Introductory Physical Chemistry Solutions

Transcription

1 Introductory Physical Chemistry Solutions Our previous work on equilibria involved gas-phase systems. Most reactions we carry out are in solution, so we would like to use the same equilibria principles as applied to solutions. To do this, we need to see how ideal solutions behave and then apply the non-ideal principles and use them to solve for equilibrium conditions in such solutions. Our objectives, therefore are as follows: Learn and understand the behavior of ideal solutions and use such behavior to connect to physical and equilibrium properties. (Most of this has been introduced in General Chemistry. We will present a bacround here. Observe the behavior and properties of non-ideal solutions and from such observations, connect non-ideal to thermodynamic criteria using activities. Apply the activities of the solutions to concentrations in order to predict the real behavior of non-ideal solutions For solutions, the activity is used to replace the concentrations in order to represent solution systems of condensed material. For these systems, we need to establish a different reference. Recall that for gas-phase systems, we defined the activity as: a γ P γx P o where P X using Dalton's Law P o For solutions, we will take the reference state to be the e fraction of the component now in solution. (We could take arity, ality, etc. However, we wil find that we can convert as needed later) a B Now γ B where X B X B lim 0 a B X B Now interpreting for solutions, we interpret this to mean that a solution will become ideal as the concentration goes to zero. We now turn our attention to determining how to measure γ, and thus connecting lab measurements to thermodynamics. We will begin by examining ideal behavior and then we will introduce the activity coefficient to correct for non-ideal behavior. For consistency, we will use component B of a solution to be the solute in solution with solvent, A Now, all we need to do is to determine what reference method we will use to measure the activity in order to connect it to the e fraction. We will take the paths of least resistance in this instance and measure what we have available. In parallel, we will develop methods of measurement for ideal and nonideal system by developing expressions for non-ideal systems, then generalizing to ideal.

2 Behavior of mixtures of volatile components Consider the following component diagram of an ideal mixture... Examination of this solution shows that the vapor pressure of the components and of total solution is linearly related to the e fractions of the respective components, i.e. P A x x X A P A and P B X B P B where the superscript "x" denotes pure vapor pressure of the component. The relationships given, are known as Raoult's Law and hold for all ideal solutions or solutions that have balanced interactions such that they behave ideally. Distinguishing.. Ideal Solution: No A-A, B-B or A-B interactions. 2

3 Ideal behaving solutions: A-A and B-B interactions are identical to A-B interactions. Now, having seen the ideal behaviour, examine a non-ideal solution Carbon Disulfide/Acetone Vapor Pressure vs. Mole Fraction Carbon Disulfide Pressure CS2 Data Acetone Data CS2 Fit to the Data Acetone Fit to the Data CS2 Henry's Law Line Acet Hentry's Law Line CS2 Raoult's Law Line Acet Raoult's Law Line Mole Fraction Carbon Disulfide 3

4 As can be seen, there are significant deviations from Raoult's Law. However, some interesting relationships may be drawn..) Each of the components approach Raoult's Law relationships as the component becomes more pure. 2.) When the component becomes low in e fraction, linearity is also seen, but not Raoult's Law linearity. The second observation allows the establishment of a second relationship for volatile substances in low concentrations. That relationship is: P A k X_A X A and P B k X_b X B These expressions are known as "Henry's Law" and the k's are Henry's Law constants in e fraction concentrations. This is distinguished from Henry's law constants based on other concentration standards. As can be seen, reference standards for such solutions must consider the nature of the component. The Raoult's Law calculation of vapor pressure is most accurate at high e fractions and can be calculated directly. The Henry's Law vapor pressures must have a constant, which must be calculated upon extrapolation of data to zero e fraction. Then, only low e fractions will produce the most accurate results. Consider the following examples... Raoults Law: For volatile components in high concentration. In the Raoults Law approach, the e fraction of a vapor over a liquid is a function of the e fraction of the substance in the liquid phase, or.. P i X i P o where X i is the e fraction of component, i and P o is the pure substance vapor pressure. Henry's Law: For volatile components in low concentration. P i k X_i X i where X i is the e fraction of component, i and P o is the pure substance vapor pressure. 4

5 Non-Ideal Systems For our ideal volatile solution, we have that: μ B_soln μ B_pure RT ln P B ideal expression. x P B NOTE: In this expression, since we are talking about solutions now, the P x B is the pure vapor pressure of component, B. Let us now replace the pressure expressions with allowances for non-ideality, giving μ B_soln μ B_pure RT ln a B non-ideal expression What do we use for the activity now? We saw that if the component is in high concentration, then it behaves closest to Raoult's Law. But if the component is in low concentration, then the behavior is closest to Henry's Law. These then form the reference standards for which the activity is calculated. If the component is volatile and in high concentration, it is most practical to establish a reference based on a Raoult's Law standard. Therefore we define the activity as... a B P B and since a γx we can now write x P B γ B X B a B P B P B X B x if solution is ideal, then and x P B P B If the solute is volatile and is in low concentration, we have found that Henry's Law is a more appropriate standard. If we replace the activity reference with Henry's Law. we have a B P B and, again since a γx k X a B γ B X B P B P B if solution is ideal, then and X B k X k X 5

6 Measurement from Freezing/Boiling Point Depression/Elevation Since it is observed that solutions with different concentrations undergo changes in the temperature of their phase transitions, this measured change can be a suitable measurement of solution activities. The expressions relative to each of these properties are nearly identical and will be produced in parallel. We begin with the observation that at the freezing point/boiling point, the pure solid/vapor is in equilibrium with the component solution, thus Freezing Point Deriv. μ B_solid μ B_soln or since the solid is in equilibrium with it's vapor μ B_solid μ B_pure RT μ B_soln lna B Solving for the ln(a B ).. lna B μ B_solid RT μ B_pure Since we measure temperatures in freezing and boiling points rather than pressures, we use the P-T relationship via the Gibbs Helmholtz equation to express the activity of the solvent in terms of temperatures. This, after some Calculus, gives... a B ΔH fus exp where T x R f the normal freezing point of the pure solvent. T f x T f By identical analysis for boiling point elevation, we find that: a B P atm exp ΔH vap R T b x T b where P is the applied pressure, usually atmospheric and where T x b the normal boiling point of the pure solvent. If the applied pressure is atm, then the pre-exponential term disappears and we have a B exp ΔH vap Important Note! Notice these are the activities R T b x T of the solvent, not the solute!! This poses a b problem as we are generally interested in the solutes. We will address this in two ways below. 6

7 Dilute Solution Approximation Let the solution be sufficiently dilute as to allow treatment as ideal. Then a B X B Also, then ln X A ln X B X A By approximation. (Note, for example) ln(.002) Now.. X A ΔH fus R T f x T f ΔH fus R ΔT f T f 2 if T f is very small X A ality 000 gm ality MW B MW B ality 000 gm if ality is small Now, combining.. 2 MW B RT f ΔT f 000 gm ality K f ality ΔH fus where K f is the freezing point constant for the solvent. Dilute SolutionEstimate Eqn. ΔT f K f al solute Example: Calculate K f for water and compare with literature ΔH f_water 600 J MW water 8.0 gm R J K T f_water 273.5K al K f 2 MW water RT f_water K 000 gm K f.859 K f_lit_water.86 K al ΔH f_water 7

8 Example: The cryoscopic constant for water is K f.86 K/al. A solution of 0.90 grams of ethylene glycol in ml of water is prepared. The freezing point is found to be -0.9 o C. a. Calculate the activity of water in this solution. b. Calculate the activity coefficient of water in this solution. T f 273.5K 0.9K T f_norm 273.5K ΔT f 0.9K ΔH f_water 600 J M L a.) Using our expression from above... a Water ΔH f_water exp a Water R T f T f_norm b.) In order to calculate the activity coefficient, we need the e fraction of water. Using the amounts delineated in the problem, I proceed.. MW Ethyl gm MW water 8.0 gm V water 0.5L mass Ethyl 0.90gm mass Water 500.0gm assuming a density of.00 gm/ml mass Water mass Ethyl calculating es... n Water n Ethyl MW water MW Ethyl calculating the e fraction of water and finally, the activity coefficient... X Water n Water n Water X Water γ Water n Ethyl a Water γ Water X Water 8

9 Measurement by Osmotic Pressure Another excellent way (particularly for large ecules) to calculate non-ideal behaviour and other properties of solutions is the calculation of an osmotic pressure. This is described as follows: The Osmotic Pressure is the pressure exerted by the excess height of the fluid column, or Π ρgh In an effort to reach an equilibrium in the Gibbs Free energy, the solvent attempts to dilute the solution. As a result, the solution level rises, producing a pressure differential. This pressure difference is known as the Osmotic Pressure and has been proven to be quite useful, particularly in the determination of the ecular weight. From previous analysis, we know that G B_soln G B_pure RT lna B The same condition applies to this system except that the Free energies are at a pressure other than standard conditions. Thus, we need to augment this expression in order to evaluate it at the osmotic pressure,. Again, applying the Calculus, we now have the result.. Dilute Solution Approximation lna B Π RT V B lnx B ln a B n ln X A A X A so n B n A n B ΠV B RT Here,as before, we assumed A was the major component. Now... n A Πn B RT V B ΠV RT where we have assumed that since B is the major component, the V B V total Finally.. n A Π V RT or Π M A RT Ideal result 9

10 Distillation and Boiling Point Diagrams We can adopt the above for use in determining distillation composition which are used to separate components of a mixture. Coonsider a mixture of n-pentane and n-hexane. We wish to separate these two components. As they are very similar in structure and polarity, we will adopt Raoult's Law for ideal solutions. Example: Distillation The pure vapor pressure of n-pentane at 25 o C is bar and n-hexane is 0.98 bar. a.) Calculate the partial pressure of each substance and the total pressure over a solution containing X pentane Assume ideal behaviour. b.) Calculate the composition of the vapor phase P pentane_pure 0.674bar P hexane_pure 0.98bar bar 0 5 Pa X pentane X hexane X pentane X hexane 0.74 P pentane P hexane P pentane_pure X pentane P pentane 0.75bar P hexane_pure X hexane P hexane 0.47bar P tot P pentane P hexane P tot 0.32bar Y pentane Y hexane P pentane Y pentane P tot P hexane Y hexane P tot Note that "Y" is used as e fraction in the vapor phase whereas "X" is the e fraction in the liquid phase. 0

11 Boiling Point Diagrams The hexane-pentane Raoults Law example above is a demonstration of a room temperature distillation. If we were to collect and condense the vapor collected, the composition would be richer in the more volatile compont. The vapor from such a solution would more rich, yet. Practically, solutions are boiled at atmospheric pressure or at reduced pressure if the components are heat sensitive. The Raoults Law vapor pressures along with the Clausius-Clapyron equation provides all we need in order to construct a BP diagram. The MCAD handout demonstrates this approach. In summary, a solution will boil when the vapor pressure of the solution is equal to the pressure above the solution, usually the atmospheric pressure. We can then write that o P tot P A P B X A P A o X B P B This is what we did in the hexane-pentane example. Our modification here is to use the C-C equation in order to include temperature dependence. P i P i o exp ΔH vapi R T o i T i Substituting for the pure vapor pressures, we get: P tot P o ΔH vapa X exp A R T j T nbpa ΔH vapb X exp B R T j T nbpb where nbp means normal boiling point.both terms have the same T j as they will be combined in the same solution and will thus boil at the same point. If the reference pressure/bp combination is atm, then this expression becomes P tot 760torrX exp ij i ΔH vapi R T j T bni If the distillation is carried out under reduced pressure, then the proper P tot must be substituted See MCAD Sheet - Boiling_Point_Diagrams

12 Non-ideal Boiling Point Diagrams If the distillation mixture does not behave ideally, then activity coefficients are used to construct the boiling point diagrams. Note the mixture fo chloroform and methanol, two components of clearly different polar properties. The calculated diagram look as follows: Note that there is a limit to the purity one can obtain via distillation.any attempt will ultimately produce a distillate of composition of X Chloroform that has a boiling point of K 2

13 Ionic Solutions Ionic solutions present for us a unique opportunity. The fact that electrostatic attractions/repulsions are so long range and well understood, a general theory can be developed that allows the calculation of the activity and activity coefficient directly! The approach uses the Debye-Huckel theory and is presented here. First, we must introduce some definitions and quantities. In solutions of ions, there are activities that must be calculated for each ion. Unfortunately, because of the fact that the ions cannot be easily altered individually in order to determine their individual activity, a quantity known as the mean ionic activity coefficient is used instead. We thus define the activity of a salt as... a a salt mi ν m ν cat m cat an ν an γ mi ν where mi is the mean ionic activity coefficient. The 's are the stoichionmetric coefficients of each ion and of the cation, anion and sum, respectively. Example: Express the activity of 0.5 m sodium sulfate in water. 2 a Na aso4 ana 2 SO 4 2 m Na mso4 γ 3 ( 2m ) 2 mγ 3 4m 2 γ 3 4( 0.5) 3 γ γ 3 The activity coefficient can be determined experimentally using methods given previously, specifically freezing point, osmotic pressure, etc. However, two researchers developed a method for calculating the coefficient which is good for dilute (<.000 m ) solutions. The theory is known as the Debye-Huckel Limiting Law and is expressed as follows... logγ mi α DH z z I cat an where DH is a parameter that is specific for a particular I solvent at a given temperature. where I is known as the Ionic Strength and is calculated as... 2 m i I z 2 i where, the z are the charges on the ions and m are the alities. i i i 3

14 The parameter, α DH, is specific for a given solvent at a given temperature. For water α DH.726 if the natural logarithm is used or α DH if a base-0 logarithm is used. (See below) Example: Calculate the activity coefficient for.035 al solutions of HCl, CaCl 2 and ZnSO 4 at 25 o C. al α DH For HCl.. I HCl 2 2 al 2 al I HCl CaCl 2 I CaCl al 2 ( 2al) I CaCl ZnSO 4 I ZnSO ( al) 2 2 ( al) I ZnSO α DH I HCl I HCl γ HCl 0 γ HCl 0.83 α DH 2 I CaCl2 I CaCl2 γ CaCl2 0 γ CaCl α DH 22 I ZnSO4 I ZnSO4 γ ZnSO4 0 γ ZnSO

15 Conventionally, the Debye-Huckel constant is expressed as a base-0 logarithmic quantity. Converting gives.. ln( γ) α DH z z cat an I I but, by the rules of logarithms.. ln( γ) log( γ) or log( e) ( log( γ) log( e) ln( γ) ) Thus log( γ) α DH z z cat an I I converting the constant.. α DH Example of it's use: Calculate the solubility of BaF 2. The K sp of BaF 2 is.7 x 0-6 BaF 2 (s) Ba +2 (aq) + 2 F - (aq) K sp.70 6 x 2x From solubility, we know that K sp a Ba a F 2 al Ba γ al o al F γ al o 2 al al 2 al 2 o al γ 3 3 al 4 γ 3 o al o therefore al K sp al o 3 4 γ 3 3 In order to solve this system, we need the coefficient. 5

16 logγ mi α DH z z I cat an α DH al o I where I z 2 i i 2 al i The problem, as before, is that we need the alities in order to calculate I and γ. So we need the answer before we can find the answer!! We take the same approach as introduced before: Solve the system assuming ideality. Using the ideal alities, calculate a new ionic strength and coefficient. Use these new values to calculate a corrected ality. Repeat 2 and 3 until the answer converges. Step : Set γ which assumes ideality. Doing so gives. K sp 3 al o al al ideal result Step 2: Use this ideal ality to calculate a new I and gamma... I 2 22 al 2 2al I I α DH 2 al o I al o γ 0 γ Step 3: Calculate a new value for the ality using this new gamma value... 3 K sp al 4γ 3 al o al

17 Step 4: Repeating,,, I 2 22 al 2 2al I α DH 2 I al o I al o γ 0 γ One more time... 3 K sp al 4γ 3 al o al I 2 22 al 2 2al I α DH 2 I al o I al o γ 0 γ K sp al 4γ 3 al o al Now we see that the answer has converged. Note: percent_error percent_error 30.6 % This is a large error!!! 7

18 Comparison with Literature When does the Debye-Huckel Limiting Law begin to fail? The Debye_Huckel Limiting Law is limited in it's scope. In order to evaluate it, we need some experimental data. Below is a well behaved set of activity coefficients for NaCl. We would expect the best possible behavior because as the ions become more charged, the deviations become greater. i 0 6 ality i γ NaCl_expi m α DH.77 α DH.77 I NaCli 2 γ NaCli exp 2 ality i α DH I NaCli 2 ality i I NaCli γ NaCl Activity Coeff Experimental and DH Activity Coefficients NaCl Exper Calculated ality 8

19 Comparison with Aluminum Sulfate I Al2SO4i 2 γ Al2SO4i exp 3 2 ality 2 i α DH ality 3 i I Al2SO4i I Al2SO4i γ Alum_Sulfate_Expi γ Al2SO4i Activity Coeff Experimental and DH Activity Coefficients Aluminum Sulfate Exper Calculated ality Notice the effect of increasing ionic strength. The deviations become more severe very quickly 9

20 Summary so far.. We wish to know how a solution system will behave. Our best criteria is still the same as it has always been..the Gibbs Free Energy determines the thermodynamic criteria for which a system will behave and under what conditions a system will achieve equilibrium. However.. the thermodynamics does not connect directly to concentrations for non-ideal behaving systems. Therefore, we introduced the activity and the activity coefficient. The Gibbs Free energy tells us the activities of the solution. The activity coefficient connects the activity to the concentrations. Thus, our objective so far was to connect the activity and the concentration. We did this by measuring solution properties.unfortunately, in doing this, we are generally measuring the properties of the solvent and thus we obtain the activities of the solvent. In order to get to the activities of the solute, we take two approaches. The first is: If the solution was sufficiently dilute, then the activities of the solute can be estimated If the solution is ionic and sufficiently dilute, then the Debye-Huckel approach may be used.. The second approach to determining the activity of a solute is given below and is presented for illustrative purposes only. It is an approach using the "Gibbs-Duhem" relation. 20

21 The Gibbs-Duhem Equation We are now in a position to calculate the activities of at least one component of the system, if not both. To complete this analysis, we must ask the question. "What happens if one of the components is a non-volatile, non-ionic solute? There is no vapor pressure, so we can't get at the activity that way. The boiling point, freezing point and osmotic pressure relate to the solvent, so what do we do. The answer lies in the Gibbs-Duhem discussion. The Gibbs Free Energy expression as we have introduced earlier is dg SdT VdP for single component systems. When we introduced additional components, we had to complete this expression for the individual components as: dg SdT VdP μ dn i i i Now, consider a system at constant T, and P. Under these conditions, the Gibbs free energy reduces to dg μ i dn i or dg μ A dn A i μ B dn B By analysis using differential Calculus, we find that the two components (or more) are related, intimately. (No real surprise.) The relationship is simple, but not obvious and is given as: n A dμ A n B dμ B 0 This is the Gibbs-Duhem expression The G-D expression has important implications. It tells us that information about the chemical potential (i.e. activity) of one component is related directly to the other. (or others for multiple component systems) They are NOT independent! This is the leverage we need in order to get at components that are difficult or impossible to measure. In practice, this is a very difficult expression with which to work. We must have a great deal of excellent data at a variety of concentrations. Execution is discussed below and examples of this being executed are shown. 2

22 After extensive mathematics, we are left with the following from the rather simple expression presented above. lnγ A X A X A 0 X A X A dln a B X B dx A d X A This expression requires that we know the activity coefficient of B as it varies with A. This, more often than not requires numerical evaluation, partially because of the lower limits. The following example illustrates the application of this process in MCAD. In general, the procedure is as follows: Construct solutions of known e fraction of solute, X A. Calculate X B as well. Measure some property of the solution such as vapor pressure, freezing point, etc for each solution. Calculate the activity of the solvent, a B. Execute the mathematical integral above. In detail, one:.) Calculates the natural log of ratio of a B /X B and fit the ln(a B /X B ) to a general equation, usually a polynomial as a function of the solute, X A 2.) Takes the derivative of the fitted function w.r.t the e fraction of the solute. 3.) Multiplies by (X A - )/X A. and integrates to produce the ln( A ) as a function of X A. 22

23 Gibbs-Duhem Applications This first example illustrates the use of the Gibbs-Duhem equation as applied to e fraction systems. Data is in the form of the vapor pressure of the solvent. Raoults Law is used to get the activities. i 0 24 Pu Bi X Bi i X Ai X B24 i P Bi P Bi Pu Bi torr a Bi Fitfact ln i F( x) x 2 x 3 x 4 x 5 fit( z) F( z) s z fit( z) 0.4 Fitfact i P B24 This is the function that I wish to fit with the linfit function s This builds the actual function. Alternatively, you can do it by hand. Plotting... a B24 i X B24 i s linfit X A FitfactF These are the newly fitted coefficients. Notice that they are very close to those exhibited inth eexample provided by Raff zx Ai 23

24 Now, here is the MCAD advantage. Raff takes the function and manually takes the derivative, then manually integrates. I choose to do all in a single MCAD step as follows: gx A x A 0 x A x A d fit x A d dx A xa expgx A γ x A Plotting... x A 00.0 To find the actual activity coefficient, simply take the exponential of the function, g(x). Note that at the limiting value of X 0, the coefficient approaches as it should. Although the solute/solvent system is not identified, it does seem to show significant deviation from ideality at high e fractions. 3 Activity Coefficient vs. Mole Fraction 2.8 Activity Coefficient of A in Solution Mole Fraction of A 24