Repeat the calculations of Illustrations 13.4-1 and 13.4-2 for 50°C. The Henry’s law constant for ammonia in water at this temperature is HHN3 = 384.5 kPa/mole fraction, and you can assume that the Henry’s constants for nitrogen and hydrogen are unchanged from the values given in Illustration 13.4-1.

One mole of nitrogen, 3 moles of hydrogen, and 5 moles of water are placed in a closed container maintained at 25°C and 13.33 kPa and, using the appropriate catalyst and stirring, allowed to attain phase and chemical equilibrium. Assuming that ammonia is formed by chemical reaction and that the liquid and vapor phases are ideal, compute the amount and composition of each phase at equilibrium (neglecting the aqueous-phase reaction of ammonia to form NH₄OH, and its subsequent ionization). The following data are available: 

(The Henry’s law constant for ammonia was computed from experimental data using the equation P_NH3 = H_NH3 x_NH3 and assuming all of the ammonia absorbed to be present as NH₃.)

Illustrations 13.4-1

Combined Chemical and Vapor-Liquid Equilibrium Repeat the calculations of the previous illustration for a range of pressures.

Transcribed Image Text:

Vapor pressure of water at 25°C = 3.167 kPa Henry's law constants: N₂ in H₂O: H₂ in H₂O: NH3 in H₂O: HN2 = 13.224 × 107 kPa/mole fraction 7.158 x 107 kPa/mole fraction HNH3 = 97.58 kPa/mole fraction HH₂ =