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Hess' Law

Find the Δ ...

Question

Find the $Δ H$ of the following reaction. $O F_{2} (g) + H_{2} O (g) ⟶ O_{2} (g) + 2 H F (g)$ . Average bond energies of $O - F$ , $O - H$ , $O = O$ and $H - F$ are $44$ , $111$ , $118$ and $135$ kcal mol $^{- 1}$ , respectively.

Δ H = - 78 k c a l

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Δ H = + 78 k c a l

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Δ H = - 498 k c a l

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Δ H = + 498 k c a l

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Solution

The correct option is A $Δ H = - 78 k c a l$
As we know,
$Δ H =$ Bond energy data of formation of bond + Bond energy data of dissociation of bond

So for $O F_{2} (g) + H_{2} O (g) ⟶ O_{2} (g) + 2 H F (g)$ ,

$Δ H_{r e a c} = Δ H_{B E r e a c t a n t} - Δ H_{B E p r o d u c t}$

$Δ H = 2 \times 44 + 2 \times 111 - 118 - 2 \times 135 = - 78 k c a l / m o l$

Option A is correct.

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