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Calculate DeltaH_(f)^(0) for the reactions. CO_(2_((g)))+H_(2_((g)))rarrCO_(2_((g)))+H_(2)O_((g)) " given that "DeltaH_(f)^(0) " for " CO_(2_((g))),CO_((g))and H_(2)O_((g)) " are " -393.5,-111.31 and -242 " KJ.mol"^(-1)respectively . |
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Answer» Solution :`DeltaH_(f)^(0)(CO_(2))=-393.5 "KJ MOL "^(-1)` `DeltaH_(f)^(0)(CO) =-111.31 " kJ.mol"^(-1)` `DeltaH_(f)^(0)(H_(2)O)=-242"kJmol"^(-1)` `CO_(2_((g)))+H_(2_((g)))rarrCO_((g))+H_(2)O_((g))` `DeltaH_(r)^(0)=? ` `DeltaH_(r)^(0)=sum(DeltaH_(r)^(0))_("product")-sum(DeltaH_(r)^(0))_("REACTANT")` `DeltaH_(r)^(0)=[DeltaH_(r)^(0)(CO)+DeltaH_(r)^(0)(H_(2)O)]-[DeltaH_(r)^(0)(CO_(2))+DeltaH_(r)^(0)(H_(2))]` `=[-111.31+(-242)]-[-393.5+(0)]` `DeltaH_(r)^(0)=[-353.31]+393.5` `DeltaH_(r)^(0)=40.19` `DeltaH_(r)^(0)=+40.19 "kJ mol "^(-1)` |
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