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Equilibrium

Question
CBSEENCH11006429

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0·78 M?
2 ICl left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon space space straight I subscript 2 left parenthesis straight g right parenthesis space plus space Cl subscript 2 left parenthesis straight g right parenthesis semicolon space space space straight K subscript straight c space equals 0.14

Solution

Let at equilibrium,
                            space open square brackets straight I subscript 2 close square brackets space equals space open square brackets Cl subscript 2 close square brackets space equals space straight x space mol space straight L to the power of negative 1 end exponent
Then                      2 ICl space space rightwards harpoon over leftwards harpoon space space straight I subscript 2 left parenthesis straight g right parenthesis space plus space space Cl subscript 2 left parenthesis straight g right parenthesis
Initial conc.            0.78 M       0            0
At eqm.                 0.78 - 2x    x            x
Applying the law of chemical equilibrium
                   straight K subscript straight c space equals fraction numerator open square brackets straight I subscript 2 left parenthesis straight g right parenthesis close square brackets space open square brackets Cl subscript 2 left parenthesis straight g right parenthesis close square brackets over denominator open square brackets ICl left parenthesis straight g right parenthesis close square brackets squared end fraction space space space space space... left parenthesis 1 right parenthesis
Putting the values in expression (1), we have
        0.14 space equals space fraction numerator straight x space cross times space straight x over denominator left parenthesis 0.78 minus 2 straight x right parenthesis squared end fraction
        or space space space straight x squared space equals space 0.14 left parenthesis 0.78 minus 2 straight x right parenthesis squared
or space space space fraction numerator straight x over denominator 0.78 minus 2 straight x end fraction space equals square root of 0.14 end root space equals space 0.374
or space space space space space space straight x space equals space 0.292 space minus space 0.748 space straight x
or space space space space space 1.748 straight x space equals space 0.292
or space space space space space space space space straight x space equals space 0.167
Hence, at equilibrium                  left square bracket straight I subscript 2 right square bracket space equals space open square brackets Cl subscript 2 close square brackets space equals space 0.167 space straight M
and    
   
 open square brackets ICl close square brackets space equals space 0.78 space minus space 2 space cross times space 0.167 space straight M space equals space 0.446 space straight M