Content

Part 1

1st Calculation

Enter the values and select the options under the following headings: Ambient Conditions Relative humidity = 0.99 (i.e., 99%). Ionic Composition in Moles Ammonium = 1.0, Nitrate = 1.0. Solid Phases There are no entries under this heading. Click on the "Run" button at the end of the page to do the calculation.

Note:  the above should be entered on the "simple" calculations page of Model III (http://www.aim.env.uea.ac.uk/aim/model3/model3a.php).

Interpreting the Results

1. The title of the results page should read "E-AIM Model Results" followed by a line starting with "Problem no.". E-AIM is able to run parametric calculations where one run will produce multiple sets of results. For your problem, E-AIM produces a single results section, which will always be given as problem number 1. "iFail = 0" (or sometimes 1 or 6) confirms that the answer was calculated successfully.
2. Other symbols and quantities at the top of the results page have the following meanings:

   System Pressure:	ambient pressure in atmospheres (usually fixed at 1 atm).
   Volume:	the assumed volume, in cubic metres, containing the aerosol mass.
   T:	ambient temperature in Kelvin. This is input by the user, except for Model III where temperature is fixed at 298.15 K.
   RH:	the ambient relative humidity, expressed as a fraction. This is either set by the user or determined from the partitioning of a known amount of water between the vapour and condensed phases. The equivalent partial pressure of water (pH2O) is also listed here.

3. The rest of the results page is divided into three sections which give information for the liquid, gas and solid phases that may be present at equilibrium:

   Aqueous Phase
   
   

   All the species present are listed here. Note that all species except H2O and NH3 are ions, for which the charges are not shown. In addition to NH4⁺, NO3⁻, and water (H2O), the output contains calculated amounts and concentrations of three other species: H⁺, OH^-, and NH3. These arise from the dissociation of water and the ammonium ion (NH4⁺). The "Grams" column shows the same quantities as in the "Moles" column, but in mass units. The output shows that 1 mole of NH4NO3 takes up 173.65 moles, or 3.128 kg, of water at 99% relative humidity and 298.15 K.

   Scroll right so you can see the "Molality" and "Mole Frac." columns. The "Molality" column is the concentration in moles of species per kg of liquid water and the "Mole Fraction" column is the concentration in moles of species per total moles. Remember that this concentration and mole fraction are within the particles and refer only to the aqueous phase portion of the system. Consequently, the concentration (molality) of water is always 55.51 which is just the number of moles of water in 1 kg of water.

   Returning now to the "Molality" column, the concentrations of ammonium (NH4⁺) and nitrate (NO3⁻) are both about 0.32 mol kg^-1, which is the same as saying that the molality of ammonium nitrate in the particle is 0.32 mol kg^-1.

   The activity coefficients for the aqueous phase species given under "Act. Coeff." are on a mole fraction basis. The reference states for these activity coefficients (f), that is the compositions for which they are unity, are pure water for water itself, and infinite dilution in water for the ions. In this example the activity coefficient of water (1.001) is very close to one, but the activity coefficients of the ions are rather lower (0.6402). You will find that the activity coefficients of ions can deviate by quite large amounts from unity even in dilute solutions.

   Gases
   
   

   Scroll left again to see that there are no species in the gas phase. This is because the model that you are running assumes that all species are in the particle phase. Other models, which we will work with in subsequent lessons, can provide more information about gas phase species.

   Although no species (here H2O, NH3 and HNO3) were allowed to partition into the vapour phase it is possible to calculate what their equilibrium partial pressures would be from the properties of the aqueous phase. For water we have:

   pH2O = fH2O × xH2O × pH2O^o,

   where fH2O is the mole fraction activity coefficient of water (1.001 in the "Act. Coeff." column), xH2O is the mole fraction of water (0.9886 in the "Mole Frac." column), and pH2O^o is the vapour pressure of pure water (0.031258 atm) which is calculated internally by the model and not shown on the output.

   Remember that the relative humidity (RH) in a vapour phase is simply equal to the actual partial pressure of water divided by the equilibrium partial pressure of water over pure water at the system temperature. Thus, from the equation above:

   RH = pH2O/pH2O^o = fH2O × xH2O,

   which tells us that, for a system at equilibrium, the aqueous phase water activity is the same as the relative humidity.

   The calculation of equilibrium partial pressures of NH3 and HNO3 from activities in the aqueous phase is straightforward:

   pHNO3 = aH⁺ × aNO3⁻ / KH(HNO3)

   pNH3 = aNH3 / KH'(NH3)

   In the above equations KH and KH' are Henry's law constants and the activity of each aqueous species, denoted by the prefix a, is equal to its mole fraction multiplied by its activity coefficient. For example, the activity of H⁺ is equal to 0.2259E-6 × 0.7246 = 1.6369E-7.

   Suppose we had not determined the concentrations of NH3 and H⁺ in the ammonium nitrate particle. In this case it would not be possible to obtain the individual equilibrium partial pressures of NH3 and pHNO3, but we can still calculate the equilibrium partial pressure product pNH3 × HNO3 (1.46E-19 atm). The equilibrium constant for this product is related to the aqueous activities and Henry's law constants of the two gases, and the acid dissociation constant (Ka) of the ion NH4⁺:

   pNH3 × pHNO3 = aNH4⁺ × aNO3⁻ × Ka(NH4⁺) / (KH'(NH3) × KH(HNO3)),

   where:

   Ka(NH4⁺) = aH⁺ × aNH3 / aNH4⁺

   KH'(NH3) = aNH3 / pNH3

   KH(HNO3) = aH⁺ × aNO3⁻ / pHNO3

   The activity of each aqueous species, denoted by the prefix a, is equal to its mole fraction multiplied by its activity coefficient, as before. Notice how the activities of neither H⁺(aq) nor NH3(aq) appear in the above equation for the partial pressure product, which is therefore independent of both quantities.

   If you do the calculation described here on the "Comprehensive" input page for Model III (http://www.aim.env.uea.ac.uk/aim/model3/model3b.php), and select the option to switch off NH4⁺ dissociation, the value of the pressure product will be listed but not the equilibrium partial pressures of NH3 and HNO3.

   Solids
   
   

   You may have to scroll down to show the "Solids" section where you see that there are no solid phases. The model permits solid phases to form, but the conditions of this run lead only to a dilute aqueous phase.

Part 2

2nd Calculation

Enter the values and select the options under the following headings: Ambient Conditions Relative humidity = 0.99 (i.e., 99%). Ionic Composition in Moles Hydrogen = 1.0, Nitrate = 1.0. Solid Phases There are no entries under this heading. Click on the "Run" button at the end of the page to do the calculation.

Note:  the above should be entered on the "simple" calculations page of Model III (http://www.aim.env.uea.ac.uk/aim/model3/model3a.php).

Interpreting the Results

Aqueous Phase

Nitric acid takes up slightly more water than NH4NO3 - 186.12 moles compared to the 173.64 moles we had before, yielding a more dilute solution with a molality of 0.2982 mol kg^-1 of HNO3. The difference between the two solutes is quite small, but it becomes greater as relative humidity decreases. In general, water uptake can vary very widely between different solutes.

The mole fraction activity coefficients of H⁺ and NO3⁻ are 0.7383. These are higher than for NH4⁺ and NO3⁻ at the same relative humidity, and most of the difference is because the HNO3 solution at 99% relative humidity is more dilute.

Gases

Remember that this model does not allow volatile species to partition into the gas phase, otherwise we would expect to see an amount of HNO3 vapour present. However, we can calculate what the equilibrium partial pressures would be. For water we again have:

pH2O = fH2O × xH2O × pH2O^o,

where fH2O is equal to 1.001 and xH2O (0.9894) is the mole fraction of water.

The output lists a partial pressure of 1.805E-8 atm for HNO3, which is calculated from the Henry's law reaction given in Part 1:

KH(HNO3) = aH⁺ × aNO3⁻ / pHNO3

The mole fraction Henry's law constant is equal to 853.1 atm^-1 at 298.15 K, thus pHNO3 = (0.005316 × 0.7383)² / 853.1 = 1.805E-8 atm. The model output shows that this partial pressure is equivalent to 0.7379E-6 moles in the system volume of 1 m³.

Solids

Again there are no solids. Nitric acid does form a series of solid hydrates, some of which are important in the atmosphere. However, in this example the temperature is too high and the solution too dilute for any of them to occur.

You should now proceed to Lesson 1b, or return to the main page for this lesson.