Calculate `DeltaH_f^@` for the reaction `CO_2(g) +H_2(g) to CO(g) + H_2O(g)` given that `DeltaH_f^0` for `CO_2(g),CO(g)` and `H_2O(g)` are -393.5 , -111.31 and `-242 "kJ mol"^(-1)` respectively.

Explanation

`Delta H_(r )^(0)(CO_(2))=-393.5 "kJ mol"^(-1)`
`Delta H_(r )^(0)(CO)=-111.31"kJ mol"^(-1)`
`Delta H_(r )^(0)(H_(2)O)=-242 "kJ mol"^(-1)`
`CO_(2(g))+H_(2(g))rarr CO_((g))+H_(2)O_((g))`
`Delta H_(r )^(0)=?`
`Delta H_(r )^(0)=sum (Delta H_(r )^(0))_(("product"))=sum (Delta H_(r )^(0))_(("reactant"))`
`Delta H_(r )^(0)=[Delta H_(r )^(0)(CO)+Delta H_(r )^(0)(H_(2)O)]-[Delta H_(r )^(0)(CO_(2))+Delta H_(r )^(0)(H_(2))]`
`= [-111.31+(-242)]-[-393.5+(0)]`
`Delta H_(r )^(0)=[-353.31]+393.5`
`Delta H_(r )^(0)=40.19`
`Delta H_(r )^(0)=+40.19"kJ mol"^(-1)`.