Question

Bromine monochloride, BrCI decomposes into bromine and chlorine and reaches the equilibrium: 2BrCl (g) ⇆ Br₂ (g) + Cl₂ (g) for which K = 32 at 500 K. If initially pure BrCI is present at a concentration of 3.3 x 10⁻³ mol L⁻¹, what is its molar concentration in the mixture at equilibrium?

Answer 1

Sorry, i can't understand this question

Answer 2

Let the amount of bromine and chlorine formed at equilibrium be x.

The given reaction is: 2BrCl (g) ↔ Br2 (g) + Cl2 (g),

Initial Conc.  3.3x10-3   0   0 ,at equilibrium 3.3x10-3 -2x x   x,

Now, we can write,,

Kc = [Br2][Cl2] / [BrCl]2,

⇒ (x) x (x) / (3.3x10-3 -2x)2 = 32 ⇒ x / (3.3x10-3 -2x) = 5.66,

⇒ x = 18.678x10-3 - 11.32x,

⇒ x + 11.32x = 18.678x10-3,

⇒ 12.32x = 18.678x10-3,

⇒ x = 1.5 x 10-3

Therefore, at equilibrium,[BrCl] = 3.3x10-3 - (2 x 1.5 x 10-3), = 3.3x10-3 - 3.0x10-3= 0.3 x 10-3 = 3.0 x 10-4 mol L-1.