The value of standard enthalpy of formation of $S F_{6} (g) S (g)$ and $F (g)$ are $- 1100, 275$ and $80 k J m o l^{- 1}$ respectively. The average $S - F$ bond ethalpy is $S F_{6}$ is :

57.5 k J m o l^{- 1}

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309.16 k J m o l^{- 1}

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- 309.16 k J m o l^{- 1}

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269.17 k J m o l^{- 1}

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Solution

The correct option is D $309.16 k J m o l^{- 1}$
$S (g) + 6 F (g) \to {S F}_{6} (g) Δ H_{f} = - 1100 k J {m o l}^{- 1}$
$S (s) \to S (g) Δ H_{f} = + 275 k J {m o l}^{- 1}$
$\frac{1}{2} F_{2} (g) \to F (g) Δ H_{f} = 80 k J {m o l}^{- 1}$
Therefore heat of formation
Bond energy of reactant – Bond energy of product
$- 1100 = [275 + 6 \times 80] - [6 \times S - F]$
Thus bond energy of $S - F = 309 k J m o l^{- 1}$

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The standard heat of formation values of $S F_{6} (g)$ , $S (g)$ and $F (g)$ are: $- 1100, 275$ and $80 K J m o l^{- 1}$ respectively. Then the average S - F bond energy in $S F_{6}$ is:

The standard heat of formation values of $S F_{6} (g), S (g)$ and $F (g)$ are $- 1100, 275$ and $80$ J mol $^{- 1}$ respecitvely. The $S - F$ bond energy in $S F_{6}$ is :