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Find DeltaH of the following reaction: ...

Find `DeltaH` of the following reaction:
`OF_(2)(g) +H_(2)O(g) rarr O_(2)(g)+2HF (g)`, average bond energies of `O -F, O-H, O =O`, and `H - F` are `44,111, 118`, and `135 kcal mol^(-1)`, respectively.

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`OF_(2)(g)+H_(2)O(g)rarr O_(2)(g)+2HF(g) " " , " " Delta_(r )H=?`
(i) Bond Breaking : (endothermic) : `Delta_(r )H_(1)`
`F-O-F(g)rarr O(g)+2F(g) " " , Delta_(r )H=+2xx44 " "` [bond energy of O - F = 44]
`H-O-H(g)rarr O(g)+2H(g)" " , Delta_(r )H=+2xx111 " "` [bond energy of O - H = 111]
`rArr Delta_(r )H_(1)=2xx44+2xx111=+310` kcal.
(ii) Bond Formation : (exothermic) : `Delta_(r )H_(2)`
`O(g)+O(g)rarrO=O(g) " " , Delta_(r )H=-118 " "` [bond energy of O = O = 118]
`2[H(g)+F(g)rarr H-F(g)] " " , Delta_(r )H=-2xx135 " "` [bond energy of H - F = 270]
`rArr Delta_(r )H_(2)=-118+(-2xx135)=-388` kcal
Now using Hess' Law , `Delta_(r )H=Delta_(r )H_(1)+Delta_(r )H_(2)=310+(-388)=-78` kcal
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