#### Question

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

2BrCl (g) ⇌ Br_{2} (g) + Cl_{2} (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?

#### Solution

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

2BrCl (g) ⇌ Br_{2} (g) + Cl_{2} (g)

Initial mole/litre 0.0033 0 0

Moles/litre at eqm point 0.0033 - x x/2 x/2

Applying Law of chemical equilibrium, K_C = `([Br_2][Cl_2])/[BrCl]^2` or `32 = ((x/2)xx(x/2))/(0.0033 - x)^2`

On taking the square root, 5.656 = `"x/2"/(0.0033 - x)`

x/(0.0033- x) = 11.31 or 12.31x = 0.037 ; x = 0.037/12.31 = 0.003

∴Molar concentration of BrCl at equilibrium point = 0.0033 - 0.003

`= 0.003 mol L^(-1) = 3 xx 10^(-4) mol L^(-1)`