Question

A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5 was found to be $0.5×10^{–1}\ mol L^{–1}$. If value of Kc is $8.3×10^{–3}$, what are the concentrations of PCl3 and Cl2 at equilibrium?
$PCl_5 (g) ⇋ PCl_3 (g) + Cl_2(g)$

Answer 1

Let the concentrations of both PCl3 and Cl2 at equilibrium be x molL-1.
The given reaction is:
$PCL_5(g) ↔ PCl_3(g) + Cl_2(g)$
At equilibrium 0.5×10^{-1} mol L^{-1}$$x \ mol L^{-1} $x \ mol L^{-1}$
It is given that the value of equilibrium constant, Kc is $8.3 × 10^{-3}$.
Now we can write the expression for equilibrium as:
$\frac {[PCl_2] [Cl_2]}{[PCl_5]} = K_c$

⇒ $\frac {x×x}{0.5×10^{-1}} = 8.3 \times 10^{-3}$
⇒ $x^2 = 4.15×10^{-4}$
⇒ $x = 2.04×10^{-2}$
⇒ $x= 0.0204$
⇒ $x= 0.02$ (approximately)
Therefore, at equilibrium
$[PCl_3] = [Cl_2] = 0.02 \ molL^{-1}$

So, the answer is $0.02 \ molL^{-1}$.