How many moles of PCl5 has to be taken in one litre to get 0.1 mole of chlorine at equilibrium ? The equilibrium constant Kc of PCl5 (g) ⇌PCl3 (g) +Cl2 (g) is 0.0414

Solution:

$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$
x 0 0 initially
(x -a) a a at equilibrium
But [Cl₂] = 0.1 mol L^-1
So [PCl₃] = 0.1 mol L^-1
and $[PCl 5] = (x - 0.1) mol L^{- 1}$
and K_c = 0.0414
Since $K_{c} = \frac{[{PCl}_{3}] [{Cl}_{2}]}{[{PCl}_{5}]}$
0.0414x-0.00414 = 0.01
0.0414 x= 0.01414
x = $\frac{0.01414}{0.0414} = 0.341$ 5 moles

How many moles of PCl₅ has to be taken in one litre to get 0.1 mole of chlorine at equilibrium ? The equilibrium constant $K_{c}$ of
${PCl}_{5} (g) ⇌ {PCl}_{3} (g) + {Cl}_{2} (g) is 0.0414$

Solution:

Related content