[h tt

iH J

If we define

then

[H+] + ^7.i23[H+] + ^7.123^7.124 [SO2-] _ 2 ^.123^7.124[S(VI)] (? m)

where A"7.i23 and ^7.124 are the equilibrium constants for the reactions

respectively. For all practical purposes X7.123 may be considered to be infinite since virtually no undissociated sulfuric acid will exist in solution, and ^7.124 = 1-2 x 10"2 M at 298 K. Therefore the equations can be simplified to

The concentration changes are computed as follows assuming that all compounds in the system remain always in thermodynamic equilibrium. At t = 0, assuming that there is an initial sulfate concentration [S(VI)] = [S(VI)]0, the electroneutrality equation is solved to determine [H+] and, thereby, the concentrations of all dissolved species. If an open system is assumed, then the gas-phase partial pressures will remain constant with time. For a closed system, the partial pressures change with time as outlined above. The concentration changes are computed over small time increments of length At. The sulfate present at any time t is equal to that of time t — At plus that formed in the interval Ai:

The sulfate formation rate d[S(VI)]/di is simply the sum of the reaction rates that produce sulfate in this system, namely, that of the H202-S(IV) and 03-S(IV) reactions. The value of S(VI) at time t is then used to calculate [HS04] and [SO2-] at time t. These concentrations are then substituted into the electroneutrality equation to obtain the new [H+] and the concentrations of other dissolved species. This process is then just repeated over the total time of interest.

Let us compare the evolution of open and closed systems for an identical set of starting conditions. We choose the following conditions at t = 0:

[S(IV)]total = 5 ppb, [HN03]total = 1 ppb, [NH3]total = 5 ppb

For the closed system, there is no replenishment, whereas in the open system the partial pressures of all species in the gas phase are maintained constant at their initial values and aqueous-phase concentrations are determined by equilibrium.

Solving the equilibrium problem at t = 0 we find that the initial pH = 6.17, and that the initial gas-phase mixing ratios are

^so2 = 3.03 ppb, £,nh, = 1.87 ppb, qHN0;, = 8-54 x 10"9ppb = 5 PPb. ^h2o2 = 0-465 ppb

We will use the initial conditions presented above for both the open and closed systems. In the open system these partial pressures remain constant throughout the simulation, whereas in the closed system they change. A comment concerning the choice of 03 concentration is in order. We selected the uncharacteristically low value of 5 ppb so as to be able to show the interplay possible between both the H202 and 03 oxidation rates and in the closed system, the role of depletion as the reaction proceeds. Figures 7.21 and 7.22 show the sulfate concentration and pH, respectively, in the open and closed systems over a 60-min period.

Time, min

FIGURE 7.21 Aqueous sulfate concentration as a function of time for both open and closed systems. The conditions for the simulation are [S(IV)]total = 5 ppb; [NH3]total = 5 ppb; [HN03]total = 1 ppb; [03]total = 5 ppb; [H202]total = 1 ppb; h<, = 10 6; pH0 = 6.17.

Time, min

FIGURE 7.21 Aqueous sulfate concentration as a function of time for both open and closed systems. The conditions for the simulation are [S(IV)]total = 5 ppb; [NH3]total = 5 ppb; [HN03]total = 1 ppb; [03]total = 5 ppb; [H202]total = 1 ppb; h<, = 10 6; pH0 = 6.17.

FIGURE 7.22 pH as a function of time for both open and closed systems. Same conditions as in Figure 7.21.

Time, min

FIGURE 7.22 pH as a function of time for both open and closed systems. Same conditions as in Figure 7.21.

More sulfate is produced in the open system than in the closed system, but the pH decrease is less in the open system. This behavior is a result of several factors:

1. Sulfate production due to H202 and 03 is less in the closed system because of depletion of H202 and 03.

2. Continued replenishment of NH3 provides more neutralization in the open system.

3. The low pH in the closed system drives S(IV) from solution.

4. The lower pH in the closed system depresses the rate of sulfate formation by 03.

5. S(IV) is continually depleted in the closed system.

6. Total HN03 decreases in the open system due to the decrease in pH.

In the open system the importance of H202 in sulfate formation grows significantly, from 20% initially to over 80%, whereas in the closed system the increasing importance of H202 is enhanced at lower pH but suppressed by the depletion of H202. In the closed system, after about 30 min, the H202 has been depleted, the pH has decreased to 4.8 slowing down the ozone reaction, and sulfate formation ceases. The fractions of S02, 03, and H202 reacted in the closed system at 60 min are 0.43, 0.23, and 0.997, respectively. The ozone reaction produces 44.4% and 53.9% of the sulfate in the open and closed systems, respectively.

Was this article helpful?

0 0
How to Improve Your Memory

How to Improve Your Memory

Stop Forgetting and Start Remembering...Improve Your Memory In No Time! Don't waste your time and money on fancy tactics and overpriced

Get My Free Ebook


Post a comment