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 28 torr , calculate the equilibrium partial pressure of SO2 (PSO2) in the system at 298 K.
SO2(g) + 2H2S (g)-------> 3S + 2H2O(g)
G =
Gf
products -
Gf reactants
= -2*228.59+0 -( -371.08 +2*-33.05)
= -20KJ/mole = -20000J/mole
G = -RTlnKp
-20000 = -8.314*298*2.303logkP
logKp = 20000/5705.8483
= 3.5051
Kp = 103.5051 = 3199.6
PH2O = 28torr = 28/760 = 0.0368atm
Kp = P2H2O/PSo2 P2H2S
3199.6 = (0.0368)2/x*x2
x3 = 0.0368*0.0368/3199.6 = 4.23*10-7
x = 7.5*10-3 atm
PSo2 = x = 7.5*10-3 atm
Assume that the partial pressure of sulfur dioxide, PSO2, is equal to the partial pressure of...
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 vaporpressure of water is 27 torr, calculate the equilibrium partial pressure of SO2 (PSO2) in the system at 298 K. Express the pressure in atmospheres to two significantfigures.PSO2= __ atmIn part A, Kp= 8*10^15:( I keep getting it wrong.
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Sulfur dioxide and oxygen react to form sulfur trioxide, like
this: 2SO2(g)+O2(g)→2SO3(g)
Also, a chemist finds that at a certain temperature the
equilibrium mixture of sulfur dioxide, oxygen, and sulfur trioxide
has the following composition:
Calculate the value of the equilibrium constant Kp for this
reaction. Round your answer to 2 significant digits.
compound pressure at equilibrium SO2 58.3 atm 02 84.7 atm SO3 66.3 atm