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

(i) H=U+PV
Considering the initial and final states of H,U and V as `H_(1) ,U_(1) ,and V_(1)and H_(2) ,U_(2) ,and V_(2)`
Change in enthalpy is
`(H_(2)-H_(1))=(U_(2)-U_(1))+P(V_(2)-V_(1))`
`DeltaH=DeltaU+PDeltaV`
`CO_(2_(g))+H_(2(g))rarrCO_(g)+H_(2)O_(g)`
`DeltaH_(f)^(@)(CO_(2))=-393.5kJ.mol^(-1)`
`DeltaH_(f)^(@)(CO_(2))=-111.31kJ.mol^(-1)`
`DeltaH_(f)^(@)(H_(2)O)=-242kJ.mol^(-1)`
`DeltaH_(f)^(@)=sumH_(f_("product"))^(@)-sumH_(f_("reactant"))^(@)`
`=[-111.31+(-242)]-[-393.5]`
`=-111.31-242+393.5`
`DeltaH_(f)^(@)=40.19kJ.mol^(-1)`