SELECT * FROM question_mgmt as q WHERE id=3046 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=48 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

Question:

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

2BrCl (g) ↔ Br2 (g) + Cl2 (g) for which Kc= 32 at 500 K.

If initially pure BrCl is present at a concentration of 3.3 × 10–3 mol L–1, what is its molar concentration in the mixture at equilibrium?

Answer:

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


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