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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 22 torr , calculate the equilibrium partial pressure of so2 (pso2) in the system at 298 k. express the pressure in atmospheres to two significant figures.

Respuesta :

shinmin
First you need to know the value of Kp for this reaction. The reaction is: SO2 (g) + 2 H2S (g)<=> 3S(s) + 2 H2O(g): Kp = PH2O^2 / PSO2(PH2O)^2 but since the problem is lacking in details.
Otherwise, let x = PSO2 = PH2S. Then:
Kp = (22/760 atm)^2 / x^3 
x = cube root (0.289)/Kp = PSO2
But if the kp is given, let us assume it is 8.0*10^15The solution would be:convert the given pressure of water to atm, which is: 0.029
Kp=P(H2O)^2/P(SO2)P(H2S)^2
8.0*10^15 = (0.029)^2/x^3
(8.0*10^15).(x^3)=8.41*10^-4
take the cube root of both sides(2.0*10^5).x = 9.439130*10^-2x = P (SO2) = 4.6*10^-7

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