S as an energy relationship The relationship between a supersaturated (or subsaturated) system and a system at equilibrium

Energy released when NaCl (S=1.5) precipitates @25C/1atm This energy release drives the precipitation reaction forward SNaCl=1.5 −91,588.4 𝑐𝑎𝑙 Energy in solution 243.4 cal released when NaCl precipitates SNaCl=1.0 −91,831.8 𝑐𝑎𝑙 Concentration Energy released when NaCl (S=1.5) precipitates @25C/1atm

Saturation Ratio and Energy There is a difference however with respect to amount of solids forming Solids, umol/kg S=1 S=2 S=10 S=100 Calcite 48.2 274.8 1,620 Barite 4.4 23.5 101.6 Saturation Ratio and Energy

The reason energy is the same for a given S 𝐾 𝑠𝑝 𝑁𝑎𝐶𝑙, ℎ𝑎𝑙𝑖𝑡𝑒 = 𝑦 𝑁𝑎 + ∗ 𝑚 𝑁𝑎 + ∗ 𝛾 𝐶𝑙 − ∗ 𝑚 𝐶𝑙 − 𝑎 𝑁𝑎𝐶𝑙 (𝑠) 𝑺= 𝑰𝑨𝑷 𝑲 𝒔𝒑 = 𝒆 − ∆𝑮 𝒊𝒏 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑹𝑻 𝒆 − ∆𝑮𝑹 𝒂𝒕 𝒆𝒒𝒖𝒊𝒍𝒊𝒃𝒓𝒊𝒖𝒎 𝑹𝑻 ∆𝑮 𝒊𝒏 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 = ∆𝐺 𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 𝑎𝑡 𝑠𝑢𝑝𝑒𝑟𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛 ∆𝑮𝑹 𝒂𝒕 𝒆𝒒𝒖𝒊𝒍𝒊𝒃𝒓𝒊𝒖𝒎 = ∆𝐺 𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑎𝑡 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 − ∆𝐺 𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 𝑎𝑡 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 The reason energy is the same for a given S

Saturation Ratio and Scale Mass There is a relationship between the energy released and the saturation ratio GS=x -GS=1 S=1 S=2 S=10 S=100 Calcite 407 1,361 2,707 Barite 408 1,362 2,724 Saturation Ratio and Scale Mass

Relationship of S to scale formation S>1, supersaturated S<1, subsaturated Relationship of S to scale formation

Three Elements to Developing Scale Risk Thermodynamic “Saturation Ratio” or “Scale Tendency” “Excess Solute” or “Scale Mass” Kinetic Nucleation induction time Crystal Growth rates Mass Transfer Transport to surface Surface Adhesion Three Elements to Developing Scale Risk

NaCl-H2O phase balance 𝑁𝑎𝐶𝑙,𝐻𝑎𝑙𝑖𝑡𝑒 ↔ 𝑁𝑎 + + 𝐶𝑙 − the rate expression for dissolving or precipitating NaCl is: 𝑟𝑎𝑡𝑒= 𝑑 𝑁𝑎 + 𝑑𝑡 = 𝑑 𝐶𝑙 − 𝑑𝑡 =− 𝑑 𝐻𝑎𝑙𝑖𝑡𝑒 𝑑𝑡 NaCl-H2O phase balance

𝑑 𝑁𝑎 + 𝑑𝑡 = 𝑑 𝐶𝑙 − 𝑑𝑡 =− 𝑑 𝐻𝑎𝑙𝑖𝑡𝑒 𝑑𝑡 <0 :𝑁𝑎𝐶𝑙 𝑖𝑠 𝑝𝑟𝑒𝑐𝑖𝑝𝑖𝑡𝑎𝑡𝑖𝑛𝑔 water is Supersaturated with NaCl 𝑑𝑡 𝑑 𝑁𝑎 + concentration at equilibrium 𝑑 𝑁𝑎 + 𝑑𝑡 = 𝑑 𝐶𝑙 − 𝑑𝑡 =− 𝑑 𝐻𝑎𝑙𝑖𝑡𝑒 𝑑𝑡 =0 water is Subsaturated with NaCl 𝑑𝑡 𝑑 𝑁𝑎 + 𝑑𝑡 = 𝑑 𝐶𝑙 − 𝑑𝑡 =− 𝑑 𝐻𝑎𝑙𝑖𝑡𝑒 𝑑𝑡 >0 :𝑁𝑎𝐶𝑙 𝑖𝑠 𝑑𝑖𝑠𝑠𝑜𝑙𝑣𝑖𝑛𝑔 t= Plotted on a time curve

Three parts of the curve @t0, rate0 @t=0, rate=0 @t=, rate=0 Scale Tendency value Excess Solute value Supersaturated Equilibrium Subsaturated t= Two boundaries and one curve At t=0, rate=0 (by definition) – we obtain the Scale tendency At t=, rate=0 (equilibrium) – we obtain the Excess Solute Between t=0 and , rate0 (reaction occurs) – nothing obtained Three parts of the curve

How is Ksp calculated? 𝐾 𝑠𝑝 = 𝑒 − ∆𝐺 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑅𝑇 The fundamental equation for Ksp 𝐾 𝑠𝑝 = 𝑒 − ∆𝐺 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑅𝑇 Definition: The Solubility Product constant of a solid that is in equilibrium with its corresponding dissolved species in water It is derived from the thermodynamics properties of the solid and of the dissolved ions ∆𝐺 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = ∆𝐺 𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 - ∆𝐺 𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 For NaCl (halite): 𝑁𝑎𝐶𝑙 = 𝑁𝑎 + + 𝐶𝑙 − ∆𝐺 𝑅,𝑁𝑎𝐶𝑙 = ∆𝐺 𝑓,𝑁𝑎 + + ∆𝐺 𝑓,𝐶𝑙 − - ∆𝐺 𝑓,𝑁𝑎𝐶𝑙,(𝐻𝑎𝑙𝑖𝑡𝑒) How is Ksp calculated?

Thermodynamic K 𝐾 𝑟𝑥𝑛 = 𝑒 − ∆𝐺 (𝑇,𝑃) 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑅𝑇 The fundamental equation 𝐾 𝑟𝑥𝑛 = 𝑒 − ∆𝐺 (𝑇,𝑃) 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑅𝑇 when used with a rigorous Equation of State, is accurate to 300C and 1500atm ∆𝐺 (𝑇,𝑃) 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 =𝑓(𝐻, 𝑆, 𝐶 𝑝 , 𝑉 𝑚 ) Log K for HCO3-1 Thermodynamic K

A simplified equation for Ksp An simplified thermodynamic equation that is useful to ~350K (75C) and 1atm. This is called the Van’t Hoff equation. ln 𝐾 𝑟𝑥𝑛, 𝑇 2 𝐾 𝑟𝑥𝑛,298 = −∆ 𝐻 𝜃 𝑅 1 𝑇 2 − 1 298 A simplified equation for Ksp

Curve fitting equation used when solubility data is available 𝑙𝑜𝑔𝐾=𝑎+ 𝑏 𝑇 +𝑐∗𝑇+𝑑∗ 𝑇 2 +𝑒∗𝑃+𝑓 𝑃 2 Useful tool and effective within the experimental region Empirical Ksp

Temperature effects on  and Ksp