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Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:
2BrCl(g)⇋Br2(g)+Cl2(g)2BrCl (g) ⇋ Br_2 (g) + Cl_2 (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)2BrCl(g) ↔ Br_2(g) + Cl_2(g)

Initial conc. 3.3×10−33.3×10^{-3} 00 0
At equilibrium 3.3×10−3−2x3.3×10^{-3}-2x xx xx

Now, we can write,
[Br2][Cl2][BrCl]2=Kc\frac {[Br_2][Cl_2]}{[BrCl]^2} = K_c

⇒ x×x(3.3×10−3−2x)2=32\frac {x×x}{(3.3×10^{-3}- 2x)^2} = 32

⇒ x(3.3×10−3−2x)=5.66\frac {x}{(3.3×10^{-3}- 2x)} = 5.66
⇒ x=18.678×10−3−11.32xx = 18.678×10^{-3 }-11.32 x
⇒ 12.32x=18.678×10−312.32 x = 18.678×10^{-3}
⇒ x=1.5×10−3x = 1.5×10^{- 3}
Therefore, at equilibrium,
[BrCl]=3.3×10−3−(2×1.5×10−3)[BrCl] = 3.3×10^{-3}-(2×1.5×10^{-3})
= 3.3×10−3−3.0×10−33.3×10^{-3}-3.0×10^{-3}
= 0.3×10−30.3 × 10^{-3}
= 3.0×10−4 molL−13.0×10^{- 4} \ mol L^{-1}