c5f28afd19
class. This means that the function can be modified depending on the inheritance. This allows for the printout of molalities and pH's for liquid phases where appropriate. Took out a duplicate function, report(), in the Cantera namespace. There was one in cxxutils.cpp and one in phasereport.cpp.
153 lines
4.1 KiB
Text
153 lines
4.1 KiB
Text
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This is the classic problem of the solubility of NaCl in water.
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The result is a saturated solution of Na+ and Cl- in water in equilibrium
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with NaCl solid. This problem also has a gas phase consisting of N2, H2O,
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H2, and O2. CO2 has been thrown in as well. However, the element abundance
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of C is zero. So, CO2 should turn out to have a zero concentration.
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The problem can be divided up into two parts: estimating the Gibbs
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reaction delta for
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Na+ + Cl- = NaCl(solid)
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(so that Del(Gf) = G(NaCl(solid)) - G(Na+) - G(Cl-))
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and estimating the activity coefficients for the electrolytes at
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the solubility limit.
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By far the most important as always is estimating the delta G
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for the reaction above. Note the electrolytes are using the
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molality basis.
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Using the NASA basis we get
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NaCl(solid) : Hf(298.15) = -411.1207 kJ/gmol
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S(298.15) = 72.1093 J/(gmol K)
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(from NIST Webbook)
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Gf(298.15) = -432.6200 kJ/gmol
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From Codata key values for Thermodynamics:
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Cl- Hf(298.15) = -167.080 kJ/gmol
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S(298.15) = 56.60 J/(gmol K)
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Gf(298.15) = -183.9552 kJ/gmol
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Na+ Hf(298.15) = -240.34 kJ/gmol
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S(298.15) = 58.45 J/(gmol K)
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-> Gf(298.15) = -257.7668 kJ/gmol
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Del(Gf)(298.15) = 9.1020 kJ/gmol
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In addition, the relative humidity of the salt solution may be compared
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to the humidity above a pure water solution in order to understand
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the effects of the lowering of the water activity.
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If you run the equilibrium calculation without salt, you get the
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equilibrium of water vapor above water, consistent with the
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current thermo.
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X_H2O(g) = 0.03169
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at 25C and 1atm
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If you then run the calculation with a saturated salt solution you
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get a smaller amount of water vapor in equilibrium with the water
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electrolyte:
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X_H2O(g) = 0.023243
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at 25V and 1atm
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The ratio of these two numbers is the relative humidity at which
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the a saturated salt solution deliqueses from a salt particle.
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rel humidity = 0.733
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This is a very important number in terms of its effect on aqueous
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corrosion.
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Next, let's compare the above calculation with the calculation presented
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in the current directory. First, getting Gabs from the cttables calculation:
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G(298.15) NaCl(solid) = -432.6201 kJ/gmol
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G(298.15) Na+ = -257.7668 kJ/gmol
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G(298.15) Cl- = -183.9552 kJ/gmol
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(note, Na+ and Cl- have screwed up H and S values. However, the
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G value seems to be correct).
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Putting this together yields:
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Delta G = 9.1019 kJ / gmol
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----------------------------------------------------------------------------
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H2O(l) = H+ + OH- Equilibrium Reaction
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------------------------------------------------------------------------
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The equilibrium constant for water is given in Robinson and Stokes, p. 363
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and p.544.
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10^(-13.996) = actCoef(H+) actCoef(OH-) Molal(H+) Molal(OH-) / activity_H2O.
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This works out to a value of
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DeltaG = 79.88936 kJ/gmol
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DeltaH = 56.576 kJ/gmol
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DeptaCp = -194.68 J/K gmol
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In the current database:
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G(298, OH-) = -226.7839 kJ / gmol
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G(298 H+) = 0.0
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G(298, H2O(L)) = -306.6858.
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-------------------------
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DeltaG = 79.902
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Therefore, there is a slight error of 0.013 kJ/gmol, but the database is roughly correct.
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Silvestre & Pitzer
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------------------------------------------------------------------------
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The equilibrium condition for this simple system comes down to the
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following equation
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Delta G = - 2 R T ln (m * actCoeff)
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This is the basic test of the system.
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Delta G = -2161 cal gmol-1 = -9.0416 kJ gmol-l
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M_sat = 6.146
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ActCoeff_mixed_molalityScale = 1.008
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Using the given files and conditions, I calculate equilibrium as:
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Delta G = 9.1019 kJ gmol-l
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M_sat = 6.193
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ActCoeff_mixed_molalityScale = 1.0132
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Relative Humidity Lowering
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------------------------------------------------------
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from the saturated NaCl calculation using HMW
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x_H2O = 0.0237641
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From the pure water equilibrium case:
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x_H2O = 0.0316882
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Therefore the relative humidity lowering is 0.7504
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