Q. No. 2: Henry's law works well for simple absorption of gases, e.g., Example 10.16 of textbook. It is much less satisfactory for solution of acid gases, which dissociate in aqueous solution. For SO2 in water the reactions seem to be those in Eq. SO2 (gas) SO2 (dissolved); +H2OH2SO3H++HSO3, Whether the free acid, H2SO3, exists or whether instead the dissolved SO2+H2O goes directly to the ions is not clear experimentally. For this treatment, we assume that the free acid does not exist, so that the equilibria that exist are SO2(gas)SO2(dissolved)SO2(dissolved)+H2OH++HSO3HSO3H++SO32 for which the individual equilibrium relationships for dilute solutions are HSO2=PySO2/[SO2(dissolved)]K1=[HSO3][H+]/[SO2(dissolved)]K2=[SO32][H+]/[HSO3] In aqueous chemical equilibria the [] symbol almost always stands for concentration in mol/L. Here the Henry's law constant, HSO2, has dimensions of atm /(mol/L).K1 and K2 are the first and second ionization constants, with dimensions (mols/L). Reference 26 shows equations for these equilibrium constants (and many of the others that are important for limestone scrubbers) as a function of temperature. The following table shows the calculated values at 68F=20C, and at 125F52C. (a) Henry's law, Eq. (10.19), is most often written using the mol fraction, x1, of the dissolved material as the liquid concentration value. In the above table, and in much of the scrubber literature, Henry's law is written with molarity ( mol/L ) or mass ratio or concentration ( mol/kg or lbmol/ft3 ) as the liquid-phase concentration value. These concentration values can all be converted one to the other, so the H for one can be converted to the H for any of the others. Estimate the value of the Henry's law constant at 20C in Eq. (10.19), which takes mol fraction as the concentration variable (and has the dimension atm) from the value in the above table. (b) Estimate the values of the concentrations at 20C of [SO2 (dissolved) ,[H+],[HSO3]and [SO32] in a solution that is in equilibrium with a gas at 1atm, with ySO2=0.001. The simplest procedure is to solve Eq. (11.24) for [SO2 (dissolved)], then assume that [H+]and [HSO3]]are equal [i.e., that all the [H+]comes from Reaction (11.22)]. This allows a direct solution of Eqs. (11.25) and (11.26). Finally, check the quality of that assumption. SO2(gas)SO2(disolvel)HSO3SO2(dissolved)+H2OH++HSO3H++SO5 for which the individual equilibrium relationships for dilute solutions are HSO1=PYSO1/[SO2(dissolved)]K1=[HSO1][H+]/[SO2(dissolved)]K2=[SO3][H+]/[HSO3](11.24)(11.25](11.26) (c) Estimate the pH of the solution in (b). (d) The Henry's law constant shown in Eq. (11.24) is based on the dissolved SO2 only, and not its ionization products. If one wrote it to take into account the ionized products, i.e., HSO2=PySO2/{[SO2(dissolved)]+[HSO3]+[SO32]} for the situation in (b), what would the value of HSO2 be, both in atm and in atm/( mol/L) ? Note: Refer to textbook for given Eqs in the problem. Reference from Text Book Example 7.10. A "typical Pittsburgh seam coal" has the following ultimate analysis by weight: hydrogen, 5.0%; carbon, 75.8%; nitrogen, 1.5%; sulfur, 1.6\%; oxygen, 7.4\%; ash, 8.7\% (see Appendix C). It is burned with 20% excess air with humidity 0.0116mol/mol dry air, and combustion is complete. Determine the amount and composition of the gas produced. Instead of choosing one mol of feed, we choose 100g of dry coal. Thenthe mols of the individual components are these: nC=75.8/12=6.32,nH=5.0/1=5.0, nN2=1.5/28=0.054,nS=1.6/32=0.050, and nO2=7.4/32=0.231. Then we can see by inspection that the mols of CO2,H2O, and SO2 formed by combustion are 6.32, 5/2=2.50, and 0.050. The nitrogen is assumed to exit as N2 (although some of it actually exits as NO or NO2, see Chapter 12) so that the mols of nitrogen, from the fuel, are 0.054. The mols of oxygen needed for combustion are those needed to oxidize the C, H, and S, less that contained in the fuel. In this case nstoict=nC+4nH+nsnO2=6.32+45+0.0500.231=7.39100gdrycoalmol and ndryntotalnN2nO2=nstoich(0.211+E)=7.390.211.2=42.23100gdrycoalmol=ndry(1+X)=42.231.0116=42.72100gdrycoalmol=0.79ndry+nfuclnatogen=0.7942.23+0.054=33.42100gdrycoalmol=Enstoich=0.27.39=1.48100gdrycoalmol