The degree of dissociation of `PCl_5` at a certain temperature and under atmospheric pressure is 0.2. Calculate the pressure at which it will be half dissociated at the same temperature.

Explanation

`PCl_5` dissociates as :
`{:(,PCl_5, If `alpha` is the degree of dissociation at certain temperature under atmospheric pressure, then total number of moles at equilibrium
`= 1- alpha + alpha + alpha = 1 + alpha`
Partial pressures of `PCl_5, PCl_3 and Cl_2` will be
`p(PCl_5) = (1 - alpha)/(1 + alpha) . P , p_(PCl_3) = (alpha)/(1 + alpha) . P,`
`p_(Cl_2) = alpha/(1 + alpha) . P`
Now `K_p = (p_(PCl_3).p_(Cl_2))/(p_(PCl_5))`
`= ((alpha/(1 + alpha) . P) xx (alpha/(1 + alpha) .P))/((1-alpha)/(1 + alpha).P) = (alpha^2)/(1 - alpha^2) P`
Putting P = 1 atm and `alpha = 0.2`
`K_P = ((0.2)^2)/(1 - (0.2)^2) xx 1 = 0.041`
when `alpha = 1/2 = 0.5`, let pressure is P.
`K_p = (alpha^2)/(1 - alpha^2) . P. implies 0.041 = ((0.5)^2 P.)/(1 - (0.5)^2)`
`implies P. = ((0.041)[1 - (0.5)^2])/((0.5)^2) = 0.123 atm`