CHEM-UA 652: Thermodynamics and Kinetics

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1 CHEM-UA 652: hermodynamics and Kinetics Notes for Lecture 6 I. SAISICAL MECHANICS OF SOLVAION: SOLVAION FREE ENERGIES We consider a solvent with coordinates r (a),...,r(a) N a in to which a solute with coordinates r (b),...,r(b) N b. he total number of particles in the system is N N a +N b. Let r (a) denote tne full set of solvent coordinates, r (b) denote the full set of solute coordinates, and r (r (a),r (b) denote the full set of coordinates in the system. he pure system of each system, before the solute is introduced into the solvent, is described by a potential energy U (r) U a (r (a) )+U b (r (b) ) () When the solute is introduced into the solvent, we change the thermodynamic state of the system, and we change the mechanical state as well. Hence, the potential energy for the solution is of the form U(r) U (r)+u ab (r (a),r (b) ) (2) where U ab is an interaction potential between solvent and solute, which we can take to be a pair potential N a N b U ab (r (a),r (b) ) u ab ( r (a) i i r (b) ) (3) In order to affect the change of thermodynamic state so that we can compute the solvation free energy, we define an arbitrary path via the introduction of a coupling parameter λ that slowly, reversibly switches on the interaction between solvent and solute: Ũ(r;λ) ( λ)u (r)+λu(r) (4) Note that Ũ(r;) U (r) is the potential of the pure substances, and Ũ(r;) U(r) is the potential energy for the solution. Here λ [,] is a continuous parameter that we slowly switch from to, thereby slowly solvating the solute in the solvent. Let A(λ) denote the Helmholtz free energy for Ũ(r;λ) at a particular value of λ. he solvation Helmholtz free energy is then given by the relation Note that solv A A() A() da dλ (5) dλ A(λ) β lnz(λ) Z(λ) dr e βũ(r;λ) da dλ βz(λ) Z(λ) dz dλ dr Ũ λ e βũ(r;λ) But Ũ λ U (r)+u (r)+u ab (r (a),r (b) ) U ab (r (a),r (b) ) (6)

2 herefore, da dλ Z(λ) Now, define the radial distribution function at a given value of λ as g ab (r;λ) N av dr e βũ(r;λ) ρ a Z(λ) r (a) r(b) r hen da dλ Z(λ) N an b dru ab ( r (a) r (b) )e βũ(r;λ) N b N b 4πN b V dr U ab (r (a),r (b) )e βũ(r;λ) (7) [ ] dr (a) dr(b) u ab( r (a) r (b) ) Na dr e βũ(r;λ) Z(λ) [ ] Na dr dr u ab ( r ) dr e βũ(r;λ) Z(λ) [ dr r 2 Na u ab (r) Z(λ) [ 4πN b ρ a dr r 2 u ab (r) N a V ρ a Z(λ) 4πN b ρ a dr r 2 u ab (r) g ab (r;λ) We now integrate over λ to obtain the solvation free energy Now define solv A 4πN b ρ a dr r 2 u ab (r) g ab (r) dr e βũ(r;λ) ] r r dr e βũ(r;λ) ] 2 (8) dλ g ab (r;λ) (9) dλ g ab (r;λ) () which is the solvation radial distribution function and measures the work to bring together solute and solvent particles at a distance r from each other. hen, solv A 4πN b ρ a dr r 2 u ab (r)g ab (r) () Now, we compute the solvation chemical potential as the solvation free energy per solute particle solv µ solva N b 4πρ a dr r 2 u ab (r)g ab (r) (2) Finally, then, the molar solvation free energy is simply obtained by multiplying this by Avogadro s number N A : solv µ solva N b 4πρ a N A dr r 2 u ab (r)g ab (r) (3) II. CHEMICAL POENIALS IN SOLUION-VAPOR EQUILIBRIA When treating the vapor as an ideal gas, it is relevant to ask what the Gibbs free energy of the vapor phase, and therefore, the chemical potential are. Recall that the partition function of an ideal gas of ust one component is Q(N,V,) ( ) N V N! λ 3 (4)

3 where λ βh 2 /2πm. he Gibbs free energy, however, comes from the isobaric partition function (N,P,), given by herefore, the Gibbs free energy is (N,P,) N!λ 3N dv e βpv Q(N,V,) (βp) N+ λ 3N e βpv V N dv (βp) N λ 3N (5) G(N,P,) β ln (N,P,) N K kg mol lnβ +N K kg mol lnp +3N K kg mol lnλ 3 Now suppose we have a process in which we change P to P 2 in a gas at fixed N and. he change in Gibbs free energy is ( ) ( ) G G(N,P 2,) G(N,P,) N K kg mol P2 P2 ln nr ln P P hus, the molar Gibbs free energy is which is also the chemical potential of the gas. Ḡ G n µ R ln ( P2 P ) In thermodynamics, we can only defines changes relative to an arbitrary reference, so we actually need to define that reference. his reference is known as the standard state. For gases, the standard state is a pressure of atm at some specified temperature, assuming ideal-gas behavior, i.e., that P V nr. he pressure of the standard state is denoted P o, and it is ust a constant having the value of atm. If we now consider a change from the standard state to an arbitrary pressure P, the change in the Gibbs free energy is then given by ( ) P Ḡ Ḡ ()+R ln P o and the chemical potential, which is the same thing, is µ µ ()+R ln ( ) P P o Let us denote a general component in our system as, i.e., A,B in a two-component solution. For each component in the vapor phase, ( ) µ (vap) µ P +R ln P (6) However, in equilibrium µ (vap). herefore, if we take P µ (vap) atm, then µ h +R lnp (7) An equivalent expression for a pure sample of component in an equilibium between its liquid and vapor would appear as µ (liquid) µ (vapor) µ +R lnp (8)

4 4 Subtracting the two expressions yields Moreover, ( ) µ (liquid)+r ln P PJ (9) µ (vap) µ +R ln(x P) µ +R lnx +R lnp (2) hus, for each of the two components, µ (vap) µ +R lnx +R lnp 2 µ (vap) 2 µ 2 +R lnx 2 +R lnp 2 µ 2 µ +R ln ( x2 x ) (2) his shows that the difference of chemical potentials of the two components in solution is related to the logarithms of the ration of the mole fractions. Due to the equilibrium condition ( ) µ (vap) 2 µ (vap) µ 2 µ +R ln x2 (22) x which relates the chemical potential difference to the same mole fraction ratio (which we recall is the ratio of gas-phase mole fractions), the same mole fraction ratio must be the same in solution in order to determine the solution-phase chemical potential difference. III. NON-IDEAL SOLUIONS AND ACIVIIES We showed that the solution chemical potential is µ (liquid)+r ln ( ) P P (23) In an ideal solution P x PJ, in which case In a non-ideal solution, we can fit the behavior of P vs. x as and µ (liquid)+r lnx (24) P x P eb( x)2 +c( x ) 3 (25) µ (liquid)+r lnx +Rb( x ) 2 +Rc( x ) 3 (26) he second ahd third and fourth terms on the right of this expression measure the deviation from ideal behavior, and we see that as x, Let us define µ (liquid)+r lnx (27) a P P J (28)

5 5 which is called the activity. For an ideal solution, a x. Generally, however, P /P J x but is rather some function f(x ) of x. hus, in the non-ideal case, we can write he ratio µ (liquid)+r lna (29) γ a x (3) is known as the activity coefficient and is equal to for ideal solutions. Activities can be defined for any state of a substance, i.e., solutions, dissolved species, gases, liquids, solids,... IV. CHARACERIZING SOLID-LIQUID SOLUIONS AND DISSOLVED SPECIES Solid-liquid solutions involved the dissolution of a solid substance in a solvent leading to a homogeneous mixture. We will discuss the microscopic nature of dissolved species later in the lecture. Before we get into that, however, let us discuss some basic measures for characterizing the composition of such a solution. he most common measure of solution composition is the molarity defined as Molarity moles of solute liters of solvent and is denoted M. A second measure is known as the molality defined as molality moles of solute kilograms of solvent (3) (32) Example: A solution has a solvent (component ) and N dissolved species indexed k 2,...,N. Let x be the mole fraction of the solvent and x 2,...,x N be the mole fractions of each of the dissolved solute species. Let M be the molar mass of the solvent. Show that x + Mm, x k M m k / + Mm where m k is the molality of the kth component and m is the total molality of the solution, i.e., (33) m k moles of component k kilograms of solvent, N m m k (34) k2 Solution: Start with the definition of the mole fractions. For component, the definition is Divide top and bottom by n to give x n n + N k2 n k x + N n k k2 n he number of moles of solvent is n mass of solvent in grams M (mass of solvent in kilogramt) M µ M

6 6 where µ is the mass of solvent in kilograms. hus, x becomes x + M N n k k2 µ + Mm where n k /µ is the molality of component k, and m N k2 m k. For component k, n k n k n + N 2 n Use the fact that n µ /M. Substituting this in, we obtain n k /n + N n 2 n n k M n k µ + M N 2 n µ Again, since n /µ m, and m N 2 m, we finally obtain x k M m k / + Mm Example: A solution is prepared by dissolving 22.4 g of MgCl 2 in.2 L of water. he resulting solution has a density of.98 g/cm 3. Calculate the mole fraction, molarity, and molality of MgCl 2 in the solution. Solution: he molar mass of MgCl 2 is g/mol. hus, the moles of MgCl 2 are moles of MgCl g.235 moles g/mol In addition ( ) cm 3 mass of water (.2 L) (. g/cm 3 ) 2 g. L he molar mass of water is 8.2 g/cm 3. Hence, the moles of water are moles of water 2 g 3. moles 8.2 g/cm he mole fraction of MgCl 2, which is component 2, becomes x moles.27 moles (.+.235) moles he total solution volume is solution volume mass of solution density of solution 22.4 g+2 g.89 g/cm 3 24 cm 3.24 L

7 7 hus, the molarity of MgCl 2 is Finally, the molality is Molarity.235 moles.24 L.5 M molality.235 moles.2 kg.8 mol/kg.8 m V. FREEZING POIN DEPRESSION At the freezing point, the solid solvent is in equilibrium with the liquid solvent. his requires where component is the solvent. Recall, however, that hus, µ (solid,soln) (P,) µ (liq,soln) (P,) (35) µ (liq,soln) µ (pure liquid)+r lna µ (solid,soln) (36) lna µ(solid,soln) µ ((pure liquid)) R How does a depend on? ake the temperature derivative of both sides: [ ] lna (µ (solid,soln) /) (µ (pure liq)/) R (37) (38) For a general chemical potential, use the fact that G(n,P,) µ(p,)n H S, (G/) G H S µ n n (µ/)/ H 2 + ( ) H S (39) However, remember that and, under isobaric conditions, H C P() (4) S C P() dqrev S dh CP ()d ds C P()d (4)

8 8 herefore, (G/) n (µ/)/ H 2 + C P() C P Applying Eq. (42) to the molar Gibbs free energy µ, we obtain where H is the molar enthalpy. hus, for the solvent H 2 (42) (µ/) (µ /) H 2 (43) H 2 (44) We can use these results to determine the activity and ultimately the amount of freezing point depression. If we differentiate Eq. (37) with respect to temperature and apply the above result, we obtain [ ] lna (µ (solid,soln) /) (µ (pure liq)/) R H (solid,soln) H (pure liq) R 2 H (solid,soln) H (pure liq) R 2 (45) Now we assume that, in a dilute solution, the solvent dominates, so that H (liquid,soln) (pure liq) H. hus, lna (liquid,soln) (solid,soln) H H R 2 fus H R 2 (46) In order to see if there is a change in the freezing point, we need to integrate both sides from fus, the freezing point of the pure solvent, to fus, the freezing point of the solution. his gives lna fus fus fus H d (47) R2 Let us assume that the solution is dilute so that a x, lna lnx ln( x 2 ) x 2, where x 2. We then have x 2 fus H fus R fus H R x 2 fus H R fus [ fus ] fus [ fus fus ] fus fus Since H fus >, it follows that fus < fus, hence, for the solution, the freezing point is lowered. Previously, we showed that (48) x 2 M m 2 / +M m/ (49)

9 For a two-component solution m 2 m, hence this becomes which we can write as x 2 M m/ +M m/ x 2 m m+ M m /M M m where we have assumed that m /M. Now, let fus fus fus. Hence, we can write x 2 fus H R fus fus fus M m Solving for the change in the freezing temperature fus, we obtain where K f is a constant known as the freezing-point depression constant. 9 (5) (5) (52) fus fus fus R fus H M m K fm (53) By a similar argument, it can also be shown that the boiling point of the solution is elevated, i.e., vap K b m (54) Boiling point elevation and freezing point depression can be used to determine molar masses. he following examples illustrate how this is done:. When 5.5 g of biphenyl (C 2 H ) is dissolved in. g of benzene, the boiling point increases by.93 o C. When 6.3 g of an unknown hydrocarbon is dissolved in 5. g of benzene, the boiling point of the solution increases by.597 o C. What is the molar mass of the solute? Solution: moles of biphenyl 5.5 g.357 mol 54.2 g/mol molality of solution herefore, we can determine the constant K b : moles of solute kg of solvent.357mol.357 mol/kg. kg K b b m.93k 2.53 K kg mol.357 mol/kg Now that we know the value of K b for benzene, we can use it to determine the molar mass of the unknown solute. Given the increase in the boiling temperature, we solve for the molality of the unknown solution: m b K b.597 K.236 mol/kg 2.53 K kg mol From this, we can determine the number of moles of the solute: moles of solute molality of solution kilograms of solvent (.236 mol/kg)(.5 kg).354mol hen, given the number of grams of solute, the molar mass is M 6.3 g 78 g/mol.354 mol

10 2. When.494 g of K 3 Fe(CN) 6 is dissolved in. g of water, the freezing point is found to be -.6 o C. What is the dissociation reaction for K 3 Fe(CN) 6? he K f for waterl is.86 kg K/mol. Solution: We first calculate the molality of the solute: hus, we have m.6 K.6 mol/kg.86 kg K/mol (.6 mol/kg)(. kg).6 mol moles of solute. he molar mass of K 3 Fe(CN) 6 is g/mol. his gives.494 g.5 mol g/mol of K 3 Fe(CN) 6. We know, however, that this solute dissociates into ions, and there must be moles of ions in the solution. Accordingly, the dissociation reaction must be K 3 Fe(CN) 6 3K + (aq)+[fe(cn) 6 ] 3 (aq) which gives 4 moles of ions per mole of undissociated solute.

Unit 7. Solution Concentrations and Colligative Properties

Unit 7. Solution Concentrations and Colligative Properties Unit 7 Solution Concentrations and Colligative Properties Molarity Most widely used concentration unit [HCl] means concentration of HCl in mol/l Notice volume is total volume of solution Molarity (M)=