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.]

Question

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.] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

SF 6( g ) arrow S ( g )+6 F ( g ) If ∈ - bond enthalpy Δ r H =6 × ϵ S - F =Δ f H ( S , g )+6 × Δ f H ( F , g )-Δ f H ( S F 6, g ) =275+6 × 80-(-1100) =1855 kJ ϵ S - F =(1855/6)=309.16 kJ / mol

Q. 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.]

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

If $\in$ - bond enthalpy

$Δ_{r} H = 6 \times ϵ_{S - F}$

$= Δ_{f} H (S, g) + 6 \times Δ_{f} H (F, g) - Δ_{f} H (S F_{6}, g)$

$= 275 + 6 \times 80 - (- 1100)$

$= 1855 k J$

$ϵ_{S - F} = \frac{1855}{6} = 309.16 k J / m o l$

Q. The average S−F bond energy in kJmol−1 of SF6​ is _____ .(Rounded off to the nearest integer) [Given : The values of standard enthalpy of formation of SF6​(g),S(g) and F(g) are −1100,275 and 80kJmol−1 respectively.]

SF6​(g)→S(g)+6F(g) If ∈ - bond enthalpy Δr​H=6×ϵS−F​ =Δf​H(S,g)+6×Δf​H(F,g)−Δf​H(SF6​,g) =275+6×80−(−1100) =1855kJ ϵS−F​=61855​=309.16kJ/mol

Q. 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.]

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

If $\in$ - bond enthalpy

$Δ_{r} H = 6 \times ϵ_{S - F}$

$= Δ_{f} H (S, g) + 6 \times Δ_{f} H (F, g) - Δ_{f} H (S F_{6}, g)$

$= 275 + 6 \times 80 - (- 1100)$

$= 1855 k J$

$ϵ_{S - F} = \frac{1855}{6} = 309.16 k J / m o l$