
Answer:
C) [PCl5] = 0.0280 M, [PCl3] = 0.222 M, [Cl2] = 0.222 M
Explanation:
Moles of [PClâ‚…] = 0.250 M
Considering the ICE table for the equilibrium as:
           PCl₅ (g)    ⇔      PCl₃ (g) +    Cl₂ (g)
t = o        0.250
t = eq         -x               x            x
--------------------------------------------- --------------------------
Moles at eq: 0.250-x            x            x    Â
The expression for the equilibrium constant is:
[tex]K_c=\frac {[PCl_3][Cl_2]}{[PCl_5]}=1.80[/tex]
So,
[tex]\frac{x^2}{0.250-x}=1.80[/tex]
x = 0.222 M
Equilibrium concentrations :
[PCl₃] = [Cl₂] = 0.222 M
[PClâ‚…] = 0.250 - 0.222 = 0.280 M
Option C is correct.