# The average S-F bond energy in

Question:

The average S-F bond energy in $\mathrm{kJmol}^{-1}$ of $\mathrm{SF}_{6}$ is (Rounded off to the nearest integer)
[Given : The values of standard enthalpy of formation of $\mathrm{SF}_{6}(\mathrm{~g}), \mathrm{S}(\mathrm{g})$ and $\mathrm{F}(\mathrm{g})$ are $-1100,275$ and $80 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively.]

Solution:

$(309)$

$\mathrm{SF}_{6}(\mathrm{~g}) \longrightarrow \mathrm{S}(\mathrm{g})+6 \mathrm{~F}(\mathrm{~g})$

$\Delta \mathrm{H}_{\text {reaction }}^{\circ}=6 \times \mathrm{E}_{\mathrm{S}-\mathrm{F}}=\Delta \mathrm{H}_{\mathrm{f}}^{\circ}[\mathrm{S}(\mathrm{g})]+6 \times \Delta \mathrm{H}_{\mathrm{f}}^{\circ}[\mathrm{F}(\mathrm{g})]-\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left[\mathrm{SF}_{6}(\mathrm{~g})\right]$

$6 \times \mathrm{E}_{\mathrm{s}-\mathrm{F}}=275+6 \times 80-(-1100)$

$=275+480+1100$

$6 \times \mathrm{E}_{\mathrm{s}-\mathrm{F}}=1855$

$\mathrm{E}_{\mathrm{s}-\mathrm{F}}=\frac{1855}{6}=309.1667$

$\simeq 309 \mathrm{~kJ} / \mathrm{molAns}$