The standard heat of formation values of `SF_(6)(g), S(g)`, and `F(g)` are `-1100, 275`, and `80 kJ mol^(-1)`, respectively. Then the average `S-F` bond enegry in `SF_(6)` (Answer in KJ/mol)

Answer

309

Explanation

The required equation is :
(i)`SF_(6)(g) to S(g) + 6F(g) DeltaH` = ?
Given equations are :
(ii)`S(s) + 3F_(2)(g) to SF_(6)(g) DeltaH = -1100 kJ "mol"^(-1)`
(iii) `S(s) to S(g) DeltaH = 275 kJ "mol"^(-1)`
(iv) `1/2 F_(2)(g) to F(g) DeltaH = 80 kJ "mol"^(-1)`
To calculate `DeltaH ` for Eq(i), carry out the following:
(iii) +6 (iv) -(ii)
`therefore DeltaH = 275 + 6 xx 80 -(-1100)`
= `1855 kJ "mol"^(-1)`
`therefore` S-F bond energy = `(1855)/(6) = 309.16 kJ "mol"^(-1)`.