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.

To estimate the enthalpy change (ΔH) of a reaction using average bond enthalpies, you need to calculate the energy required to break the bonds and the energy released when new bonds form.

First, let's identify the bonds in the reactants and products:

Reactants:
2 SF4 molecules: 2 (S-F) bonds

1 O2 molecule: 1 (O=O) bond

Products:
2 OSF4 molecules: 2 (O=S) bonds and 4 (S-F) bonds

Now, use the given average bond enthalpies to calculate the energy required to break the bonds:

For SF4, we have 2 (S-F) bonds:
2 × 327 kJ/mol = 654 kJ/mol

For O2, we have 1 (O=O) bond:
1 × 495 kJ/mol = 495 kJ/mol

Next, calculate the energy released when new bonds are formed:

For OSF4, we have 2 (O=S) bonds:
2 × 523 kJ/mol = 1046 kJ/mol

And we have 4 (S-F) bonds:
4 × 327 kJ/mol = 1308 kJ/mol

Now, calculate the total energy change:

Energy required to break bonds in the reactants: 654 kJ/mol + 495 kJ/mol = 1149 kJ/mol

Energy released when new bonds are formed in the products: 1046 kJ/mol + 1308 kJ/mol = 2354 kJ/mol

ΔH (rxn) = Energy of products - Energy of reactants = 2354 kJ/mol - 1149 kJ/mol = 1205 kJ/mol

Therefore, the estimated enthalpy change (ΔH) of the reaction 2SF4 + O2 → 2OSF4 is 1205 kJ/mol.