Please help!! thank you!! The reaction SO_2(g) + 2H_2S(g)^reversible3S(s) + 2H_2
ID: 966609 • Letter: P
Question
Please help!! thank you!!
The reaction SO_2(g) + 2H_2S(g)^reversible3S(s) + 2H_2O(g) is the basis of a suggested method for removal of SO_2 a pollutant that irritates airways causing coughing, from power-plant stack gases. The values below may be helpful when answering questions about the process. Assume that the partial pressure of sulfur dioxide, P_so_2>is equal to the partial pressure of dihydrogen sulfide, P_H_2S, and therefore P_so_2 = P_H_2S- If the vapor pressure of water is 27 torr, calculate the equilibrium partial pressure of SO_2 (P_so_2)in the system at 298 K.Explanation / Answer
Given reaction is SO2 (g) + 2H2S (g) <-----------> 3S (s) + 2 H2O (g)
Gorxn = Gfo(products) - Gfo( reactants)
= 2 x -228.6 - [ -300.4 + 2 x -33.01]
= - 90.78 kJ/mol
= - 90780 J/mol
Gorxn = - 90780 J/mol
G = -RT ln Kp
Kp = P2H2O / PSO2 P2H2S
GIven that
PSO2 = PH2S
PH2O = 27 torr = 27/760 atm= 0.035 atm
Hence,
Kp = P2H2O / P3SO2
G = -RT ln [P2H2O / P3SO2 ]
- 90780 J/mol = - 8.314 J/K/mol x 298 K x In[ ( 0.035 )2 / P3SO2 ]
In[P3SO2 / ( 0.035)2] = (- 90780) / ( 8.314 x 298)
[P3SO2 / ( 0.035)2] = 1.22 x 10-16
P3SO2 = 1.22 x 10-16 x (0.035)2
On solving,
PSO2 = 6.13x 10-7 atm