The standard heats of formation at 298K for `C Cl_(4)(g),H_(2)O(g),CO_(2)(g) and HCl(g)` are `-25.5,-57.8,-94.1 and -22.1` kcal `mol^(-1)` respectively. Calculate `DeltaH_(298)^(@)` for the reaction.
`C Cl_(4)(g)+2H_(2)O(g)toCO_(2)(g)+4HCl(g)`.

Explanation

Given that,
(i) `C(s)+2Cl_(2)(g) to C Cl_(4) (g),DeltaH^(0)=-25.5kcal`
(ii) `H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(g),DeltaH^(0)=-57.8kcal`
(iii) `C(s)+O_(2)(g) to CO_(2)(g),DeltaH^(0)=-94.1kcal`
(iv) `(1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g) to HCl(g),DeltaH^(0)=-22.1kcal`
`CO_(2)` and HCl in eqn. (iii) and (iv) respectively are on the same sides as in the required equation, while `c cl_(4)` in Eqn. (i) and `H_(2)O` in Eqn. (ii) are on the opposite sides in the required equation. Hence applying,
`[-Eqn. (i)-2xxEqn.(ii)+Eqn.(iii)+4xxEqn.(iv)]`, we get,
`C(s)-2Cl_(2)(g)-2H_(2)(g)-O_(2)(g)+C(s)+O_(2)(g)+2H_(2)(g)+2Cl_(2)(g)`
` to -c cl_(4)(g)-2H_(2)O(g)+CO_(2)(g)+4HCl(g),`
`DeltaH^(0)=-(-25.5)-(2xx-57.8)+(-94.1)+(4xx22.1)`
or `c cl_(4)(g)+2H_(2)O(g) to CO_(2)(g)+4HCl(g),Delta^(0)=-41.4kcal`