Precipitation Size! Shape! Size distribution! Agglomeration!
Precipitation Four major questions: 1. Why do molecules/ions precipitate? 2. What determines the size? 3. What determines the size distribution? 4. What determines the shape? We have to have a closer look at the precipitation process!
1. Why do molecules/ions precipitate? Supersaturation as a First and Simple Thermodynamic Approach The molar Gibbs Free Energy ΔG and the Supersaturation S are related as follows (T, p are constant): S = Δ G = - RT ln S Actual concentration in solution K sp Equilibrium concentration = (c i )v i c i : Concentration of species i v i : Stoichiometric index (number of mols) Δ G = 0 for S = 1 Δ G < 0 for S > 1 ΔG (kj/mol) The larger the supersaturation S, the larger the driving force for precipitation! The thermodynamic driving force for crystallization is the supersaturation of the solution! 8 6 4 2 0-2 -4-6 Driving force for precipitation 0 1 2 3 4 5 Saturation, S (a (c act act /a /c equ )
Precipitation Four major questions: 1. Why do molecules/ions precipitate? 2. What determines the size? 3. What determines the size distribution? 4. What determines the shape?
Precipitation 1. Nucleation Nucleation and growth determine size, size distribution, shape and agglomeration! 2. Growth Let s have a closer look at nucleation and growth!
1. Nucleation Homogeneous nucleation seed Heterogeneous nucleation
ΔG (kj/mol) 8 6 4 2 0-2 1.1 Homogeneous Nucleation: Thermodynamics Equilibrium T = 25 o C P = 1 atm 0 1 2 3 4 5 Δ G = - RT ln S Driving force for precipitation -4-6 is ΔG < 0 (S > 1). Saturation, S (a act /a equ ) 10 ΔG (10-18 J) 5 0-5 -10 V = 25 cm 3 /mol T = 25 o C S = 0.5 0 10 20 30 S = 2 S = 100 S = 10 However: Often S >> 1, but still no precipitation! Why? Radius, r (A o )
1.1 Homogeneous Nucleation: Thermodynamics Why? Because surface free energy is not considered. r During nucleation, small clusters are formed with a large fraction of the molecules/ions at the surface. The free energy for the surface molecules/ions is larger than the free energy for the interior molecules/ions. This surface free energy has to be taken into account when calculating ΔG.
1.1 Homogeneous Nucleation: The Classical Crystallization Model r Nucleus r > r*: Growing Δ G = - RT ln S 8 6 4 V = 25 cm 3 /mol T = 25 o C Surface (r 2 ) Embryo r < r*: Dissolving Δ G = - v RT ln S + γ A V v = Volume of the aggregate V= Molar volume of the precipitate ΔG (10-18 J) 2 0-2 r* ΔG max 0 10 20 30 γ = Surface free energy A= Area of the aggregate -4-6 -8 Bulk (r 3 ) Radius, r (A o ) Total If the particle/aggregate is a sphere: 4πr Δ G = - 3 RT ln S + 4πr 2 γ 3V S < 1: ΔG always positive S > 1: ΔG with maximum at r * (critical size) Bulk ( Interior ) Surface
1.1 Homogeneous Nucleation: Thermodynamics r Nucleus r > r* Small supersaturation is not enough to overcome the surface free energy! Nucleation only with large S! Embryo r < r* ΔG (10-18 J) 4 3 2 1 0-1 -2-3 -4 V = 25 cm 3 /mol T = 25 o C γ = 0.15 J/m 2 S = 2 0 10 20 30 S = 10 S = 20 S = 100 Radius, r (A o ) Δ G = - 4πr 3 RT ln S + 4πr 2 γ 3V Bulk ( Interior ) Surface One possibility to lower the activation energy ΔG max (or to decrease r*) is to increase the saturation S.
1.2 Heterogeneous Nucleation: Thermodynamics High surface free energy: r * and ΔG max are large. Low surface free energy: r* and ΔG max are small ΔG (10-18 J) 4 V = 25 cm 3 3 /mol T = 25 o C 2 S = 10 1 γ = 0.20 J/m 2 0 0 10 20 30-1 γ = 0.15-2 γ = 0.10-3 γ = 0.05-4 Radius, r (A o ) Δ G = - r Embryo r < r* Nucleus r > r* 4πr 3 RT ln S + 4πr 2 γ 3V Bulk ( Interior ) Surface Another possibility to lower the activation energy ΔG max (or to decrease r*) is to lower the surface free energy γ.
1.1 Homogeneous Nucleation: Thermodynamics 4πr Δ G = - 3 RT ln S 3V Determination of ΔG max and r*: + 4πr 2 γ dδg = 0 dr 4 3 2 r * = 2γV RT ln S r * : Small for high saturation and/or small surface energy ΔG (10-18 J) 1 0-1 -2-3 ΔG max 0 10 20 30 r* ΔG max = 16 π V2 γ 3 3 (RT ln S) 2-4 Radius, r (A o ) Free energy necessary to form a stable nucleus!
1.1 Homogeneous Nucleation: Exercise r * = 2γV RT ln S ΔG max = 16 π V2 γ 3 3 (RT ln S) 2 Suppose we want to produce Mg(OH) 2 by mixing MgCl 2 and NH 4 OH. Assuming that we start with a solution containing 0.25 M of MgCl 2 and 0.55 M of NH 4 OH at 100 C. Data for Mg(OH) 2 : Solubility in water at 100 C = 0.04 g/l, M w = 58.33 g/mol, ρ = 2.36 g/cm 3, γ SL = 0.12 J/m 2, R= 8.31 J/Kmol, N A = 6.02 10 23 mol -1. 1) What is the critical nuclei size? Assumption: S n n lim, sp = e lim, sp ν Actual concentration of limiting specie Equilibrium concentration of limiting specie ν: stoichiometric index (number of mols) 2) What is the number of molecules in a critical nuclei?
1.1 Homogeneous Nucleation: Thermodynamics is not enough! 4πr Δ G = - 3 RT ln S + 4πr 2 γ 3V dδg Determination of ΔG max and r*: dr No indication about the time! = 0 4 3 2 r * = 2γV RT ln S r * : Small for high saturation and/or small surface energy ΔG (10-18 J) 1 0-1 -2-3 ΔG max 0 10 20 30 r* ΔG max = 16 π V2 γ 3 3 (RT ln S) 2-4 Radius, r (A o ) Free energy necessary to form a stable nucleus!
1.2 Homogeneous Nucleation: Kinetics How fast does nucleation occur? r Embryo Nucleus Ion Diffusion Embryo + Reaction Embryo Nucleus Step 1 Step 2 Rate of nucleation can be expressed in a simple Boltzmann approach: dn(r*) J = = K exp ΔG max - dt kt K = Kinetic prefactor; difficult to determine, often empirical! The nucleation rate J (number of nuclei formed per unit volume per unit time) is dependent on the activation energy ΔG max, i.e., on the difficulty for embryos to reach the critical radius of a stable nuclei!
1.2 Homogeneous Nucleation: Kinetics How fast does nucleation occur? J = dn(r*) dt = K exp ΔG max - kt Kinetic factor (~ 10 10-10 35 cm -3 sec -1 ) J max = Maximum nucleation rate J = J max exp - ΔG max kt ln(j/j max ) = - ΔG max kt
1.2 Homogeneous Nucleation: Kinetics How fast does nucleation occur? ln(j/j max ) = - ΔG max kt ΔG max = 16 π V2 γ 3 3 (RT ln S) 2 R = k N A ln(j/j max ) = - 16 π V 2 γ 3 1 3 N A2 (kt) 3 (ln S) 2 A = - (ln S) 2 A A γ 3
1.2 Homogeneous Nucleation: Kinetics How fast does nucleation occur? ln(j/j max ) = - 16 π V 2 γ 3 1 3 N A2 (kt) 3 (ln S) 2 A = - (ln S) 2 Decreasing nucleation rate ln(j/jmax) 0-2 -4-6 -8-10 -12 1 2 10 20 100 A = 200 400 1 10 100 1000 If A is small (i.e., surface free energy γ is small), then a small saturation S and also a small increase of S leads to fast nucleation. If A is large (i.e., the particles exhibit a high surface free energy), then even a large saturation S does not result in fast nucleation. Saturation ratio, S
1.3 Heterogeneous Nucleation: Kinetics Decreasing A: From homogeneous to heterogeneous nucleation One possibility to get nucleation at lower saturation S is to lower the surface energy γ (and thus A): hom J = J max BaSO 4 Heterogeneous nucleation Het J max Concentration of seeds < Homo J max hom J << J max 1) Heterogenous nucleation takes only place as long as there are seeds present 2) Heterogeneous nucleation (A=2) takes place at lower saturation ratio ( J max is 10 5 ) 3) At higher saturation (A=200), homogeneous nucleation occurs with J max = 10 30 4) Homogeneous nucleation: Very sensitive to slight changes in S, difficult to control 5) Heterogeneous nucleation: Better control over particle size distribution, because the nucleation rate is almost independent of S
2. Growth Cylinder r d Additional surface energy New nucleus (nucleation): ΔG = - 4π r 3 3V RTln S + 4πr 2 γ 4 Bulk Surface ΔG (10-20 J) 3 2 1 0-1 -2-3 -4 V = 25 cm 3 /mol T = 25 o C d = 10 A o ΔG max Surface 0 2 4 6 8 10 Bulk Total New layer on the surface (growth): ΔG = - π dr 2 V S, γ RTln S Bulk + 2πdr γ Surface ΔG max Similar to nucleation: S and γ are the main parameters to be considered! Radius, r (A o ) Main difference: Surface energy r!
2. Growth Two borderline cases: : 1) rough and 2) smooth surfaces 1. Case: Assumption that surface energy γ is low, and saturation ratio S is high 2. Case: ΔG max is small Formation of rough surfaces Assumption that surface energy γ is high, and saturation ratio S is low ΔG max is large Formation of smooth surfaces
1. Case: Assumption that surface energy γ is low, and saturation ratio S is high ΔG max is small Formation of rough surfaces Growth rate T. Ring p. 197-198 Diffusion = x rate For spherical coordinates C eq dr dt r Diffusion controlled growth 2. Growth Probability factor =1 C C eq δ Fick s law: J = - D Diffusion Δ Every ion/molecule that hits the surface is attached to the nuclei (because of the small ΔG max ). Probability = 1 C Growth rate only dependent on diffusion rate Diffusion rate dependent on C eq and r (concentration at the crystal surface) Diffusion controlled growth (when γ low, S high) dependent on i) equilibrium concentration at the surface of the nucleus, and ii) radius of the nucleus
2. Case: Assumption that surface energy γ is high, and saturation ratio S is low 2. Growth ΔG max is large (nucleation controlled!) Formation of smooth surfaces growth Mononuclear growth growth Polynuclear growth
2. Case (large γ, low S): 2. Growth Growth rate Diffusion = x rate Probability factor growth Mononuclear growth dr dt r 2 x exp - ΔG max RT Nucleation controlled growth! The high surface energy ( large ΔG max ) leads to the situation that only very few molecules/ions are able to overcome the large activation energy (nucleation controlled), and thus are able to attach to the growing nuclei. The slow process does not result in a concentration gradient. Diffusion growth Polynuclear growth dr dt C eq x exp - ΔG max RT Nucleation controlled growth! Due to the large number of nucleation sites on the surface, thermal fluctuations and Fick s diffusion play a role: There is a concentration gradient. The growth rate depends on thermal and Fick s diffusion [(C eq ) comes into play] and on the activation energy ΔG max, but it does NOT depend on the radius r of the nuclei!
2. Growth: Summary 1. Case: Diffusion Controlled (low γ, high S) C C eq δ Diffusion dr dt r -1 ΔG max does not play a big role! Growth rate is inversely proportional to r! Nucleation controlled (high γ, low S) 2. Case: Mononuclear Polynuclear Diffusion Diffusion growth growth dr dt r 2 Growth rate is directly proportional to r 2! dr dt r 0 Growth rate does not depent on r!
3. What determines the size distribution? Polynuclear growth Diffusion growth dr dt r 0 What is the prediction regarding size distribution? Time Growth rate is not dependent on r, i.e., all nuclei grow with the same speed. Initial absolute size distribution remains constant over time (0.1 μm) (but relative size distribution decreases).
3. What determines the size distribution? Mononuclear growth dr dt r 2 Diffusion growth + Polynuclear growth dr dt r 0 Diffusion growth What is the prediction regarding size distribution? Time Absolute particle size distribution broadens with growth time, i.e, larger particles grow faster (growth rate proportional to r 2 ) than the small ones.
3. What determines the size distribution? C Diffusion controlled C eq δ Diffusion dr dt r -1 What do you expect regarding size distribution (small ones grow fast, large ones slow down)? Time Particle size distribution narrows with growth time, i.e., smaller particles grow faster than the larger ones.
Nucleation & Growth: Examples Homogeneous Nucleation for Monodisperse Particles: Hot Injection Method The method involves the injection of a cold (room temperature) solution of precursor molecules into a hot liquid (300 C). Hot injection leads to instantaneous nucleation (very high S), quenched by fast cooling of the reaction mixture and fast decrease of the supersaturation by the nucleation burst. Further growth of the nuclei into mature nanocrystals occurs at a lower temperature, such that new nucleation events do not occur: Separation of nucleation and growth Monodisperse nanoparticles Small 2005, 1, 1152; Angew. Chem. In. Ed. 2007, 46, 4630
Nucleation & Growth: Examples Homogeneous Nucleation for Monodisperse Particles: Hot Injection Method CdSe Small 2005, 1, 1152
Nucleation & Growth: Examples Heterogeneous Nucleation for Monodisperse Particles: 1) Synthesis of monodisperse iron nanoparticles (seeds) with sizes of 4, 8, and 11 nm from reaction mixtures containing 1:1, 1:2, and 1:3 molar ratios of pentacarbonyliron and oleic acid. 2) Solutions containing 1.5, 3.0, and 4.5 mmol of the iron oleate complex (low concentration to suppress homogeneous nucleation!) were prepared by heating appropriate amounts of pentacarbonyliron and oleic acid in dioctyl ether at 403 K for 12 h. After refluxing the mixtures generated from the various combinations of the iron nanoparticles and the iron oleate solutions, it was possible to synthesize monodisperse iron nanoparticles with particle sizes of 6, 7, 9, 10, 12, 13, and 15 nm. TEM images of a) 6-, b) 7-, c) 8-, d) 9-, e) 10-, f) 11-, g) 12-, and h) 13-nm-sized air-oxidized iron oxide nanoparticles showing the one nanometer level increments in diameter. Angew.Chem. Int. Ed. 2005, 44, 2872
4. What determines the shape? What determines the crystal morphology/shape? (a) Equilibrium shape: thermodynamics-determined - Low saturation ratio, S (b) Kinetic shape: kinetics-determined - High saturation ratio, S
4. What determines the shape? (a) Equilibrium shape: thermodynamics-determined Low saturation ratio, S What if the surface energy (γ) depends on the crystallographic orientations? Face with lowest surface energy determines the crystal shape! F2 γ F1 h 1 h 2 γ F2 F1 γ F1 < γ F2 Wulff diagram Aspect ratio = b c γ F1 = γf2 The growth rate h i of each face is proportional to its surface energy γ i. Thermodynamic equilibrium shape: Minimization of the surface free energy!
2. What determines the shape? (b) Kinetic shape: kinetics-determined What if the growth rate depends on the crystallographic orientations? High saturation ratio, S Face with slowest growth rate determines the crystal shape! F2 F1 m2 b dr dt F 2 k2s Aspect ratio = = = m1 c dr dt k S F1 1
4. What determines the shape? Surface energy can be tuned through the adsorption of ions, surfactants, proteins! Example:
4. What determines the shape? Surface energy tuning through surfactants! Surface modulation effects induced by surface-selective surfactants on either a) anisotropic rod or b) disc growth. When surfactant molecules specifically bind to the {100} and {110} surfaces of a hexagonal structure, preferential growth along the <001> directions and therefore rod growth is facilitated (a). In contrast, when surfactant molecules bind to the {001} surfaces of a hexagonal structure, it prevents growth along the <001> direction and therefore disc shapes are obtained (b). Angew. Chem. Int. Ed. 2006, 45, 3414
4. What determines the shape? Surface energy tuning through polymers! 2μm BaSO 4 Morphogenesis at ph 5 0.5μm PEG-b-PEI-SO 3 H PEG-b-PMAA-Asp 0.5μm No additive 2μm 2μm PEG-b-PEI PEI-COOH Dr. Cölfen, MPI Potsdam PEG-b-PMAA PMAA-POPO 3 H 2
: Summary... involve the precipitation of ions from a supersaturated solution and can be used for: Synthesis of powders (ex: Al 2 O 3, ZrO 2 ) Construction materials (ex: concretes) Fabrication of complex microstructures (ex: nacre)... occur through nucleation and growth mechanisms which ultimately control: Particle size distribution Particle shape Aggregation state
: Summary Nucleation and growth are thermally activated processes (ΔG max ), which can occur by a number of different mechanisms: Nucleation: Homogenous Heterogeneous (seeds) Growth: Diffusion-controlled Nucleation-controlled The saturation ratio (S) and the surface energy (γ) are the main factors determining the nucleation and growth mechanisms Therefore, can be tailored to control the particle size distribution and crystal morphology of precipitates
: Summary Saturation ratio, S High nucleation rate Diffusioncontrolled growth Size distribution narrows during growth Kinetic shape 1 μm Ex: Precipitated submicron-sized SiO 2 1 -- Low nucleation rate Nucleationcontrolled growth Size Equilibrium distribution shape broadens during growth Ex: Big quartz crystals