1 Answer | Add Yours
Let us consider the following equation.
`aA+bB harr cC`
The equilibrium constant for the following equation can be given by;
`K_p = (P_C)^c/((P_a)^axx(P_b)^b)`
P denotes partial pressure.
P = (mole fraction)x(Total Pressure)
This equation for `K_P` only accounts when all the substances in the medium are gasses. When we have solids or liquids we will ignore those terms from the `K_P` equation.
`P_4+6Cl_2 harr 4PCl_3`
`Cl_2` is a gas. But `P_4` is a solid with higher boiling point. `PCl_3` is a gas at about 80C which is very lower than `P_4` boiling point.
So most likely the reaction would be as follows.
`P_(4(s))+6Cl_(2(g)) harr 4PCl_(3(g))`
As above explanations there would be no term for `P_4` in the `K_p` equation.
`K_P = (P_(PCl_3))^4/(P_(Cl_2))^6`
We’ve answered 318,966 questions. We can answer yours, too.Ask a question