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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 22 of 73
Marks: +1, -0
Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium :
2BrCl(g)2\mathrm{BrCl}_{(g)} ⇌ Br2(g)+Cl2(g)\mathrm{Br}_{2(g)} + \mathrm{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−110^{-3}\,\text{mol L}^{-1}, what is its molar concentration in the mixture at equilibrium?
Solution:  
2BrCl(g)2\mathrm{BrCl}_{(g)} ⇌ Br2(g)+Cl2(g)\mathrm{Br}_{2(g)} + \mathrm{Cl}_{2(g)}
Initial3.30×10−3 mol L−100At eqm.(3.30×10−3−x)x2x2\begin{array}{lccc} \text{Initial} & 3.30 \times 10^{-3}\,\text{mol L}^{-1} & 0 & 0 \\ \text{At eqm.} & (3.30 \times 10^{-3} - x) & \frac{x}{2} & \frac{x}{2} \end{array}
KcK_c = (x2)(x2)(3.30×10−3−x)2\frac{\left(\frac{x}{2}\right)\left(\frac{x}{2}\right)}{\left(3.30 \times 10^{-3} - x\right)^2} = 32 (Given)
∴ x24(3.30×10−3−x)2\frac{x^2}{4\left(3.30 \times 10^{-3} - x\right)^2} = 32 or , x2(3.30×10−3−x)\frac{x}{2\left(3.30 \times 10^{-3} - x\right)} = 32\sqrt{32} = 5.66
or, x = 11.32 (3.30 × 10−310^{-3} – x) or, 12.32x = 11.32 × 3.30 × 10−310^{-3} or, x = 3.0 × 10−310^{-3}
∴ At eqm., [BrCl] = (3.30 × 10−310^{-3} – 3.0 × 10−310^{-3}) = 0.30 × 10−310^{-3} = 3.0 × 10−4 mol L−110^{-4}\,\text{mol L}^{-1}
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