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Equilibrium

Question
CBSEENCH11006434

The value of Kc = 4·24 at 800 K for the reaction:
CO left parenthesis straight g right parenthesis space plus space straight H subscript 2 straight O left parenthesis straight g right parenthesis space space space rightwards harpoon over leftwards harpoon space space space CO subscript 2 left parenthesis straight g right parenthesis space plus space straight H subscript 2 left parenthesis straight g right parenthesis
Calculate equilibrium concentration of CO2, H2, CO and H2O at 800K, if only CO and H2O are present initially at concentrations of 0·10M each. 

Solution
The reaction is
             space space space space space space space space space space space space space space space space CO left parenthesis straight g right parenthesis space plus space straight H subscript 2 straight O left parenthesis straight g right parenthesis space space space rightwards harpoon over leftwards harpoon space space space space CO subscript 2 left parenthesis straight g right parenthesis space plus space straight H subscript 2 left parenthesis straight g right parenthesis
Initial conc          0.1 M          0.1 M             0              0
At eqm.          (0.1 - x)M       (0.1 -x)M         xM           xM
where x mole of each of the product be formed.
 Applying law of chemical equilibrium, we have
                 straight K subscript straight c space equals space fraction numerator open square brackets CO subscript 2 left parenthesis straight g right parenthesis close square brackets space open square brackets straight H subscript 2 left parenthesis straight g right parenthesis close square brackets over denominator open square brackets CO left parenthesis straight g right parenthesis close square brackets space open square brackets straight H subscript 2 straight O left parenthesis straight g right parenthesis close square brackets end fraction space space space space space... left parenthesis 1 right parenthesis
Putting the values in expression (1), we have
                    4.24 space equals space fraction numerator straight x squared over denominator left parenthesis 0.1 minus straight x right parenthesis squared end fraction space equals space fraction numerator straight x over denominator 0.1 space minus straight x end fraction space equals space 2.06
  or space space space space space space space straight x space equals space 2.06 space left parenthesis 0.1 minus straight x right parenthesis
or space space space space space space space straight x space equals space 0.206 minus 2.06 space straight x
or space space space space 3.06 space straight x space equals space 0.206
or space space space space space space space space straight x space equals space fraction numerator 0.206 over denominator 3.06 end fraction space equals space 0.067

therefore space space space space open square brackets CO subscript 2 close square brackets subscript eq space equals space open square brackets straight H subscript 2 close square brackets subscript eq space equals space 0.067 space straight M
or space space space space open square brackets CO close square brackets subscript eq space equals space open square brackets straight H subscript 2 straight O close square brackets subscript eq space equals space 0.1 minus 0.067
space space space space equals space 0.033 space straight M