrate is exposed to air in which the partial pressure of water vapor is less than its vapor pressure, the reaction will proceed to the right and the hydrate will lose moisture. Vapor pressures always increase with temperature, so any of these compounds can be dehydrated by heating. Loss of water usually causes a breakdown in the structure of the crystal; this is commonly seen with sodium sulfate, whose vapor pressure is sufficiently large that it can exceed the partial pressure of water vapor in the air when the relative humidity is low. What one sees is that the well-formed crystals of the decahydrate undergo deterioration into a powdery form, a phenomenon known as efflorescence. When a solid is able to take up moisture from the air, it is described as hygroscopic. A small number of anhydrous solids that have low vapor pressures not only take up atmospheric moisture on even the driest of days, but will become wet as water molecules are adsorbed onto their surfaces; this is most commonly observed with sodium hydroxide and calcium chloride. With these solids, the concentrated solution that results continues to draw in water from the air so that the entire crystal eventually dissolves into a puddle of its own making; solids exhibiting this behavior are said to be deliquescent. | name | formula | vapor pressure, torr | | 25°C | 30°C | | sodium sulfate decahydrate | Na2SO4·10H2O | 19.2 | 25.3 | | copper(II) sulfate pentahydrate | CuSO4·5H2O | 7.8 | 12.5 | | calcium chloride monohydrate | CaCl2·H2O | 3.1 | 5.1 | | (water) | H2O | 23.5 | 31.6 | Problem Example 2 At what relative humidity will copper sulfate pentahydrate lose its waters of hydration when the air temperature is 30°C? What is Kp for this process at this temperature?
Solution: From the table above, we see that the vapor pressure of the hydrate is 12.5 torr, which corresponds to a relative humidity (you remember what this is, don?t you?) of 12.5/31.6 = 0.40 or 40%. This is the humidity that will be maintained if the hydrate is placed in a closed container of dry air For this hydrate, Kp = P(H2O)0.5, so the partial pressure of water vapor that will be in equilibrium with the hydrate and the dehydrated solid (remember that both solids must be present to have equilibrium!), expressed in atmospheres, will be (12.5/760)5 = 1.20E-9. One of the first hydrates investigated in detail was calcium sulfate hemihydrate (CaSO4·½ H2O) which LeChâtelier (he of the ?principle?) showed to be the hardened form of CaSO4 known as plaster of Paris. Anhydrous CaSO4 forms compact, powdery crystals, whereas the elongated crystals of the hydrate bind themselves into a cement-like mass. How can we find the equilibrium constant for a series of reactions? Many chemical changes can be regarded as the sum or difference of two or more other reactions. If we know the equilibrium constants of the individual processes, we can easily calculate that for the overall reaction according to this rule: The equilibrium constant for the sum of two or more reactions
is the product of the equilibrium constants for each of the steps. Problem Example 3 Given the following equilibrium constants: | CaCO3(s) ? Ca2+(aq) + CO32?(aq) | K1 = 10?6.3 | | HCO3?(aq) ? H+(aq) + CO32?(aq) | K2 = 10?10.3 | Calculate the value of K for the reaction CaCO3(s) + H+(aq) ? Ca2+(aq) + HCO3?(aq) Solution: The net reaction is the sum of reaction 1 and the reverse of reaction 2: | CaCO3(s) ? Ca2+(aq) + CO32?(aq) | K1 = 10?6.3 | | H+(aq) + CO32?(aq) ? HCO3?(aq) | K?2 = 10?(?10.3) | | CaCO3(s) + H+(aq) ? Ca2+(aq) + HCO3?(aq) | K = K1/K2 = 10(-8.4+10.3) = 10+1.9 | Comment: This net reaction describes the dissolution of limestone by acid; it is responsible for the eroding effect of acid raine on buildings and statues. This an example of a reaction that has practically no tendency to take place by itself (small K1) being "driven" by a second reaction having a large equilibrium constant (K?2). From the standpoint of the LeChâtelier principle, the first reaction is "pulled to the right" by the removal of carbonate by hydrogen ion. Coupled reactions of this type are widely encountered in all areas of chemistry, and especially in biochemistry, in which a dozen or so reactions may be linked The individual coupled reactions do not need to actually take place in the overall process; very frequently we use their equilibrium constants in order to calculate the equilibrium constant of another reaction. Problem Example 4 The synthesis of HBr from hydrogen and liquid bromine has an equilibrium constant Kp = 4.5¥1015 at 25°C. Given that the vapor pressure of liquid bromine is 0.28 atm, find Kp for the homogeneous gas-phase reaction at the same temperature. Solution: The net reaction we seek is the sum of the heterogeneous synthesis of HBr and the reverse of the vaporization of liquid bromine: | H2(g) + Br2(l) ? 2 HBr(g) | K p = 4.5 10 15 | | Br2(g) ? Br2(l) | Kp = (0.28)?1 | | H2(g) + Br2(g) ? 2 HBr(g) | Kp = 1.61019 | Measuring and calculating equilibrium constants Clearly, if the concentrations or pressures of all the components of a reaction are known, then the value of K can be found by simple substitution. Observing individual concentrations or partial pressures directly may be not always be practical, however. If one of the components is colored, the extent to which it absorbs light of an appropriate wavelength may serve as an index of its concentration. Pressure measurements are ordinarily able to measure only the total pressure of a gaseous mixture, so if two or more gaseous products are present in the equilibrium mixture, the partial pressure of one may need to be inferred from that of the other, taking into account the stoichiometry of the reaction. Problem Example 1 In an experiment carried out by Taylor and Krist ( J. Am. Chem. Soc. 1941: 1377), hydrogen iodide was found to be 22.3% dissociated at 730.8°K. Calculate K c for 2 HI(g) H 2(g) + I 2(g). Solution: No explicit molar concentrations are given, but we do know that for every n moles of HI, 0.223n moles of each product is formed and (1?0.233)n = 0.777n moles of HI remains. For simplicity, we assume that n=1 and that the reaction is carried out in a 1.00-L vessel, so that we can substitute the required concentration terms directly into the equilibrium expression. Problem Example 2 Ordinary white phosphorus, P4, forms a vapor which dissociates into diatomic molecules at high temperatures: P 4(g) 2 P 2(g) A sample of white phosphorus, when heated to 1000°C, formed a vapor having a total pressure of 0.20 atm and a density of 0.152 g L?1. Use this information to evaluate the equilibrium constant Kp for this reaction. Solution: Before worrying about what the density of the gas mixture has to do with Kp , start out in the usual way by laying out the information required to express Kp in terms of an unknown x. | | | | | initial moles: | 1 | 1 - x | Since K is independent of the number of moles, assume the simplest case. | | moles at equilibrium: | 1 - x | 2x | x is the fraction of P4 that dissociates. | | eq. mole fractions: |  |  | The denominator is the total number of moles:
(1-x) + 2x = 1+x. | | eq. partial pressures: |  |  | Partial pressure is the mole fraction times the total pressure. | Expressing the equilibrium constant in terms of x gives Now we need to find the dissociation fraction x of P4, and at this point we hope you remember those gas laws that you were told you would be needing later in the course! The density of a gas is directly proportional to its molecular weight, so you need to calculate the densities of pure P4 and pure P2 vapors under the conditions of the experiment. One of these densities will be greater than 0.152 gL?1 and the other will be smaller; all you need to do is to find where the measured density falls in between the two limits, and you will have the dissociation fraction. The molecular weight of phosphorus is 31.97, giving a molar mass of 127.9 g for P4. This mass must be divided by the volume to find the density; assuming ideal gas behavior, the volume of 127.9 g (1 mole) of P4 is given by RT/P, which works out to 522 L (remember to use the absolute temperature here.) The density of pure P4 vapor under the conditions of the experiment is then d = m/V = (128 g mol ?1) (522 L mol ?1) = 0.245 g L ?1. The density of pure P 2 would be half this, or 0.122 g L ?1. The difference between these two limiting densities is 0.123 g L ?1, and the difference between the density of pure P 4 and that of the equilibrium mixture is (.245 ?.152) g L ?1 or 0.093 g L ?1. The ratio 0.093  0.123 = 0.76 is therefore the fraction of P 4 that remains and its fractional dissociation is (1 ? 0.76) = 0.24. Substituting into the equilibrium expression above gives K p = 1.2. How can we predict equilibrium compositions? This is by far the most common kind of equilibrium problem you will encounter: starting with an arbitrary number of moles of each component, how many moles of each will be present when the system comes to equilibrium? The principal source of confusion and error for beginners at this relates to the need to determine the values of several unknowns (a concentration or pressure for each component) from a single equation, the equilibrium expression. The key to this is to make use of the stoichiometric relationships between the various components, which usually allow us to express the equilibrium composition in terms of a single variable. The easiest and most error-free way of doing this is adopt a systematic approach in which you create and fill in a small table as shown in the following problem example. You then substitute the equilibrium values into the equilibrium constant expression, and solve it for the unknown. This very often involves solving a quadratic or higher-order equation. Quadratics can of course be solved by using the familiar quadratic formula, but it is often easier to use an algebraic or graphical approximation, and for higher-order equations this is the only practical approach. There is almost never any need to get an exact answer, since the equilibrium constants you start with are rarely known all that precisely. Problem Example 3 Phosgene (COCl2) is a poisonous gas that dissociates at high temperature into two other poisonous gases, carbon monoxide and chlorine. The equilibrium constant Kp = 0.0041 at 600°K. Find the equilibrium composition of the system after 0.124 atm of COCl2 is allowed to reach equilibrium at this temperature. Solution: Start by drawing up a table showing the relationships between the components: | COCl 2 | | | | | | 0 | 0 | | | | | | | | | | | Substitution of the equilibrium pressures into the equilibrium expression gives This expression can be rearranged into standard polynomial form x2 +.0041 x ? 0.00054 = 0 and solved by the quadratic formula, but we will simply obtain an approximate solution by iteration. Because the equilibrium constant is small, we know that x will be rather small compared to 0.124, so the above relation can be approximated by which gives x = 0.0225. To see how good this is, substitute this value of x into the denominator of the original equation and solve again: This time, solving for x gives 0.0204. Iterating once more, we get and x = 0.0206 which is sufficiently close to the previous to be considered the final result. The final partial pressures are then 0.104 atm for COCl2, and 0.0206 atm each for CO and Cl2. Note: using the quadratic formula to find the exact solution yields the two roots ?0.0247 (which we ignore) and 0.0206, which show that our approximation is quite good. Problem Example 4 The gas-phase dissociation of phosphorus pentachloride to the trichloride has Kp = 3.60 at 540°C: PCl 5 PCl 3 + Cl 2 What will be the partial pressures of all three components if 0.200 mole of PCl5 and 3.00 moles of PCl3 are combined and brought to equilibrium at this temperature and at a total pressure of 1.00 atm? Solution: The partial pressures in the bottom row were found by multiplying the mole fraction of each gas by the total pressure: Pi = XiPt. The term in the denominator of each mole fraction is the total number of moles of gas present at equilibrium: (0.200 ? x) + (3.00 + x) + x = 3.20 + x.
Substituting the equilibrium partial pressures into the equilibrium expression, we have whose polynomial form is 4.60 x2 + 13.80 x ? 2.304 = 0. Plotting this on a graphical calculator yields x = 0.159 as the positive root:
Substitution of this root into the expressions for the equilibrium partial pressures in the table yields the following values: P(PCl5) = 0.012 atm, P(PCl3) = 0.94 atm, P(Cl2) = 0.047 atm. Effects of dilution on an equilibrium In the section that introduced the LeChâtelier principle, it was mentioned that diluting a weak acid such as acetic acid CH3COOH (?HAc?) will shift the dissociation equilibrium to the right: HAc + H 2O H 3O + + Ac ? Thus a 0.10M solution of acetic acid is 1.3% ionized, while in a 0.01M solution, 4.3% of the HAc molecules will be dissociated. This is because as the solution becomes more dilute, the product [H3O+][Ac?] decreases more rapidly than does the [HAc] term. At the same time the concentration of H2O becomes greater, but because it is so large to start with (about 55.5M), any effect this might have is negligible, which is why no [H2O] term appears in the equilibrium expression. For a reaction such as CH 3COOH(l) + C 2H 5OH(l) CH 3COOC 2H 5(l) + H 2O(l) (in which the water concentration does change), dilution will have no effect on the equilibrium; the situation is analogous to the way the pressure dependence of a gas-phase reaction depends on the number of moles of gaseous components on either side of the equation. Problem Example 5 The biochemical formation of a disaccharide (double) sugar from two monosaccharides is exemplified by the reaction fructose + glucose-6-phosphate sucrose-6-phosphate (Sucrose is ordinary table sugar.) To what volume should a solution containing 0.050 mol of each monosaccharide be diluted in order to bring about 5% conversion to sucrose phosphate? Solution: The initial and final numbers of moles are as follows: Substituting into the expression for Kc in which the solution volume is the unknown, we have Solving for V gives a final solution volume of 78 mL. Phase distribution equilibria It often happens that two immiscible liquid phases are in contact, one of which contains a solute. How will the solute tend to distribute itself between the two phases? One?s first thought might be that some of the solute will migrate from one phase into the other until it is distributed equally between the two phases, since this would correspond to the maximum dispersion (randomness) of the solute. This, however, does not take into the account the differing solubilities the solute might have in the two liquids; if such a difference does exist, the solute will preferentially migrate into the phase in which it is more soluble. For a solute S distributed between two phases a and b the process Sa = Sb is defined by the distribution law in which Ka,b is the distribution ratio (also called the distribution coefficient) and [S]i is the solubility of the solute in the phase. The transport of substances between different phases is of immense importance in such diverse fields as pharmacology and environmental science. For example, if a drug is to pass from the aqueous phase with the stomach into the bloodstream, it must pass through the lipid (oil-like) phase of the epithelial cells that line the digestive tract. Similarly, a pollutant such as a pesticide residue that is more soluble in oil than in water will be preferentially taken up and retained by marine organism, especially fish, whose bodies contain more oil-like substances; this is basically the mechanism whereby such residues as DDT can undergo biomagnification as they become more concentrated at higher levels within the food chain. For this reason, environmental regulations now require that oil-water distribution ratios be established for any new chemical likely to find its way into natural waters. The standard ?oil? phase that is almost universally used is octanol, C8H17OH. In preparative chemistry it is frequently necessary to recover a desired product present in a reaction mixture by extracting it into another liquid in which it is more soluble than the unwanted substances. On the laboratory scale this operation is carried out in a separatory funnel in which the two phases are brought into intimate contact by shaking. After the two liquids have separated into layers, the bottom layer is drawn off. If the distribution ratio is too low to achieve efficient separation in a single step, it can be repeated; there are automated devices that can carry out hundreds of successive extractions, each yielding a product of higher purity. In these applications our goal is to exploit the LeChâtelier principle by repeatedly upsetting the phase distribution equilibrium that would result if two phases were to remain in permanent contact.  | This separatory funnel is used to extract a substance from one liquid into another in which it is more soluble. The two immiscible liquids are poured into the funnel through the opening at the top. The funnel is then shaken to bring the two phases into intimate contact, and then set aside to allow the two liquids to separate into layers, which are then separated by allowing the more dense liquid to exit through the stopcock at the bottom. | Problem Example 6 The distribution ratio for iodine between water and carbon disulfide is 650. Calculate the concentration of I2 remaining in the aqueous phase after 50.0 mL of 0.10M I2 in water is shaken with 10.0 mL of CS2. Solution: The equilibrium constant is Let m1 and m2 represent the numbers of millimoles of solute in the water and CS2 layers, respectively. Kd can then be written as (m2/10 mL) ÷ (m1/50 mL) = 650. The number of moles of solute is (50 mL) × (0.10 mmol mL?1) = 5.00 mmol, and mass conservation requires that m1 + m2 = 5.00 mmol, so m2 = (5.00 ? m1) mmol and we now have only the single unknown m1. The equilibrium constant then becomes ((5.00 ? m1) mmol / 10 mL) ÷ (m1 mmol / 50 mL) = 650. Simplifying and solving for m1 yields (0.50 ? 0.1)m1 / (0.02 m1) = 650, with m1 = 0.0382 mmol. The concentration of solute in the water layer is (0.0382 mmol) / (50 mL) = 0.000763 M, showing that almost all of the iodine has moved into the CS2 layer.
|
About the Author:
|
this article: 44 points
(with 8 
in 10 votes ) [?]
|
|
You have to be logged on to rate
|
|
(posted on 11 Feb 2008 23:18:48 IST)
|
| oooha...nice work done..! |
|
(posted on 11 Feb 2008 23:19:33 IST)
|
| gr88 job really |
|
|
(posted on 11 Feb 2008 23:21:27 IST)
|
| ooha..nice article..! |
|
(posted on 11 Feb 2008 23:22:47 IST)
|
| nice article!! |
|
(posted on 11 Feb 2008 23:26:41 IST)
|
| Classic |
|
(posted on 11 Feb 2008 23:33:34 IST)
|
i need these ,..............
cause i m weak in physical chem and dont find that subj interesting at all |
|
(posted on 12 Feb 2008 09:14:52 IST)
|
| superb.... |
|
(posted on 12 Feb 2008 15:30:47 IST)
|
| excellent!! |
|
|
|
|
|
|
|
Back to Community Shelf
Awaiting Review for Nickels![[Post New]](/templates/default/images/icon_minipost_new.gif){kind=icon}
