Question

Calculate the enthalpy change for the following reaction:
`CH_(4)(g)+2O_(2)(g)toCO_(2)(g)+2H_(2)O(l)`
given, enthalpies of formation of `CH_(4),CO_(2) and H_(2)O` are `-74.8kJ`
`mol^(-1),-393.5kJ" "mol^(-1) and -286.2kJ" "ml^(-1)` respectively.

Answer 1

`DeltaH^(@)=DeltaH_(f("products"))^(@)-DeltaH_(f("reactants"))^(@)`
`=[DeltaH_(f(CO_(2)))^(@)+2DeltaH_(f(H_(2)O))^(@)]-[DeltaH_(f(CH_(4)))^(@)+2DeltaH_(f(O_(2)))^(@)]`
`=[-393.5+2xx(-286.2)]=[-74.8+2xx0]`
`=-393.5-572.4+74.8`
`=891.1kJ`