Calculate delta H_(r)^(0) for the reaction CO_(2)(g) + H_(2)(g) rarr CO(g) + H_(2)O(g) given that Delta H_(r)^(0) for CO_(2) (g), CO (g) and H_(2)O (g) are - 393.5, -111.31 and - 242 KJ mol^(-1) respectively.

Calculate delta H_(r)^(0) for the reaction CO_(2)(g) + H_(2)(g) rarr

1.	Calculate delta H_(r)^(0) for the reaction CO_(2)(g) + H_(2)(g) rarr CO(g) + H_(2)O(g) given that Delta H_(r)^(0) for CO_(2) (g), CO (g) and H_(2)O (g) are - 393.5, -111.31 and - 242 KJ mol^(-1) respectively.
Answer» Solution :GIVEN: `Delta H_(F)^(0) CO_(2) = -393.5kJ mol^(-1)` `DeltaH_(f)^(0) CO = -111.31kJ mol^(-1)` `DeltaH_(f)^(0) (H_(2)O) = -242kJ mol^(-1)` `CO_(2)(g) + H_(2)(g) rarr CO(g) + H_(2)O(g)` `DeltaH_(r)^(0) =?` `DeltaH_(r)^(0) = Sigma(DeltaH_(f)^(0))_("products")- Sigma (DeltaH_(f)^(0))_("REACTANTS")` `DeltaH_(r)^(0) = [DeltaH_(f)^(0) (CO) + DeltaH_(f)^(0) (H_(2)O)]- [DeltaH_(f)^(0) (CO_(2)) + DeltaH_(f)^(0) (H_(2))]` `DeltaH_(r)^(0) = [-111.31 + (-242)] -[-393.5+ (0)]` `Delta H_(r)^(0) = [-353.31] + 393.5` `DeltaH_(r)^(0) = 40.19` `DeltaH_(r)^(0) = +40.19 kJ mol^(-1)`