The standard enthalpies of formation of SF6(g), S(g) and F(g) are -1150, +280 and + 85 kJ mol-1. Thus, average bond energy of (S - F) in SF6 is

Question

The standard enthalpies of formation of SF6(g), S(g) and F(g) are -1150, +280 and + 85 kJ mol-1. Thus, average bond energy of (S - F) in SF6 is (A) 309.16 kJ mol-1 (B) 1855 kJ mol-1 (C) 11.130 × 103 kJ mol-1 (D) 323.33 kJ mol-1

Tardigrade Admin · Accepted Answer

S(g)+6F(g)→ SF6(g) Based on BE values ΔfH°(SF6)=-1100=ΔfH°(S,g)+6ΔfH°(F, g)-6(B E)S-F -1150=280+6×85-6(BE)S-F ∴ (BE)S-F=(280+510+1150/6)=323.33 kJ mol-1

Q. The standard enthalpies of formation of $S F_{6} (g), S (g)$ and $F (g)$ are $- 1150, + 280$ and $+ 85 k J m o l^{- 1}$ . Thus, average bond energy of $(S - F)$ in $S F_{6}$ is

$309.16 k J m o l^{- 1}$

$1855 k J m o l^{- 1}$

$11.130 \times 1 0^{3} k J m o l^{- 1}$

$323.33 k J m o l^{- 1}$

$S (g) + 6 F (g) \to S F_{6} (g)$
Based on $BE$ values
$Δ_{f} H^{\circ} (S F_{6}) = - 1100 = Δ_{f} H^{\circ} (S, g) + 6 Δ_{f} H^{\circ} (F, g) - 6 (BE)_{S - F}$
$- 1150 = 280 + 6 \times 85 - 6 (BE)_{S - F}$
$∴ (BE)_{S - F} = \frac{280 + 510 + 1150}{6} = 323.33 k J m o l^{- 1}$

Q. The standard enthalpies of formation of SF6​(g),S(g) and F(g) are −1150,+280 and +85kJmol−1. Thus, average bond energy of (S−F) in SF6​ is

309.16kJmol−1

1855kJmol−1

11.130×103kJmol−1

323.33kJmol−1