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)rarrF_(2)O(g), Delta H=5.5 kcal`
(ii) `H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(g), Delta H= -57 kcal`
(iii) `(1)/(2)H_(2)(g)+(1)/(2)F_(2)(g)rarr(g)rarrHF(g), Delta H= -64 kcal`
`F_(2)O` and `H_(2)O` in eqns. (i) and (ii) and in the eqn. given in the problem are on the oppsite sides, while `HF` in eqn. (iii) and in the eqn. give in the problem is on the same sides.
Thus applying , `[-Eqn.(i) +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)rarr-F_(2)O(g)-H_(2)O(g)+2HF(g)`,
`Delta H= -5.5-(-57)+2xx(-64)`
or `F_(2)O(g)+H_(2)O(g)rarrO_(2)(g)+2HF(g), Delta H= -76.5 kcal`.