| Lesson 9a: gas-aerosol partitioning of H+/NH4+/NO3−/H2O |
| 1st Calculation |
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| Note: the above should be entered on the
"variable temperature" parametric calculations page of
Model II |
First determine the amount of solid material in the particle by plotting the amount of NH4NO3(s):
| 1st Graph: select the variables and enter the options as given below. | |
| X Variable: "temperature" Range: leave blank Scale: linear (the default) |
Y Variable: "moles NH4NO3" Range: leave blank Scale: linear (the default) |
| Click on the "Draw the Graph" button at the end of the page, and the plot will appear in the right frame. | |
The graph shows that about 1.0E-6 moles of NH4NO3(s) - containing almost all of the ammonium initially present in the system - exists up to about 284 K. At higher temperatures no solid is present, presumably because the particle becomes aqueous. Check this by plotting the amount of NH4+(aq) present:
| 2nd Graph: select the variables and enter the options as given below. | |
| X Variable: "temperature" Range: leave blank Scale: linear (the default) |
Y Variable: "moles of NH4+(aq)" Range: leave blank Scale: linear (the default) |
| Click on the "Draw the Graph" button at the end of the page, and the plot will appear in the right frame. | |
At 284 K the amount of NH4+(aq) rises from zero to about 1.0E-6 moles, confirming the transition from solid to aqueous particle at this temperature. At higher temperatures there is a steady decline followed by disappearance at about 315 K which we will investigate later.
What is happening at the 284 K transition? We saw in Lesson 8a that the deliquescence relative humidity of NH4NO3 falls sharply with temperature, from about 76% at 273.15 K to 62% at 298.15 K. In the present calculation the ambient relative humidity is 70%: thus at temperatures below 284 K it is below the deliquescence relative humidity of the salt and the particulate ammonium nitrate exists as a solid. At 284 K the ambient and deliquescence relative humidities are equal and the particle takes up water to become an aqueous droplet.
Next, we look at what is happening to the species NH3(g) and
HNO3(g) in the gas phase. Plot the moles of
NH3(g) (make sure you select a logarithmic scale):
| 3rd Graph: select the variables and enter the options as given below. | |
| X Variable: "temperature" Range: leave blank Scale: linear (the default) |
Y Variable: "moles of NH3(g)" Range: leave blank Scale: log10 |
| Click on the "Draw the Graph" button at the end of the page, and the plot will appear in the right frame. | |
At low temperatures, for which the particle is solid, there is a steadily rising amount of NH3(g). This can be explained in terms of the following reaction:
NH4NO3(s) = NH3(g) + HNO3(g)
Kp = pNH3 × pHNO3 = 4.356E-17 atm2
The value of the equilibrium constant Kp increases with temperature, rising from about 8.36E-20 atm2 at 275 K to 10.7E-19 atm2 at 284 K. Thus some of the solid particle volatilises as temperature rises, increasing the amount of NH3(g) in the gas phase in order to maintain equilibrium. The change in the amount of solid particle is almost negligible, however, as the amounts of NH3(g) below 284 K are about three orders of magnitude less than the amount of solid.
At temperatures above 284 K the particle exists as a liquid droplet. Some of the 1.0E-6 moles of HNO3 in the system dissolves into the droplet and the equilibrium of NH3(g) between the gas phase and the liquid droplet is described by the reaction:
pNH3 = (Ka(NH4+) / KH'(NH3)) × aNH4+ / aH+
where the equilibrium constants are defined in Lesson 1a and Lesson 3a. The value of the equilibrium constant quotient rises with increasing temperature leading to a rise in the amount of NH3(g) with temperature for those conditions in which a liquid particle exists. This is what the graph shows. You may also notice a small inflection in the curve at the transition temperature between solid and liquid particle, although this is quite small.
Last, we interpret the transition at about 315 K. We see from the third graph
that above this temperature the amount of gas phase NH3 present
is equal to 1.0E-6 moles, which is equivalent to all the ammonium present in
the system. Plot the amount of liquid phase water
in the system against temperature:
| 4th Graph: select the variables and enter the options as given below. | |
| X Variable: "temperature" Range: leave blank Scale: linear (the default) |
Y Variable: "moles of H2O(aq)" Range: leave blank Scale: linear (the default) |
| Click on the "Draw the Graph" button at the end of the page, and the plot will appear in the right frame. | |
Between 285 K, when the solid NH4NO3(s) particle deliquesces, and just under 300 K the particle grows by a very small amount. This is due to non-ideal solution effects as the amounts of both H+(aq) and NH4+(aq) decline with temperature. You can verify this by plotting these quantities.
Above 300 K the amount of water present in the particle declines steeply. This is because the dissolved NH4+(aq), H+(aq), and NO3−(aq) ions partition increasingly into the gas phase (as HNO3(g) and NH3(g)) as temperature rises, leaving smaller and smaller quantities left in the particle. The amount of liquid water therefore also decreases, to maintain equilibrium with the ambient relative humidity (which requires that the ion concentrations in the particle remain about the same). Eventually, at 315 K, all of the ions have transferred to the gas phase leaving a system that contains no particle at all.
You should now review the conclusions on the main page of this lesson.