The average S-F bond energy in kJ mol^{-1} o | vedclass

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(A) The reaction for the atomization of $SF_{6(g)}$ is: $SF_{6(g)} \rightarrow S_{(g)} + 6F_{(g)}$The enthalpy of reaction $(\Delta_{r}H)$ is given by

Vedclass · Accepted Answer

$309$

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(A) The reaction for the atomization of $SF_{6(g)}$ is: $SF_{6(g)} ightarrow S_{(g)} + 6F_{(g)}$The enthalpy of reaction $(\Delta_{r}H)$ is given by: $\Delta_{r}H = \Delta_{f}H(S, g) + 6 	imes \Delta_{f}H(F, g) - \Delta_{f}H(SF_{6}, g)$Substituting the given values: $\Delta_{r}H = 275 + 6 	imes 80 - (-1100)$$\Delta_{r}H = 275 + 480 + 1100 = 1855 \ kJ \ mol^{-1}$Since there are $6$ $S-F$ bonds in $SF_{6}$,the average bond energy $(\epsilon_{S-F})$ is: $\epsilon_{S-F} = \frac{1855}{6} \approx 309.16 \ kJ \ mol^{-1}$Rounding off to the nearest integer,we get $309 \ kJ \ mol^{-1}$.

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

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