Question

Question 7.19 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?

Class XI Equilibrium Page 226

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: PCl_5(g) < = > PCl_3(g) + Cl_2(g) \\ at \: eqlm \: \: \: \: \: \: \: \: \: \: \: 0.05 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x$

Given,
[PCl5] at equilibrium = 0.05 M
Kc = 8.3 × 10^-3

but we know,
Kc = [PCl3][Cl2]/[PCl5]

8.3 × 10^-3 = x . x /0.05
8.3 × 10^-3 × 0.05 = x²
x² = 0.415 × 10^-3 = 4.15 × 10^-4
now, take square root both sides,
x = 2.04 × 10^-2 M

hence, [PCl3 ] = [Cl2] = 2.04 × 10^-2 M