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 .

Explanation

`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)`