Calculate the average S-F bond energy in SF
6(g) using the bond data in the table, the ΔHf of SF
6(g) which is -1100 kJ mol-1, and
S(s) → S(g) ΔHº = 223 kJ mol-1
The average S-F bond energy is one sixth of the process:
SF
6(g) --> S(g) + 6F(g)
Solution
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You are given:
1. F
2 --> 2F(g) in the table (+158 kJ)
and you are told that:
2. S(s) → S(g) ΔHº = 223 kJ mol-1
Enthalpy of formation is represented by the equation:
3. S(s) + 3F
2(g) --> SF
6(g) ........... -1100 kJ
So you use equations 1, 2 & 3 to construct the equation you need.
This is really application of Hess's law.
You can't go from:
SF
6(g) --> S(g) + 6F(g)
directly, so you have to construct a way starting with SF
6(g).
Reverse equation 3:
SF
6(g) --> S(s) + 3F
2(g) ..... +1100 kJ
Then use equation 2.
SF
6(g) --> S(s) + 3F
2(g) --- > S(s) → S(g) + 3F
2(g) ....... ΔHº = (223 + 1100) kJ
Then use 3x equation 1: 3F
2 --> 6F(g) ........ +474 kJ
SF
6(g) --> S(s) + 3F
2(g) --- > S(s) → S(g) + 3F
2(g) --> S(g) + 6F(g) ....... ΔHº = (223 + 1100 + 474 ) kJ
So, you have got an alternative route from SF
6(g) to S(g) + 6F(g)
Then just divide the sum by 6 ... chachi