Chapter 6, Problem 6.45P

Sulfur trioxide (SO₃) dissolves in and reacts with water to form an aqueous solution of sulfuric acid (H₂SO₄). The vapor in equilibrium with the solution contains both SO₃ and H₂O. If enough SO₃ is added, all of the water reacts and the solution becomes pure H₂SO₄. If still more SO₃ is added, it dissolves to form a solution of SO₃ in H₃SO₄, called oleum or fuming sulfuric acid. The vapor in equilibrium with oleum is pure SO₃. Twenty percent oleum by de?nition contains 20 kg of dissolved SO₃ and 80 kg of H₂SO₄ per hundred kilograms of solution. Alliteratively, the oleum composition can be expressed as % SO₃ by mass, with the constituents of the oleum considered to be SO₃ and H₂O.

(a) Prove that a 15.0% oleum contains 84.4% SO₃.

(b) Suppose a gas stream at 40°C and 1.2 atm containing 90 mole% SO₃ and 10% N₂ contacts a liquid stream of 98 wt% H₂SO₄ (aq), producing 15% oleum. Tabulated equilibrium data indicate that the partial pressure of SO₃ in equilibrium with this oleum is 1.15 mm Hg. Calculate (i) the mole fraction of SO₃ in the outlet gas if this gas is in equilibrium with the liquid product at 40°C and 1 atm, and (ii) the ratio (m³ gas feed)/(kg liquid feed).