Calculate heat of the following reaction at constant pressure,
`F_(2)O(g)+H_(2)O(g)rarrO_(2)(g)+2HF(g)`
The heats of formation of `F_(2)O(g),H_(2)O(g)` and `HF(g)` are `5.5 kcal-57kcal` and `-64 kcal` repectively.

Explanation

Given that,
(i) `F_(2)(g)+(1)/(2)O_(2)(g) to F_(2)O(g),DeltaH=-5.5kcal`
(ii) `H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(g),DeltaH=-57.8kcal`
(iii) `(1)/(2)H_(2)(g)+(1)/(2)F_(2)(g) to HF(g),DeltaH=-64.2kcal`
`F_(2)O` and `H_(2)O` in Eqns. (i) and (ii) and in the equation given in the problem are on the opposite sides, while HF in Eqn. (iii) and in the equation given in the problem is on the same sides.
Thus applying, `[-Eqn. (i) - Eqn.(ii)+2xxEqn. (iii)]`, we get
`-F_(2)(g)-(1)/(2)O_(2)(g)-H_(2)(g)-(1)/(2)O_(2)(g)+H_(2)+F_(2)(g) to `
`-F_(2)O(g)-H_(2)O(g)+2HF(g),`
`DeltaH=-5.5-(-57.8)+2xx(64.2)`
or `F_(2)O(g)+H_(2)O(g) to O_(2)(g)+2HF(g),DeltaH=-76.1kcal.`