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.

Calculate DeltaH_f^@ for the reaction CO_2(g) +H_2(g) to CO(g) + H_2O(

1.	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.
Answer» Solution :`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)`.