Calculate the enthaply change in the reaction, `CH_(4)(g) + 2O_(2)(g) to CO_(2)(g) + 2H_(2)O` (i)The enthalpy change of formation of `CH_(4) (g), CO_(2)(g) and H_(2)O` (i) are - 74.8 kj `mol^(-1), -393.5 KJ mol^(-1) and - 286 KJ mol^(-1)` respectively.

Explanation

The aim is `CH_(4) (g) + 2O_(2) (g) to CO_(2)(g) + 2H_(2)O(l) Delta ""_(r)H=` ?
Given :
`Delta H^(@) ""_(f)(CH_(4))_((g)) = - 74.8 kj mol^(-1)`
`DeltaH^(@)""_(f)(CO_(2))_((g))=-393.5kjmol^(-1)`
`DeltaH^(@) ""_(f)(H_(2)O)_((l))=285.8kjmol^(-1)`
`Delta " "_(r)H=sumDelta H^(@) " "_(f)("Product")-sumDeltaH6(@) " "_(f)("Reactant")`
`={(-393.5)+2(-285.8)}-(-74.8)`
`Delta " "_(r)H= -890.30 kj mol^(-1)`