is answer +103 KJ.. plz chk

Plz explain in detail....

i tried to search on internet but i am confused with co-efficents.. how to set up the equaiton .. plz help me ???

Bond enthalpy is the energy required to break a mole of a certain type of bond.

O=O = 495 kj/mol
S-F = 327 kj/mol
S=O = 523 kj/mol

Use average bond enthalpies to estimate the enthalpy delta H (rxn) of the following reaction:

2SF4 + O2 ---- 2OSF4

Express your answer numerically in kilojoules.

Thanks in Advance

How did you get 103?

2*[(4*SF4)+O=O] - 2*[(4*SF4)+S=O] = ??
I get something in the order of -550 or so.

To estimate the enthalpy change (ΔH) of a reaction using bond enthalpies, you need to consider the bonds broken and the bonds formed in the reaction.

First, write down the balanced chemical equation for the reaction:

2SF4 + O2 → 2OSF4

Next, identify the bonds broken and bonds formed in the reaction. In this case, we have the following bonds:

Bonds broken:
- 2 S-F bonds (2 × 327 kJ/mol)

Bonds formed:
- 2 O-S bonds (2 × unknown bond enthalpy)
- 4 S-F bonds (4 × 327 kJ/mol)

Now, we need to find the unknown bond enthalpy for the O-S bond. Unfortunately, since the O-S bond is not given in the provided bond enthalpy values, we have to use an average value.

To find the average bond enthalpy for the O-S bond, you can calculate it by taking the difference between the given bond enthalpies of O=O and S=O, and adding/subtracting the appropriate values.

Average O-S bond enthalpy = O=O - S=O

Using the given bond enthalpy values:
Average O-S bond enthalpy = 495 kJ/mol - 523 kJ/mol
Average O-S bond enthalpy = -28 kJ/mol

Now we can calculate the total energy change using the bond enthalpies:

ΔH = sum of bonds broken - sum of bonds formed

ΔH = (2 × 327 kJ/mol) + (2 × average O-S bond enthalpy) - (4 × 327 kJ/mol)

ΔH = (2 × 327 kJ/mol) + (2 × -28 kJ/mol) - (4 × 327 kJ/mol)

Simplifying:

ΔH = 654 kJ/mol - 56 kJ/mol - 1308 kJ/mol

ΔH = -710 kJ/mol

Therefore, the estimated enthalpy change (ΔH) for the reaction 2SF4 + O2 → 2OSF4 is -710 kJ/mol.

Note: The negative sign indicates that the reaction is exothermic (releases heat).