The average S–F bond energy in $k J m o l^{- 1}$ of $S F_{6}$ is______. (Rounded off to the nearest integer)
[Given : The values 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.]

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Solution

$S F_{6} (g) \to S (g) + 6 F (g)$

$Δ H_{reaction}^{0} = 6 \times E_{S - F} = Δ H_{f}^{0} [S (g)] + 6 \times Δ H_{f}^{0} [F (g)] - Δ H_{f}^{0} [S F_{6} (g)]$
$6 \times E_{S - F} = 275 + 6 \times 80 - (- 1100)$
$6 \times E_{S - F} = 1855$
$E_{S - F} = \frac{1855}{6}$
$E_{S - F} = 309.166 \approx 309$

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