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JEE MAIN 2021
26.02.2021_S2
Question
The average $\mathrm{S}-\mathrm{F}$ bond energy in $\mathrm{kJ} \mathrm{mol}^{-1}$ 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.]
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$\begin{aligned} & \mathrm{SF}_6(\mathrm{~g}) \rightarrow \mathrm{S}(\mathrm{g})+6 \mathrm{~F}(\mathrm{~g}) \\ & \text { If } \in-\text { bound enthalpy } \\ & \Delta_{\mathrm{r}} \mathrm{H}=6 \times \epsilon_{\mathrm{S}-\mathrm{F}} \\ & =\Delta_{\mathrm{r}} \mathrm{H}(\mathrm{S}, \mathrm{g})+6 \times \Delta_{\mathrm{f}} \mathrm{H}(\mathrm{F}, \mathrm{g})-\Delta_{\mathrm{f}} \mathrm{H}\left(\mathrm{SF}_6, \mathrm{~g}\right) \\ & =275+6 \times 80-(-1100) \\ & =1855 \mathrm{~kJ} \\ & \epsilon_{\mathrm{S}-\mathrm{F}}=\frac{1855}{6}=309.16 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$
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