The reaction, `PCl_3(g)+Cl_2(g)toPCl_5(g)` is started with 1 mol of `PCl_3(g)` and 2.5 mol of `Cl_2` at `300^(@)`C in a vessel of volume 2 L . At equilibrium of the reaction `PCl_5(g)` is found to have an amount of 0.65 mol . Calculate `K_c`, for the reaction.

Explanation

`PCl_3+Cl_2 Initial number of mol: 1 2.5 0
Number of moles at `1-0.65" "2.5-0.65" "0.65`
Equilibrium: =0.35 =1.85
At equilibrium: `[PCl_3]=(0.35)/(2)mol*L^(-1)`
`[Cl_2]=(1.85)/(2)mol*L^(-1)` and `[PCl_5]=(0.65)/(2)mol*L^(-1)`
`K_c=([PCl_5])/([PCl_3][Cl_2])=(((0.65)/3))/(((0.35)/2))((1.85)/(2))=2.008mol*L^(-1)`