Question

In principle, is this reaction a feasible method of removing SO2 from power-plant emissions?

Answer 1

SO₂(g) + 2H₂S(g) ⇋ 3S(s) + 2H₂O(g)

Assume that the partial pressure of sulfur dioxide, PSO₂, is equal to the partial pressure of dihydrogen sulfide, PH₂S, and therefore PSO₂=PH₂S. If the vapor pressure of water is 24 torr , calculate the equilibrium partial pressure of SO₂ (PSO₂) in the system at 298 K.

Using relative enthalpy and entropy values, determine how the process is affected after each of the following temperature or pressure changes. Consider that a more effective reaction produces more product or more product in a shorter amount of time.
Kp = P(H₂O)²/P(SO₂)×P(H₂S)² = 8×10¹⁵ so at equilibrium the P(SO₂) and P(H₂S) must be extremely small to give a Kp value of 8×10¹⁵

Yes; the reaction is highly spontaneous at 298 K and almost no SO2 will remain at equilibrium.

PSO₂ = 5.0×10⁻⁷ atm

If For example, P(H2O) = 1.0 atm then P(SO₂) and P(H₂S) ~2×10^-5.

Yes in principle with this reaction which is a highly spontaneous at 298 K and almost no SO₂ will remain at equilibrium​ therefore, feasible method of removing SO₂ from power-plant emissions.