Lesson 7b: aqueous sulphuric acid at a fixed relative humidity


Content

The water uptake of sulphuric acid particles as a function of temperature is examined. The behaviour of this electrolyte differs from the salts studied in Lesson 7a, due to the equilibrium between aqueous hydrogen (H+), sulphate (SO42−) and bisulphate (HSO4−) ions.


Entering the Data

In previous runs we calculated the water uptake of NH4NO3 and (NH4)2SO4 particles. Here we do a similar calculation for H2SO4. Use the back button on the other browser window to return to the data input page for "variable temperature" calculations using Model II, or select this link. Fill in the input form as follows:

1st Calculation
  1. Select "Graph" as the form of output, and then enter the values and options under the following headings:

    Ambient Conditions
    (1) For temperature, enter: Start Value = 275, End Value = 300, Number of points = 25.
    (2) Select a fixed relative humidity (the default), and enter 0.90 (i.e., 90%).

    Ionic Composition in Moles
    Hydrogen = 2.0, Sulphate = 1.0.

    Trace Gases
    Click on "H2SO4" to prevent this gas being partitioned into the vapour phase.

    2. Solid Phases
    There are no entries under this heading.


  2. Click on the "Run" button at the end of the page to do the calculation.
Note:  the above should be entered on the "variable temperature" parametric calculations page of Model II (http://www.aim.env.uea.ac.uk/aim/model2/mod2t.php).


Viewing and Interpreting the Results

A page will appear in the other browser window which enables you to plot various quantities against each other by choosing the X and Y variables, their ranges, and scales (linear or log10) from the drop down lists. Instructions, and details of the variables, are given in the right frame.

First we plot the moles of liquid water associated with the particle against temperature:

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 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.

The water content of the particle has decreased by about 2% over the temperature range, the opposite change to that found previously for the two salts.

Next, examine the change in the activity coefficient of the water:

2nd Graph:  select the variables and enter the options as given below.
X Variable: "temperature"

Range: leave blank

Scale: linear (the default)

Y Variable: "fH2O(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.

The plot shows that the water activity coefficient fH2O falls by about 0.2% from low to high temperature. This appears to be both the wrong magnitude and direction considering our understanding from the equation relating water activity (equilibrium relative humidity) to the water mole fraction and activity coefficient:

water activity = xH2O × fH2O

As the water content of the particle has decreased from low to high temperature then the water mole fraction would be expected also to decrease. In order to maintain the fixed water activity of 0.90 an increase in the water activity coefficient by the same proportion would be required.

The explanation for this lies in the changing composition of the sulphuric acid solution, according to the equilibrium reaction:

HSO4−(aq) = H+(aq) + SO42−(aq)

The equilibrium between these ions is temperature dependent, shifting towards HSO4− as temperature increases. This decreases the total number of ions present, and appears to be main factor controlling the water uptake of the acid. In this example the water activity coefficient is nearly constant (only a 0.2% decline) so the decrease in the total number of ions requires a similar decrease in the number of moles of water to maintain the water mole fraction in solution and the constant water activity ( = xH2O × fH2O) of 0.90.


You should now review the conclusions on the main page of this lesson.