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. Express the pressure in atmospheres to two significant figures.

Kp= 8*10^15

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

Explanation:

;D just to let you know you are an amazing human YOU are on this earth to do something great, God has a purpose for you and just have an amazing day c;;