PHYSICAL CHEMISTRY PHYSICAL CHEMISTRY, Cheat Sheet of Physical Chemistry

Download PHYSICAL CHEMISTRY PHYSICAL CHEMISTRY and more Cheat Sheet Physical Chemistry in PDF only on Docsity! A deeper look 1 The Debye–Hückel theory The strategy for the development of the Debye–Hückel theory of electrolyte solutions is to establish the rela- tion between the work needed to charge an ion and its chemical potential. Then that work is related to the ion’s interaction with the atmosphere of counter ions that has assembled around it as a result of the competition of the attraction between oppositely charged ions, the repulsion of like-charged ions, and the distributing effect of thermal motion. Step 1 Calculate the work of charging an ion Imagine a solution in which all the ions have their actual positions, but in which their Coulombic interactions have been turned off and so are behaving ‘ideally’. The differ- ence in molar Gibbs energy between the ideal and real solutions is equal to we, the electrical work of charging the system in this arrangement. For a salt MpXq, µ µ µ µ= + − ++ − + −w p q p q( ) ( )e ideal ideal       p q( ) ( )ideal idealµ µ µ µ= − + −+ + − − (1) From eqn 5F.26 (µi = µiideal + RT ln γ±) the terms in paren- theses are written as µ+ idealµ− + = µ− idealµ− − = RT ln γ± so it follows that γ γ γ= + = +± ± ±w pRT qRT p q RTln ln ( ) lne     and therefore w sRT s p qln eγ = = +± (2) This equation implies that the final distribution of the ions and then the work of charging them in that distribution must be found. Step 2 Calculate the Coulomb potential experienced by an ion The Coulomb potential at a distance r from an isolated ion of charge zie in a medium of permittivity ε is φ ε= = π Z r Z z e 4i i i i (3a) However, the ionic atmosphere causes the potential to decay with distance more sharply than this expression implies. Such shielding is a familiar problem in electro- statics, and its effect is taken into account by replacing the Coulomb potential by the shielded Coulomb potential, an expression of the form Z r ei i r r/ Dφ = − Shielded Coulomb potential (3b) where rD is called the Debye length. When rD is large, the shielded potential is virtually the same as the unshielded poten- tial. When rD is small, the shielded potential is much smaller than the unshielded potential, even for short distances (Fig. 1). To calculate rD, it is necessary to know how the charge density, ρi, of the ionic atmosphere, the charge in a small region divided by the volume of the region, varies with distance from the ion. This step draws on another standard result of electrostatics, in which charge density and poten- tial are related by Poisson’s equation: 2φ ρε∇ = − poisson’s equation (4a) where ∇ = + +x y z∂ /∂ ∂ /∂ ∂ /∂2 2 2 2 2 2 2. Because the ionic atmosphere is spherically symmetric the charge density varies only with distance from the central ion and eqn 4a becomes r r r r 1 d d d d i i 2 2 φ ρ ε     = − (4b) Substitution of the expression for the shielded potential (eqn 3b) results in r i i D 2 εφ ρ= − (5) Step 3 Calculate the Debye length To solve eqn 5 for rD, ρi must be related to ϕi. For this step, note that the energy of an ion depends on its closeness to the central ion, and then use the Boltzmann distribution Figure 1 The variation of the shielded Coulomb potential with distance for different values of the Debye length, rD/a. The smaller the Debye length, the more sharply the potential decays to zero. In each case, a is an arbitrary unit of length. 0 0.2 0.4 0.6 0.8 1 0 0.5 1 Distance, r/rD Po te n ti al , ϕ /( Z /r D ) 1 3 ∞ 0.3 Gm, charged Gm ideal, uncharged µ+ − µ+ideal µ− − µ−ideal A Deeper look 1 The debye–hüCkel Theory2 to evaluate the probability that an ion will be found at each distance. The energy of an ion j of charge zje at a distance where it experiences the potential ϕi of the central ion i relative to its energy when it is far away in the bulk solution is its charge zje times the potential ϕi. Therefore, according to the Boltzmann distribution, the ratio of the molar concen- tration, cj, of ions at a distance r where their electrostatic energy is z ej iφ and the molar concentration in the bulk, c j°, where their electrostatic energy is zero, is c c ej j z e kT/j i ° = φ− (6) The charge density, ρi, at a distance r from the ion i is the molar concentration of each type of ion multiplied by the charge per mole of ions, zieNA = ziF (F is Faraday’s constant, F = eNA). It follows that c z F c z F c z F c z Fe ei z e kT z e kT/ /i iρ = + = ° + °φ φ+ + − − + + − − − −+ − (7) Because the average electrostatic interaction energy is small compared with kT it is permissible to use = + +xe 1x and to write the charge density as c z F z ekT c z F z e kT1 1i i i ρ φ φ( ) ( )= ° − + + ° − ++ + + − − − φ= ° + ° − ° + ° ++ + − − + + − −    c z c z F c z c z FekT( ) ( ) i2 2 The term in blue is zero because it is the charge density in the bulk, uniform solution, and the solution is electrically neutral. The replacement of e by F/NA and NAk by R results in the following expression: c z c z FRT( )i i2 2 2 ρ φ= − ° + °+ + − − (8) The higher-order unwritten terms are assumed to be too small to be significant. This equation can be expressed in terms of the ionic strength by noting that in the dilute aqueous solutions being considered c ≈ bρ, where ρ is the mass density of the solvent. Therefore c z c z b z b z Ib( ) 22 2 2 2    ○ρ ρ° + ° ≈ ° + ° =+ + − − + + − − −− With these approximations, eqn 8 becomes Ib F RT 2 i i 2○ ρ ρ φ= − −− (9) When this expression is substituted into r /i iD 2 εφ ρ= − , the ϕi cancel and the result is r RT F Ib2D 2 1/2 ○ ε ρ =   −− debye length (10) Step 4 Calculate the work of charging the ion in the pres- ence of the ionic atmosphere To calculate the work of charging the central ion it is necessary to know the potential at the ion due to its atmosphere, ϕatmos. This potential is the difference between the total potential, given by eqn 3b, and the potential due to the central ion itself: Z r r e 1 i r r atmosphere central ion / D φ φ φ= − = −   − (11a) The potential at the central ion (at r = 0) is obtained by tak- ing the limit of this expression as r → 0 and is Z r rr r Z r(0) lim 1 / 1 i r i atmosphere 0 D D φ = − + −   = −→ (11b) This expression shows that the potential due to the ionic atmosphere is equivalent to the potential arising from a single charge of equal magnitude but opposite sign to that of the central ion and located at a distance rD from the ion. Therefore, if the charge of the central ion were Q and not zie, then the potential due to its atmosphere would be φ ε( )= − π Q r0 4atmosphere D (11c) The work of adding a charge dQ to a region where the elec- trical potential is ϕatmosphere(0) (from dw = ϕdQ), is dwe = ϕatmosphere(0)dQ (12) It follows that the total molar work of fully charging the ion i in the presence of its atmosphere is ∫ ∫φ ε= = − πw N Q N r Q Q(0)d 4 di z e z e e, A atmosphere0 A D 0 i i ε ε= − π = − π N z e r z F N r8 8 i iA 2 2 D 2 2 A D (13) Step 5 Evaluate the activity coefficient The total work of charging p cations and q anions in the presence of their atmospheres is we = pwe,+ + qwe,−, and therefore the mean activity coefficient of the ions is γ ε= + = − +π± + − + −pw qw sRT pz qz F N sRT rln ( ) 8 e, e, 2 2 2 A D (14a) However, for neutrality pz+ + qz− = 0; therefore + = + = − + =+ − + + − − + − + −pz qz pz z qz z p q z z s z z( ) | | 2 2   It then follows that γ ε= − π± + −z z F N RT rln | | 8 2 A D (14b) 0 2 Ib⦵ − qz− − pz+ −|z+ z−|s