Question: Sulfur trioxide (SO 3 ) dissolves in and reacts with water to form an aqueous solution of sulfuric acid (H 2 SO 4 ). The
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 SO and H2O. If enough SO3 is added, all of the water reacts and the solution becomes pure H2S04. If still more SO3 is added, it dissolves to form a solution of SO3 in H2SO4, called oleum or fuming sulfuric acid. The vapor in equilibrium with oleum is pure SO3. A 20% oleum by definition contains 20 kg of dissolved SO3 and 80 kg of H2SO4 per hundred kilograms of solution. Alternatively, the oleum composition can be expressed as % SO3 by mass, with the constituents of the oleum considered to be SO, and H2O.
(a) Prove that a 15.0% ole urn 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% H2SO3 (aq), producing 15% oleum at the tower outlet. 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 SO, in the outlet gas if this gas is in equilibrium with the liquid product at 40°C and 1 atm. and (ii) the ratio (m’ gas feed)/(kg liquid feed).
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a 15 oleum Basis 100kg 85 kg HSO4 15 kg SO3 844 SO 3 1 kmol SO3 1 kmol HSO4 98... View full answer
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