CHEM 331 Physical Chemistry Fall 017 Activities and Activity Coefficients We now finish answering the question we asked during our last lecture, what is the form of the chemical potential i (T,P,x i ) for components of condensed phase (liquid/solid) solutions. Previously we determined that the chemical potential for the components of these solutions are given by: i = i * + RT ln x i (Ideal Solution) j = j ** + RT ln x j (Ideal-Dilute Solution) j = j *** + RT ln (Ideal-Dilute Solution) j = + RT ln (Ideal-Dilute Solution) provided the solution is Ideal or Ideal-Dilute. We now push toward forms for i when the solution is Real. In the case where the solution is Real, we introduce an "effective" concentration, much as we introduced the fugacity as an "effective" pressure for Real gases, which we call the Activity; denoted as a i. We also retain the previous definitions of the Reference States i *, j **, j ***, and. This then gives us: i = i * + RT ln a i I = i ** + RT ln a i
i = i *** + RT ln i = + RT ln It should be noted that we have really defined four different activities a * i, a ** i, a *** i and ; each denoted commonly as a i. For this reason, the Reference State being used must be identified when working with a given activity. Notice also that the distinction between the solvent i and solutes j has been dropped, as any one of these forms will work over the entire concentration range of the solution. However, having said this, it is still useful to use the Ideal Reference State * i when working with the solvent and the others when working with the solutes. We now introduce a new parameter, the Activity Coefficient i. This parameter is incorporates all the non-idealities of the solution. Depending on the Reference State chosen, it is defined according to: a i * a i ** a i *** = i * x i = i ** x i = i *** m i = c i Again, each is more generically denoted i. Because our solutions are expected to behave ideally at the limit x i 1 or as ideal-dilute in the limit x i 0, it is expected i should approach 1 in these limits. The limiting behavior of each i is given: i i * i i ** i i *** as x i 1 so a i x i and i 1 as x i 1 as x i 0 so a i x i and i 1 as x i 0 as m i 0 so a i m i and i 1 as m i 0 i as c i 0 so a i c i and i 1 as c i 0 Now, how do we determine a i or i? There are a number of methods. Some rely on colligative property data, others electrochemical cell data, etc. We will illustrate the determination of i from vapor pressure data. So, assume our condensed phase is in equilibrium with its vapor and that it is Real, therefore the vapor pressure of the i th component P i will deviate from either Raoult's or Henry's Law.
Now do the usual bit, equate the chemical potentials of component i across the phase boundary. i,gas (T,P i,y i ) = i,liq (T,P i,x i ) If we are working with Ideal Reference State, then: i,liq (T,P i,x i ) = i * + RT ln a i = i o + RT ln P i * + RT ln a i and: i,gas (T,P i,y i ) = i o + RT ln P i Inserting these forms into the above equation gives: RT ln a i = RT ln Or, simplifying: a i =
This, then, allows for the determination of i : i = = If the Reference State in use is the Ideal-Dilute Reference State, then: And so on. A couple of examples. a i = and i = For Hg(l) at 35 o C, P Hg * = 55.48 kpa. An amalgam of Hg and Tl at this same temperature, composed of 18.70 Hg (solvent) and 1.163g Tl (solute), has a vapor pressure of P Hg = 5.04 kpa. What is Hg for the Hg at this solution composition? Mole Fractions N Hg = 18.70g / (00.59 g/mol) = 0.093 mol Hg N Tl = 1.163g / (04.37 g/mol) = 0.005691 mol Tl x Hg = = = 0.94 Activity a Hg = = = 0.938 Activity Coefficient Hg = = = 0.996 Notice that since x Hg ~ 1, Hg ~ 1. At 5 o C, K N = 8.9 x 10 9 Pa for N in Water. What is N when a partial pressure of Nitrogen P N = 101.35 kpa of N is applied above the Water? At this vapor pressure, N N = 6.40 x 10-4 mol N. Mole Fraction Activity x N = = = 1.153 x 10-5 a N = = = 1.136 x10-5
Activity Coefficient N = = = 0.985 Again, notice that since x N ~ 0, N ~ 1. Keep in mind, this is only one method for determining activity coefficients. Now to one last detail concerning the form of i. The behavior of Electrolyte Solutions is skewed by the fact that the electrolyte disassociates in Water. So, we now turn to deal with aqueous electrolyte solutions. First, some background. Consider reasonably dilute aqueous solutions of HCl. According to Henry's Law: P HCl should = K HCl m HCl In other words, P HCl should be linear in m HCl. However, a plot of the data demonstrates this is not the case. In fact, a plot of P HCl vs. m HCl is instead linear in the dilute limit. This suggests that Henry's Law for aqueous HCl solutions should be written as: P HCl =
Physical Chemistry, nd Ed. J. Philip Bromberg What is going on? Consider a generic electrolyte that is fully soluble in Water: M p N q (aq) p M x+ (aq) + q N y- (aq) Some specific examples: HCl(aq) H + (aq) + Cl - (aq) p = 1 q = 1 CaCl (aq) Ca + (aq) + Cl - (aq) p = 1 q = Na SO 4 (aq) Na + (aq) + SO 4 - p = q = 1 If the solution molality is m, then the molality of the cations m + and molality of the anions m - is given by: m + = p m m - = q m Now, we will formally write and *** as: = p + + q - *** = p + *** + q - ***
where + and - are the chemical potentials of the cations and anions respectively. Also note that the subscript i has been dropped, so as to not muddy up the notation too much. Then, according to our usual forms: + = + *** + RT ln a + - = - *** + RT ln a - with a + = + m + and a - = - m -. Now, the problem is, we cannot measure + and - simultaneously. (Think about why this is!) So, instead of trying to indentify + and -, we instead introduce a Mean Ionic Chemical Potential: ± = ± *** = where s = p + q. And, we have a ± = ± m ±. The following relationships also follow: a s ± = a p q + a - m s ± = m p q + m - ± = p q + - Now we can write: = s ± = s ± *** + RT ln ± s + RT ln m ± s Further, m ± can be determined from m: m ± = (m + p m - q ) 1/s = ( (pm) p (qm) q ) 1/s = (p p q q m s ) 1/s = (p p q q ) 1/s m A single measurement of can indeed yield the single value of ±, making this form of quite useful for electrolyte solutions. Now back to the HCl problem. When dissolved in Water, HCl will dissociate completely. HCl(aq) H + (aq) + Cl - (aq)
with p=1, q=1, and s=. Thus, liq,hcl = ± *** + RT ln a ± ± *** + RT ln m ± (if reasonably dilute and ± ~ 1) = ± *** + RT ln m HCl Now, to consider the vapor-liquid equilibrium: HCl,liq = HCl,gas HCl,gas = o + RT ln P HCl HCl,liq = ± *** + RT ln m HCl = o + RT ln K HCl + RT ln m HCl This gives: o + RT ln K HCl + RT ln m HCl = o + RT ln P HCl resulting in a P ~ m dependence: P HCl = K HCl m HCl And, now we see why Henry's Law must be modified for the electrolyte. P HCl = Theoretical estimates of ± are extremely difficult to come by because the intermolecular interactions between the cations and anions are so extremely strong in the solution. One method for estimating ±, developed by Peter Debye and Erich Huckel, works in the extremely dilute electrolyte limit; m < 0.001m.
Peter Debye Erich Huckel Using a statistical mechanical model, Debye and Huckel showed that: log ± = 0.0509 z + z - I c 1/ where z + and z - are the charges of the cation and anion respectively and I c is the Ionic Strength. I c = c i is the concentration of each of the ionic species present in the solution. It should be noted that this form of the Debye-Huckel Law only works for aqueous solutions at 5 o C. Further note that only the charges of the ions are required to estimate ±. This model can be extended to slightly higher concentrations by including the hydrodynamic radius of each ion. We will not consider this Extended Debye-Huckel Law further. Below is a graph showing the validity of the Debye-Huckel Law for several actual electrolytes; the dashed lines are the Debye-Huckel predictions for various z + and z - values.
Walter J. Moore Physical Chemistry, 4 th Ed. As an example: Use the Debye-Huckel Law to estimate ± for 0.001m BaCl in 0.000m aqueous HNO 3. BaCl (aq) Ba + (aq) + Cl - (aq) I c = (c Ba+ z Ba+ + c Cl- z Cl- + c H+ z H+ + c NO3- z NO3- ) = ((0.001m)( ) + (0.00m)(1 ) + (0.000m)(1 ) + (0.000m)(1 )) = 0.003m 1/ log ± = 0.0509 z Ba+ z cl- I c = 0.0509 (+) (-1) (0.003) 1/ = - 0.0058 ± = 10-0.00576 = 0.987