Teaching problems in presenting the law of chemical equilibrium to

Teaching problems in presenting the law of chemical equilibrium to students in analytical chemistry. C. S. Adams. J. Chem. Educ. , 1940, 17 (1), p 22...
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TEACHING PROBLEMS in PRESENTING the LAW of CHEMICAL EQUILIBRIUM to STUDENTS in ANALYTICAL CHEMISTRY C. S. ADAMS Antioch College, Yellow Springs, Ohio INTRODUCTION the author's explanations which he has found to be of HEMICAL equilibrium has become an important value. Since most freshmen are not familiar with the calsubject in the study of pure and applied chemisculus, the derivation must be confined to algebra and try. A knowledge of its principles is essential to the mathematics of variation. Moreover it is assumed the student for understanding the kinetics of chemical reactions in general as well as for furnishing a theoreti- in the following discussion that the general character cal basis for separations in analytical chemistry. In of physical and chemical systems in equilibrium has fact, it has been said that one of the fundamental prob- been presented by the teacher; that the effects of conlems in theoretical chemistry is the study of equilibrium centration, temperature, and catalysts and the meaning states. The growing importance of the subject is of "rate of reaction" have been explained. The evidenced by the additional space given to it in recent concepts of "activities" and "fugacities," and the texts on general chemistry, qualitative and quantita- expression of the equilibrium law in terms of mole fractions or partial pressures are purposely omitted. tive analysis, and physical chemistry. The subject of chemical equilibrium is usually intro- Owing to lack of space only generalized equilibrium duced along with qualitative analysis, either during equations will be presented. The teacher should illusthe second semester of the freshman year or the early trate these generalized equations with suitable exampart of the sophomore year. Freshmen and sopho- ples. A certain amount of repetition of traditional mores in general seem to have difficulty in under- presentations for illustrative purposes is unavoidable. standing certain steps in the derivation of the Law of . ., Chemical Equilibrium and its application to certain DERIVATION OF THE EQUILIBRIUM L A W problems in analytical chemistry. Evgn seniors and Let us start with a simple, bimolecular, homograduate students, who have had the benefit of a more geneous, chemical-equilibrium system involving unlike detailed treatment of chemical kinetics and equilibrium molecules in which one molecule of A reacts with one systems in general, often admit they still have difficulty B to form one molecule of C and one molemolecule of with certain phases of the subject. cule of D, expressed as follows: When one examines the various textbooks which oreA + B a C + D sent this subiect to the beccinnine student the reasons for some of tliese difficultiesseem'to be apparent. The with which A reacts with B may depend Th, competition for space in most textbooks forces the upon (is a function of) a number of independent authors to condense the treatment of this subject to a variables which can be expressed as follows: point where the student is left in a state of mental conR~~~fomrd is a function of concentration2 of A , fusion. In other books the derivation of the equilibconcentration of B, temperature, catalysts, nature rium law is incomplete and frequently mathematically of A , nature of B, nature of solvent or medium in inaccurate, or based on implied assumptions not recogwhich the reaction takes place, mechanism of the nized by the student. reaction; electrical and magnetic effects, light, energy I t is not the object of this paper to discuss the various factors, ,,tivation, and a ~ ~ l i c a t i o nofs the law of chemical eauilibrium or its -lkitations, but rather to point out some of the dif1 ~y rate we mean the number of moles of A or B reacting in ficulties encountered in teaching its derivation to aria- unit time. Time is usually expressed in seconds. Rate of a is also called velocity or speed of a reaction. the teaching ap- reaction lytical students, to review a The concentration of A means moles of A in a liter of the proaches to these difficulties, and to present some of reaction mixture. 22

C

-

The velocity with which C reacts with D may also depend upon a similar number of factors, or Rate h c h r d is a function of concentration of C, concentration of D, temperature, catalysts, nature of C, nature of D, and the other factors mentioned above. At equilibrium the forward and backward rates are equal, therefore the two functions are equal and we may say,

Ratel = (A) X Rate, = K16 ( A ) X Rateae = (C) X Ratea = K 8 (C) X

Rate (forward) = J[(A), (B)], all other factors constant2 Rate (backward) = f[(C), (Dl], all other factors constant

It is now important to know the relationship of.these rates to the concentration terms. Guldberg and Waage enunciated this relationship in 1867 when they stated. "The rate of a chemical reaction is proportional to the active masses of the reacting substances present a t that time." This is known as The Law of Mass Action, also called The Law of Molecular Concentration. They defined the term "active masses" as the molecular concentrations of the reacting substances. By concentration they meant moles $er unit volume. From this law we may state with respect to the forward rate, Rate,' = (A), when (B) and other factors are constant. Ratel = (B), when (A) and other factors are constant.

It may he difficult for the student to see bow ( B ) remains constant in the above expression. Experimentally this would be difficult since i t would necessitate adding B as fast as it reacted with A. However, unless ( B ) or ( A ) in the above expressions remains constant, the rates will not, in general, be directly proportional to their respective concentration terms. This point is stressed, since few texts that .present the above step emphasize this fact, and the average student does not recognize the implied assumptions. In actual practice both ( A ) and (B) usually vary simultaneously, so that many teachers do not present the above step but state that, since the rate varies as the concentration of A and as the concentration of B, i t will vary as the firodzrc6 of these urncentrations provided that all other factors that affect the velocity are constant, or Varentheses enclosing a term mean the concentration of that term. Rate1 means rate forward.

'

(B). (D) (D)

therefore Similar reasoning shows therefore

At equilibrium, when the forward and backward rates are equal, the concentrations of A , B , C, and D remain constant and we have (C) X (D) X K 2 z 1, therefore (A) X (B) X KI ()' = 5 = K , (equilibrium constant) K2 (A) X (B)

J(Conc. C. Conc. D. temperature, catalysts, other factors) . - 1 J(Conc. A, Conc. B, tmrpcraulrc, catalysts, other factors)

When the temperature is constant it will be the same inboth numerator and denominator; the effect of any catalysts on forward or backward rates will be the same, otherwise, a catalyst would shift the position of equilibrium and this does not happen. In general, the effect of all the other factors, excluding the concentration terms, may be considered constant in the ordinary study of this system. It is obvious, however, that the ratio of the concentration terms in both numerator and denopinator will not, in general, be equal to l. We may say, therefore,

(B),

The teacher then goes on to show that for the halanced general equation for chemical equilibrium systems, namely, nA

+ mB.. . . . 8oC + rD.. . . ..

the equilibrium law specifies that (C). X (D)'. . . . (A)"X (B)'". . . .

-

The above presentation introduces three teaching problems.' (1) Why does the rate vary as the product of the con-

centrations (other factors constant) ? (2) Why do these products have to he raised to the

pmuer of their respective coeficients? (3) What is the physicnl significance of the constants, &, K2, Kc? These questions will be taken up in the order named. (1) THE VELOCITY VARIESAS THE PRODUCT OF THE CONCENTRATIONS.-Iu answering this question many ttachcrs rtsort to analogies. For example, it is pointed out that thc arra ul a rrctancular field \.ariesus//z~ lpneth and as the widtk it will als&ary as the length times 'ihe width. Many other analogies involving the concept of speed are used. Each teacher appears to have his own favorite analogy or approach in answering this question. Belcher and Williams (1)have used the following apt analogy to explain this point. "Suppose a hiker contacts an automobile a t the rate of one each hour; then the chances for a ride are one each hour. If traffic speeds up so that he contacts ten automobiles each hour, then his chances for a ride are ten per hour. If nine more hikers join the first, then each of the ten hikers will contact each of the ten automobiles, or there will be 10 X 10 or 100 contacts or meetmgs per hour. "But every meeting between hiker and automobile d w s not result in a ride. If one considers millions of such meetings i t will be found that a constant fraction K (proportionality constant) of such meetingsxill result in a ride, or Hikers X automobiles = meetings Meetings X K = rides "In a similar way a constant fraction K of the number of collisions of A molecules and B molecules will result in the formation of C and D, or K , and KSare known as oelocity proportionality constants or specific reaction rates. Rates means rate backward. Other problems such as an explanation of "Why is the concentration of a solid (and sometimes the Liquid and gaseous solvent) is heterogeneous (and homogeneous) equilibrium systems considered constant?" could well have been added t o the list which has been given.

' '

( A ) x ( B ) = chance of collision Chance of collision X K = speed of reaction, or ( A ) X ( B ) X K = speed of forward reaction."

concentrations as in the case of the unlike molecules aboue. The author has found the following proof of value in explaining why the velocity of the above reaction will vary as the product of the concentrations of A and B.

I t has been the author's experience that, while many students may carelessly accept these analogies as satisfactory explanations, there are always a few who feel Let be some concentration of A at the beginning the analogies are irrelevant and they want a more of the reaction. rigorous proof. Of course, the teacher can avoid the Let (BI) be some concentration of at the beginning, issue by sending these students back to their matheLet (A') be some concentration of A after a time or matics to review the laws of variation and functions, and at the meaning of the proportionality constant. UnforLet be some concentration of after a time or tunately, most mathematics books do not help the at student much in this particular problem. Another effective way, used by many teachers, of Then i t follows from the Law of Mass Action, if (A,) explaining the significance and meaning of this product changes to (A*) and the (BI) is held constant, that (A-) X (B) is the following. *R = Imaeine two molecules of A and three of B in a con,-., when (Bi) and other factors are constant. tainer defmite volume (see Figure 1). Each A moleRate, = some intermediate rate between initial and final rates. cule has a chance of colliding and reacting with each Now let (B,) change to its final (B.) while (A*)is held constant, then

of

A.-LL=

Rate.

?,!I

=

(B.) when (A.) and other factors are constant. (BJ'

Eliminating the intermediate rate (Rate+) from the above two equations and rearranging terms, we have

Rate, = ( A ) X ( B )

Similar reasoning will show that B molecule. Then two A molecules will have 2 X 3 or 6 chances of collision with the B molecules. If we double the concentration of the A molecules (4A), we OF THE COEPFICIENTS.-When more (2) PROBLEM will double the number of chances (12), of A mole- than one molecule of A, B, C, and D are required to cules colliding with B molecules. While i t does not balance the equation, then certain additional difficulties necessarily follow that all of the collisions will be &ec- in explanation are introduced. Thus, if tive and result in a reaction, yet the number of molecules reacting will either be equal tdor vary as the number of collisions if the temperature remains c ~ n s t a n t . ~ It also follows that the number of molecules of A and then the problem is to show why the equilibrium law B will be proportional to their concentrations expressed, specifies that say, in moles per liter. We may state, therefore, that ( C ) . X (D)'. . . . = Kc Rate of A reacting with B

a

number of collisions 2 X 3

(A)" X (B)".

a

(A) X (B)

...

Most authors of analytical iexts either do not atThrough the use of diagrams of this kind the number tempt to explain how this equation is derived or their of possible collisions between any number of A and B attempts are inadequate and often misleading. I t has been the author's experience that the difficulty which molecules becomes apparent. It is important to point out to the student that in the students have in applying the Equilibrium Law to cerexample given only collisions between unlike molecules tain types of problems originates from the fact that the result in a reaction, i. e., collisions between the A mole- derivation of the law itself was not made clear. The cules or between the B molecules are not counted since following examples illustrate typical presentations of they do not lead to a reaction. Later we shall dis- this derivation. Consider the following equilibrium cover that the number of collisions between two 2ik system molecules that produce a reaction varies as their total I, HI -F? 2HI concentration squared, and not as the product of their It is then pointed out that this equation may be reIn some reactions (ionic and some molecular reactions) each collision appears to result in a reaction, while in others only a written in the following form

+

small fraction of the collisions result in a reaction. Effective collisions are those that produce a reaction.

Applying the Equilibrium Law to this equation we get

The student very naturally believes, from the above derivation, that (HZ)2 is one-half the total concentration of HI squared whereas i t actually represents the total concentration of HI squared. Wben he is told this he is further confused by the following presentation: Let "a" equal the initial concentration of 12, "b" equal the initial concentration of HZ,and "x" equal the amount of l a and Hz which have reacted when equilibrium is attained, then

The student feels in this case that the concentration of HZ is both doubled and squared. The above derivations are therefore misleading and do not bring out the fact that the speed or rate of the above bimolecular reaction between like molecules of HI varies as the total concentration of the undissociated HZ molecules spared. Another method used quite generally by teachers (2) to make clear the use of coefficients in the general equilibrium equation is the combination of the separate equilibrium equations expressing the step-wise or bimolecular mechanism of the reaction. This method is effective and gives little trouble to the student, providing the real meaning of the separate concentration terms in the separate equilibrium equations is defined. Its weakness is that it does not enable the student to visualize the collision mechanism of the reaction and how this mechanism justifies the use of coefficients as powers in equilibrium equations. For example take the step-wise ionization of hydrogen sulfide, which may be represented as follows HB HS-

$ Hf

F;?

HC

+ HS-

+ S--

(primary ionization) (secondary ignization)

Applying the Equilibrium Law to these two steps, we have "Hf)

(HS-) = K I = primary ionization constant, and (Hi3

It is then pointed out that since the value of (HS-) is common to and therefore the same for both expressions, we may eliminate it by multiplying the two equations, or

Since the (H+)Z in this latter equation actually means (3) the total hydrogeu-ion concentration squared, the student is confused. He reasons that since the primary ionization is greater than the secondary ionization then the value of (H+) from the primary 'The more recent texts have adopted the hydronium ion &Of instead of H+.

ionization is larger and dserent from the value of (H+) from the secondary ionization; therefore the square of these terms should mean (H+), X (H+). which, of course, is not the total concentration of H + squared. The latter may be expressed as follows: [(H+),

+ (H+).I2

= W+)'

This picture may be simplified if we consider the fact that the H2S molecules, HS-, S-- and H + ions are all present in equilibrium in the same solution. This means there can be but one concentration of hydrogen ion, or H%SF?

+H+ HS-

~3H+

+ S--

Therefore the (H+) in both primary and secondary equilibrium equations is the same and represents the total concentration of H + in the equilibrium mixture. It may also help to point out (4) that in the case of weak dibasic acids such as H2S the secondary ionization takes place to a very small extent as compared to the primary ionization. It will therefore follow that the total concentration of H + will be practically equal to the concentration of HS-, i. e., practically all of the H + comes from the primary ionization. Using this approximation in the secondary ionization equilibrium equation we note that the concentration of the bivalent ion (S--) is numerically equal to the secondary ionization constant. The difficulty of the coefficients appears frequently in dealing with solubility product problems. For example, the solubility product of a saturated solution .. of Mg(0H)e is defined as (Mgc+) X ( 0 H - p =

K.,

If the concentration of Mg++ and OH- are required and the solubility product constant is given, the student is instructed to let x = (Mgtf) and 2 x = (OH-), then Here again the student feels he has both doubled and squared the concentration of OH-. It should be pointed out that 2 x i s the total concentration of OH- and, as subsequently shown, the Mass Law requires that this concentration be squared. The author has found the following approach to this general difficulty to be more easily understood by the student since it draws his attention to the important fact that the rate of a reaction between two unlike molecules, atoms, or ions varies as their product, whereas, if they are alike, it varies as their total concentration squared. (1) Mechanism of the Reaction. Several mechanisms for the general reaction

may be postulated: ( a ) (1) Let n A molecules react to form one molecule of the polymer A,, or

nA - A .

(Equation 1 )

Since, as a rule, only two molecules collide a t any instant (because the chance of a three-body collision is statistically improbable) we may imagine the formation of A. to take place in bimolecular steps, or 2A

+ A2, then A2

+A

-

A3 and so forth until A , is formed.

(2) Let mB molecules react'in a similar manner to form one molecule of the polymer B,, or nzB

B,

(Equation 2 )

(3) Now let A. react with B, to form the product o molecules of C and r molecules of D, or A.+B,-oCfrD

(Equation 3 )

Case I Let n = 2, then 2A + Ax; let N = 5 molecules of A; N ar ( A ) . Let us bave, say, five molecules of A in a given container. The problem is to determine the number of chances of collision between any two (=n) molecules of A, not counting the same chance twice. A1 molecule (see Figure 2) will have four chances of collision with the other A molecules, namely, 1-2, 1-3, 1 4 , and 1-5. A2 will have three further chances, namely, 2-3, 2-4, and 2-5. A3 will have two further chances, namely, 3 4 and 3-5. A4 will bave one further chance, namely, 4 5 . The total number of chances of two A molecules colliding will be 4 3 2 1 or ten chances.

+ + +

I t is apparent that the sum of the three equations above gives us our original equation, namely,

(b) If one objects to the assumption that the intermediate molecules A. and B, are formed we may assume the formation of intermediate compounds between A and B taking place in bimolecular steps, or A +B-AB AB + A - A I B AIBs and so forth, nntil AIB B A d , is formed. then

+

-

The result will be the same as above. (c) While it is statistically improbable, we may cousider the simultaneous collisiou of nA molecules to form A, and the simultaneous collision of mB molecules to form B,, or the simultaneous collision of nA molecules and mB molecules to form the product. The mathematical treatment of these possibilities will lead to the same conclusion as above. One of the inferences of t h e second law of thermodynamics is that for every possible mechanism of transformation of reactants A and 8 into the products. C and D, there is a reverse mechanism. Any chemical equation which represents the forward reaction will, when read backwards, represent the reverse reaction. When equilibrium has been reached the rate in terms of a given mechanism will be equal to the rate of the reverse of this mechanism. This statement is often called the Principle of Detailed Balancing; i t is also known as the Principle of Entire Equilibrium or of Microscopic Equilibrium (5). (2) Derivation of General Equation. Now in order to see how the Law of Mass Action applies to a chemical equilibrium involving more than one molecule of A, B, C, and D for the balanced equation, let us consider the reaction in terms of the mecbanism postulated under (c) above. Of the various ways of treating this problem the following one, expanded from a suggestion by Hogness and Johnson (6),seems to be the easiest for the student to follow. This treatment involves a study of the chance of collision between like molecules.

If N is the total number of molecules of A in the container (example, N = 5 ) , the series which expresses the total number of chances of collision is ( N - 1) ( N - 2). . . . . 1. This sum may also be expressed by the formula N(N - 1)/2, or

+

+

We may say, therefore, that the total number of chances of any two A molecules colliding, when there are N molecules of A in the container, is N ( N - 1)/2. I t follows from this that the total number of chances of collision is proportional to N(N. - 1). (The proportionality constant '/z establishes the equality of the above equation.) In all actual experiments we are dealing with millions of molecules of A, so that N is always a very large number. This means that ( N - 1) is, for all practical purposes, equal to N. We may say, therefore, -Total number of cbances of collisions oc N2. Since N will be proportional to the concentration of A, expressed in moles per liter, we obtain the following relationship : Rate of 2A forming Az number of chances of collision oc actual number of collisions oc NZ a (A)%, then Rate, = K1(A)2,other factors constant. Case II Let n = 3, then 3A +Ax If again we have five molecules of A in the given container and we consider the chance of collision of any three A molecules, not counting the same chance twice, the total number of collisions will be ten. They are (see Figure 2) : 1-23, 1-2-4, 1-2-5, 2-34, 2-3-5,

N(N- l)(N - 2 ) . . . . I N - ( n - 1 ) ) Constant

3-4r5, 3-4-1, 4-5-1, 4-5-2, and 5-3-1. The mathematical formula which expresses the total number of chances of collision possible when three A molecules farm A3, and N molecules of A are present in the container, is: Total numher of chances of collision =

Since there are "nu terms involving N in the numerator of this equation, therefore Rate of n A forming A. a P a (A)", other factors constant." This statement means that the velocity of a reaction N ( N - 1)(N - 2). or between like molecules or ions will uary as the total con6 centration of these molecules or ions raised to the power of Total number of chances of collision N ( N - 1) its respectiue coeficient in the balanced equation. ( N - 2). Now since we assumed the formation of the interSince N is always a very large number ( N - 1) and mediate molecules A , and B, and when one molecule ( N - 2) are nearly equal to N. Therefore, we may of each of these unlike molecules react they form the state product, namely, o molecules of C and r molecules of Total numher of chances of collision o:N 3 o: ( A ) 3 ,or D, it follows from our previous discussion (on unlike Rate of 3A forming A J o: ( A ) a ,therefore molecules) that Rate, = K1(A)3. Rate, a (A") X (B,), other factors constant. Note: While N is always a very large numher it is Since the numher of collisions between n A moleinteresting to speculate on the condition in this system cules is proportional to (A)" and between m B moleif we had but a few molecules of A present. In this cules to (B)m,then the number of collisions between case we cannot assume ( N - 1) and ( N - 2) to be n A molecules and m B molecules will be proportional nearly equal to N and the rate will not be proportional to (A)" X (B)". Therefore to N3. This means that the Law of Mass Action is a Rate, = K1(A)" X (B)" statistical law, true in actual cases because we are Similar reasoning will show that the velocity of the dealing with a large number of molecules. Even under reverse reaction will be high vacuum conditions we still have billions of moleRatea = K2(C)0 X (D)' cules per cubic centimeter.I0 At equilihrium when the two rates are equal, we Case 111 nave Let n = 4, or 4A + Aa (')" If four A molecules collide to form one molecule of (Dp = % = K , (equilibrium constant) Kx (A)" X ( B ) , the polymer Aa, the mathematical formula which expresses the total number of chances of ,collision, when This is a quantitative expression of the Law of Mass N molecules of A are present in the container is: Action under equilibrium co?&tions and is called the Total number of chances of collision = Law of Chemkal Equilibrium. I t may be stated thus: I n any system i n chemical equilibrium the product of the N ( N - 1)(N - 2) ( N - 3). or concentrations of the substances formed i n the reaction, 24 raised to the power of their respective coeficients, divided Total numher of chances of collision a N4 o: ( A ) 1 Although the chance of a foupbody collision is im" Kinetic formulas expressing thefiequency of collisions heprobable, the mathematical treatment of this possi- tween unlike and like molecules also corifirm this conclusion (7). learn. for enamole. that the number of collisions ( Z L A ) bility leads to the same conclusion afif we considered We between ljke molecules per cc. of a gas is given by the expr&& the reaction to take place in himolecular steps. We are now able to generalize from the discussion above that when n A molecules react to form one mole- where o is the effective diameter, Na is the numher of atoms or molecules per cubic centimeter and ua the average velocity. cule of the polymer A., and there are N molecules of A When the temperature is constant in a given reaction all of the present, then factors on the right of this equation but N A may ~ be considered constant or Total number of chances of collision = Z A A = Nn'

10 Some apparently anomalous cases will be found among heterogeneous systems assumed to be in dynamic equilibrium involving such types as saturated solutions of difficultly soluble salts, solution pressures of the free elements, electrode potential systems, and vapor pressure of solids. For example, the average solubility-product constant of mercuric sulfide is about 1 X lo-". A simple calculation shows there are lo-= ions each of mercury and sulfur per liter in equilibrium (?) with the solid phase mercuric sulfide. This really means there will be a mean probability of one ion of each in one thousand liters of water. The behavior of these ions becomes indeterminate and whether the statistical law of chemical equilibrium holds in these cases is very dubious. For a fuller discussion of this question, consult a paper by G. N. Lewis, entitled, "Generalized Thermodynamics Including the Theory of Fluctuations" ( I . Am. C h a . Sac.. 53, 2578-88 (1931)); see also G. N. Lewis and M. Randall, "Thermodynamics," McGraw-Hill Book Co., Inc.. New York City, 1923, p p 127-8.

The numher of collisions Z n s between unlike molecules per cc. is given by the expression

where NA and NB are the numbers of each kind of molecule or atom per cc., respectively,

is the average effective

diameter and p is the reduced mass which equals (mn X ma)/ (mr ma) where m~ and ma are the masses of the respective molecules. I f the temperature is constant for a particular reaction involving unlike molecules then all the factors on the right of this equation but NA and NB may be considered constant, or Z A B O: NA X N B

+

by the product of the concentrations of the reactants, raised to the power of their respective coeficients, i s equal to a contant, when other factors such as temperature are constant. (3) PI~YSICAL SIGNIFICANCE OF THE SPECIFIC REACTION RATESKI AND K2, AND THE EQUILIBRIUM CONSTANT,K,.-One of the difficult problems which the chemistry teacher encounters is'getting the student to visualize the meaning of proportionality constants. The use of a set of experimental data often helps to clarify their meaning. Also the problem may be simplifiedif we reduce these constants to their dimensions. The dimensions of the specific reaction rates Kl and Ka will depend upon the order of the reaction. These questions and that of the equilibrium constants will be considered in order. (a) Specific Reaction Rates. (1) Monomolecular reaction (first order). When experiment shows that the rate of decomposition of a substance is directly proportional to the concentration of that substance, it is called a monomolecular reaction, or if (for example) An

-

ZA, then

Raty = KI(A2),other factors constant

The dieusions of rate are moles per liter per unit of time and the dimensions of concentration are moles per liter, therefore

value of K, and "a" are known the concentration of As a t any time "t" may be calculated. (2) Bimolecular reactions. When two molecules of the same or different kinds react, the reaction is known as a bimolecular one, or A

+ B e C + D,

then

rate, = KI(A) X ( B ) , or

rate = (A) X (B)

moles time =

- liters X -

a number concentration X time

KI, in a bimolecular reaction, is not independent of the units in which the concentration is expressed. This constant may be interpreted as the number of moles of A or B reacting each second in each liter of the reaction mixture when the coucentration of A and B are both one mole per liter ( 8 ) . It is sometimes pointed out in connection with the above bimolecular reaction that when (A) and (B) both equal one, then K, = rate. However, K1 does not have the dimensions of rate and is numerically equal to rate in this particular case only. For example, if KI equals the rate when the concentrations are unity, the rate when (A) = 3 and (B) = 4 is, rate = Kl X 3 X 4 = 12K1, or twelve times the rate in the first case. When A and B are present in equivalent quantities, then from calculus we learn that z

K' = l

a 5 moles X time a number liters rate When Kl has been determined and "a" is given, the = XI=(A21 moles time rate of the reaction a t any time may be calculated. liters Since the rate of a chem.ica1reaction taking place in bimolecular iteps is controlled by the actual mechanism The student should note that the dimensions of KI by which the reaction is achieved, the value of Kl will This means do not include the concentratiou term. depend on this particular mechanism. In reality it that the value of Kl. is independent of the units in means that Kl will depend on the slowest in the sequence which the concentration is expressed. KI, is, thereof steps making up this mechanism. fore, the fraction of Az decomposing per unit of time. One of the principal objectives of chemical kinetics Thus, if K1 = 0.001 per second, then AZis decomposing is the evaluation of the velocity constant of a given a t the rate of 0.1 per cent. per seconz reaction. From the value of this constant the order Since the fraction of Az decomposing per unit of of the reaction can be determined. time is constant irrespective of its concentration i t (b) Equilibrium Constant Kc. follows that the rate of its decomposition must be Prom the nature of the velocity constants of the independent of the number of molecular collisions. forward and backward reactions in a system in chemical This suggests that the decomposition of AZ is a pheequilibrium and from other considerations not meunomenon originating within the molecule.1z tioued above, we can call the students' attention to the During the course of the decomposition the concenfollowing summary of properties of the equilibrium tration of Az varies from its initial concentration to constant. that a t equilibrium. Then the rate of the decomposi(1) The equilibrium constant is the ratio of any tion will decrease regularly to that a t equilibrium. pair of rate constants (K1/K2) selected to represent When this varying rate is treated by calculus we obtain the mechanism of the reaction. the following expression ( 2 ) I t is independent of the mechanism by which 1 the equilibrium is achieved or of the catalysts that Kt = - log .Q where t a-z' affect the rate of the reaction. Experiment shows that, whereas different catalysts may change the "a" equals the initial concentration of A2, ( a - X ) mechanism of a given reaction, the same equilibrium equals its concentration after time "1." When the state is reached, providing the properties of the "There is some experimental evidence that some monoreacting and resulting substances are not altered molecular reactions really do involve collisions with other chemically. molecules (8).

(3) It is dependent on the properties of the reacting and resulting substances, the temperature and the medium in which the reaction takes place. (4) When the ratio of the dimensions of K,/Ka gives a number free from the concentration term (which results when the total number of molecules on the left equals the total number of molecules on the right in the balanced equation) then the equilibrium constant is independent of the units used to express the concentration. (5) From the value of the equilibrium constant the yield of a chemical reaction may be determined from the initial concentrations of the reacting substances. It is a measure of the completion of a reaction; of the chemical affinities involved and the free energy change. (6) If a reaction happens to be monomolecular in one direction and bimolecular in the other, the equilibrium constant still holds because the effective concentrations occur with the same exponent in both rate and equilibrium formulas. (7) Strictly speaking, the equilibrium constant holds only for perfect gases and ideal solutions. It does not hold for strong electrolytes. In the latter case the equilibrium constant increases regularly with increase in concentration. If activities, determined by experiment, are used instead of the analytical determined concentration terms, the constant

holds for strong electrolytes. Hnnever, most of the reactionsof analvtical chemistrvdeal either with weak electrolytes or with the ions of precipitates which are present in such great dilution that calculations based on the equilibrium law are, in general, valid. (8) Equilibrium constants may be calculated from entropy and free energy data, but thermodynamics cannot give us the values of the velocity constants. SUMMARY

An attempt has been made to review and clarify some of the teaching difficulties encountered in presenting the Law of Chemical Equilibrium to the beginning student. While analogies may have their place in explaining some of the steps in the derivation of this law, a more convincing approach involves a more detailed and rigorous treatment. This treatment can be confined to the mathematics of college algebra, the teacher supplementing or clarifying the more difficult concepts. Most teachers of the author's acquaintance have their own favorite methods of presenting this law. It would seem profitable if these approaches were made available to chemistry teachers in general. It is hoped, therefore, that the above treatment of the subject may be of some value to both teachers and students and that i t may suggest other discussions of this subject.

LITERATURE CITED

(1) BELWR AND WILLIAMS."A course in qualitative analysis," Houghton Mifflin Co., Boston, Massachusetts, 1938, p. 42. (2) RBEDY,"Theoretical qualitative analysis," McGraw-Hill Book Co., Inc., New York City, 1938, p. 70. (3) Ibid.. p. 80;, (4) C ~ ~ T M AQualitative N, analysis," The Macmillan Co., New York City, 1938, p. 63. (5) WEBB,"Elementary principles in physical chemistry," D. Appleton-Century Co.. New York City, 1936, p. 83.

(6) HOGNESS a m JOHNSON, "Qualitative analysis and chemical equilibrium." Henty Holt and Co., New York City, 1937, p p 50-3. (7) TAYLORAND TAYLOR,"Elementmy physical chemistry." 2nd ed., D..Van Nostrand 'Co.. Inc.. New York City, 1937, p. 231. (8) DANIELS,"Chemical kinetics," Cornell University Press, Ithaca. New York. 1938, p. 14.