Chemical Equilibrum

If it is.. then i think it is like this....

k_p=px²/(1-x²)
p₁x₁²/(1-x₁²)=p₂x₂²/(1-x₂²)
Therefore:

(1*0.2)/0.96=(p₂*0.5)/0.75
so we get

0.15/0.48=0.3 atm

PCl_{5 (s)}\LeftrightarrowPCl_{3 (g)} +Cl_{2 (g)}

1 0 0
1-α α α Total moles=1+α

Partial pressure(PCl₃)=\frac {\alpha }{1+\alpha}P_0=Partial pressure(Cl₂)

where P₀=1 atm ,α=0.2

Now ,K_P=({\frac {\alpha}{1+ \alpha}}P_0)^2

Solving, K_P=4/144

Now K_P is dependent on temp only

=> K_P=4/144=({\frac {\alpha '}{1+ \alpha '}}P_{new})^2

where P_new is to be found out and α'=0.5

Solving P_new =0.25