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Equilibrium constant for the following reactions at 1200 K are given : 2H_(2)O_((g))iff2H_(2(g))+O_(2(g)),K_(1)=6.4xx10^(-8) 2CO_(2(g))iff2CO_((g))+O_(2(g)),K_(2)=1.6xx10^(-6) The equilibrium constant for the reaction H_(2(g))+CO_(2(g))iffCO_((g))+H_(2)O_((g)) at 1200 K will be |
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Answer» Solution :Given : `2H_(2)O_((G))iff2H_(2(g))+O_(2(g))` `K_(1)=6.4xx10^(-8)""...(i)` `2CO_(2(g))iff2CO_((g))+O_(2(g))` `K_(2)=1.6xx10^(-6)""...(ii)` Required equation is, `H_(2(g))+CO_(2(g))iffCO_((g))+H_(2)O_((g)),K=?` By reversing equation (i) and by multiplying it with 1/2, we get `H_(2(g))+(1)/(2)O_(2(g))iffH_(2)O_((g)),K._(1)=sqrt((1)/(6.4xx10^(-8)))""...(iii)` And by multiplying equation (ii) with 1/2, we get `H_(2(g))+CO_(2(g))iffCO_((g))+H_(2)O_((g)),K.=K._(1)xxK._(2)` `K.=sqrt((1.6xx10^(-6))/(6.4xx10^(-8)))=sqrt(25)=5` |
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