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Equilibrium

Question
CBSEENCH11006433

Bromine monochloride (BrCl) decomposes into bromine and chlorine and attains the equilibrium:
space 2 BrCl space rightwards harpoon over leftwards harpoon space space space space Br subscript 2 left parenthesis straight g right parenthesis space plus space Cl subscript 2 left parenthesis straight g right parenthesis
for which Kc = 32 at 500 K. If initially pure BrCl is present at a concentration of 3·30×10–3 mol L–1 what is its molar concentration in the mixture at equilibrium?

Solution
The reaction is
                      space space space space space space space space space space space space space space space space space space space space space 2 BrCl left parenthesis straight g right parenthesis space space space space space space space space space space rightwards harpoon over leftwards harpoon space space space space space space space space space space Br subscript 2 left parenthesis straight g right parenthesis space plus space Cl subscript 2 left parenthesis straight g right parenthesis
Initial moles/litre          
            3.30 x 10-3 mol-1      0              0
At equilibrium
           (3.30 x 10-3 -x)        x/2           x/2

where x moles of BrCl decompose in order to attain equilibrium.
Applying the law of chemical equilibrium
       straight K subscript straight c space equals space fraction numerator open square brackets Br subscript 2 close square brackets space open square brackets Cl subscript 2 close square brackets over denominator open square brackets BrCl close square brackets squared end fraction space space space space space... left parenthesis 1 right parenthesis

Putting the values in expression (1), we have
      space straight K subscript straight c space equals space fraction numerator open parentheses begin display style straight x over 2 end style close parentheses space open parentheses begin display style straight x over 2 end style close parentheses over denominator left parenthesis 3.30 space cross times space 10 to the power of negative 3 end exponent space minus straight x right parenthesis squared end fraction space equals space 32 space left square bracket Given right square bracket
space therefore space space space space space space space fraction numerator straight x squared over denominator 4 left parenthesis 3.30 space cross times space 10 to the power of negative 3 end exponent minus straight x right parenthesis end fraction space equals space 32
On taking the square root, we have,
     fraction numerator straight x over denominator 2 left parenthesis 3.30 space cross times space 10 to the power of negative 3 end exponent minus straight x right parenthesis end fraction space equals space square root of 32 space equals space 5.66
                                space straight x space equals 11.32 space left parenthesis 3.30 space cross times space 10 to the power of negative 3 end exponent minus straight x right parenthesis
space or space space space space space space 12.32 straight x space equals space 11.32 space cross times space 3.30 space cross times space 10 to the power of negative 3 end exponent
or space space space space space space space space space space space straight x space equals space 3.0 space cross times space 10 to the power of negative 3 end exponent
therefore   At equilibrium
space open square brackets BrCl close square brackets space equals space left parenthesis 3.30 space cross times space 10 to the power of negative 3 end exponent space minus 3.0 space cross times space 10 to the power of negative 3 end exponent right parenthesis
space space space space space space space space space equals space 0.30 space cross times space 10 to the power of negative 3 end exponent
space space space space space space space space space space equals 3.0 space cross times space 10 to the power of negative 4 end exponent mol space straight L to the power of negative 1 end exponent