Question

Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:

\[\ce{2BrCl (g) ⇌ Br2 (g) + Cl2 (g)}\]

for which K_c= 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10^–3 molL^–1, what is its molar concentration in the mixture at equilibrium?

Answer 1

Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:

	2BrCl(g)	↔	Br2(g)	+	Cl2(g)
Initial conc.	3.3 × 10-3		0		0
At equilibrium	3.3 × 10-3 - 2x		x		x

Now, we can write,

`(["Br"_2]["Cl"_2])/["BrCl"]^2 = "K"_"C"`

`=> (x xx x)/(0.0033 - 2x)^2 = 32`

`=> x/(3.3 xx 10^-3 - 2x)` = 5.66

⇒ x = 18.678 × 10^-3 - 11.32 x

⇒ 12.32 x = 18.678 × 10^-3

⇒ x = 1.5 × 10^-3

Therefore, at equilibrium,

[BrCl] = 3.3 × 10^-3 - (2 × 1.5 × 10^-3)

= 3.3 × 10^-3 - 3.0 × 10^-3

= 3.0 × 10^-4 mol L^-1