To estimate the enthalpy change (ΔH) of the reaction using average bond enthalpies, you need to consider the bonds broken and formed in the reaction and calculate the difference between the bond energies.
Given that the bond enthalpies are:
O=O = 495 kJ/mol
S-F = 327 kJ/mol
S=O = 523 kJ/mol
Let's analyze the reaction:
2SF4 + O2 → 2OSF4
In this reaction, we have bonds being broken in SF4 and O2, and bonds being formed in OSF4.
First, calculate the energy required to break the bonds. The bonds being broken are:
4 S-F bonds in 2SF4
1 O=O bond in O2
Energy required to break S-F bonds:
4 S-F bonds × 327 kJ/mol = 1308 kJ/mol
Energy required to break O=O bonds:
1 O=O bond × 495 kJ/mol = 495 kJ/mol
Next, calculate the energy released when new bonds are formed. The bonds being formed are:
4 O-S bonds in 2OSF4
8 S-F bonds in 2OSF4
Energy released when O-S bonds are formed:
4 O-S bonds × 523 kJ/mol = 2092 kJ/mol
Energy released when S-F bonds are formed:
8 S-F bonds × 327 kJ/mol = 2616 kJ/mol
Now, calculate the overall energy change:
Total energy required to break bonds = Energy required to break S-F bonds + Energy required to break O=O bond
Total energy required to break bonds = 1308 kJ/mol + 495 kJ/mol = 1803 kJ/mol
Total energy released when bonds are formed = Energy released when O-S bonds are formed + Energy released when S-F bonds are formed
Total energy released when bonds are formed = 2092 kJ/mol + 2616 kJ/mol = 4708 kJ/mol
Finally, calculate the ΔH (enthalpy change) of the reaction using the equation:
ΔH = Energy released - Energy required
ΔH = Total energy released when bonds are formed - Total energy required to break bonds
ΔH = 4708 kJ/mol - 1803 kJ/mol
ΔH = 2905 kJ/mol
Therefore, the estimated enthalpy change (ΔH) of the reaction 2SF4 + O2 → 2OSF4 is 2905 kJ/mol.