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Chemistry 7 Online
OpenStudy (vera_ewing):

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OpenStudy (photon336):

I believe that the Kp will only depend on the concentration of the gaseous substances so you can basically ignore the solids. I believe

OpenStudy (sweetburger):

Yes solids will always fall out of the equilibrium constant. So will liquids.

OpenStudy (photon336):

Kp = [HI]^2/[H2S]

OpenStudy (photon336):

Qp is the concentration of your reactants products at any point in your chemical reaction, before equilibrium has been established

OpenStudy (taramgrant0543664):

At equilibrium, Q = K (the equilibrium constant for the reaction) Q = K is referred to as the equilibrium condition. If Q > K, the reverse reaction is favoured, the reaction moves from right to left, until equilibrium is established. If Q < K, the forward reaction is favoured, the reaction moves from left to right, until equilibrium is established.

OpenStudy (photon336):

Qp = (0.43)^2/(5.4) = 0.034 Kp if I remember was .130

OpenStudy (photon336):

It's gotta go to the right because Qp <Kp which means that the reaction will need to proceed to the right.

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