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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium5 Marks
Question
Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:$2\text{BrCl (g)}\rightleftharpoons\text{Br}_2\text{ (g) + Cl}_2\text{ (g)}$for which $K_c= 32$ at $500K$. If initially pure BrCl is present at a concentration of
$3.3 \times 10^{–3} mol L^{–1}$, what is its molar concentration in the mixture at equilibrium?
✓

Answer

  $2\text{BrCl}_{\text{(g)}}$ $\rightleftharpoons$ $\text{Br}_{2\text{(g)}}$ $+$ $\text{Cl}_{2\text{(g)}}$
Initial $3.30\times10^{-3}\text{mol L}^{-1}$   $0$   $0$
At eqm. $(3.30\times10^{-3}-\text{x})$   $\frac{\text{x}}{2}$   $\frac{\text{x}}{2}$
$\text{K}_{\text{c}}=\frac{(\text{x/2})(\text{x/2})}{(3.30\times10^{-3}-\text{x})^2}=32\text{ (Given)}$
$\therefore\ \frac{\text{x}^2}{4(3.30\times10^{-3}-\text{x})^2}=32$
$\text{or, }\frac{\text{x}}{2(3.30\times10^{-3}-\text{x})^2}=\sqrt{32}=5.66$
$\text{or, }\text{x=11.32(3.30}\times10^{-3}-\text{x})$
$\text{or, }12.32\text{x}=11.32\times3.30\times10^{-3}$
$\text{or, }\text{x}=3.0\times10^{-3}$
$\therefore\ \text{At eqm., [BrCl]}=(3.30\times10^{-3}-3.0\times10^{-3})$
$=0.30\times10^{-3}$
$=3.0\times10^{-4}\text{mol L}^{-1}$

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