`P_4+6Cl_2 harr 4PCl_3` Phosphorus reacts with chlorine as shown. What is the equilibrium constant expression, Kp, for this reaction?



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Posted on (Answer #1)

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`




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