Question

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

2BrCl(g) ⇌ Br₂(g) + Cl₂(g)

for which K_C = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.30 x 10^-3 mol L^-1, what is its molar concentration in the mixture at equilibrium.

Answer 1

On taking the square not

5.656 = \(\frac{x/2}{(0.0033-x)}\)

\(\frac{x}{(0.0033-x)}\) = 11.31 or 12.31, x = 0.0037

x = \(\frac{0.037}{12.31}\) = 0.003

∴ Molar concentration of BrCl at equilibrium point

= 0.033 - 0.003 = 0.0003 mol L^-1

= 3 x 10^-4 mol L^-1

Therefore, [BrCl] equilibrium = 3 x 10^-4 mol L^-1