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

Given
`DeltaH_f^0 CO_2 =-393.5 "kJ mol"^(-1)`
`DeltaH_f^0CO=-111.31 "kJ mol"^(-1)`
`DeltaH_f^0(H_2O)=-242 "kJ mol"^(-1)`
`CO_2(g) + H_2(g) to CO(g)+ H_2O(g)`
`DeltaH_r^0`=?
`DeltaH_r^0=sum (DeltaH_f^0)_"products" -sum(DeltaH_f^0)_"reactants"`
`DeltaH_r^0=[DeltaH_f^0 (CO)+DeltaH_f^0 (H_2O)]-[DeltaH_f^0 (CO_2)+DeltaH_f^0 (H_2)]`
`DeltaH_r^0`=[-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)`