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 the given reaction using average bond enthalpies, we need to calculate the total energy required to break the bonds in the reactants and the total energy released when the bonds are formed in the products.

First, let's break down the given reaction and identify the bonds involved:

Reactants:
2SF4 (contains 2 S-F bonds)
O2 (contains 1 O=O bond)

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

Next, we need to find the bond enthalpy values for the bonds involved in the reaction. Given average bond enthalpies are as follows:
O=O = 495 kJ/mol (oxygen-oxygen bond)
S-F = 327 kJ/mol (sulfur-fluorine bond)
S=O = 523 kJ/mol (sulfur-oxygen bond)

Now, we can calculate the energy required to break the bonds in the reactants:

For 2SF4:
Energy required = 2 × (4 × S-F bond enthalpy)

For O2:
Energy required = 1 × (1 × O=O bond enthalpy)

Now, we can calculate the energy released when the bonds are formed in the products:

For 2OSF4:
Energy released = 2 × (2 × S=O bond enthalpy) + 2 × (4 × S-F bond enthalpy)

Finally, we can calculate the enthalpy change (ΔH) of the reaction:
ΔH = (Energy required for reactants) - (Energy released for products)

Let's now substitute the given bond enthalpies and calculate:

Energy required for reactants:
= 2 × (4 × 327 kJ/mol) + 1 × (1 × 495 kJ/mol)
= 2616 kJ/mol + 495 kJ/mol
= 3111 kJ/mol

Energy released for products:
= 2 × (2 × 523 kJ/mol) + 2 × (4 × 327 kJ/mol)
= 2092 kJ/mol + 2616 kJ/mol
= 4708 kJ/mol

ΔH = (Energy required for reactants) - (Energy released for products)
= 3111 kJ/mol - 4708 kJ/mol
= -1597 kJ/mol

So, the estimated enthalpy change (ΔH) of the given reaction is -1597 kJ/mol.

To estimate the enthalpy change (ΔH) of the reaction:

2SF4 + O2 → 2OSF4

We can use the average bond enthalpies to calculate the total bond energy of the reactants and products and then find the difference.

Reactant Bonds:
2SF4 - In SF4 molecule, there are 4 S-F bonds.
Total bond energy of SF4 = 4 × (327 kJ/mol) = 1308 kJ/mol

1 O2 - There is a double bond in O2 molecule.
Total bond energy of O2 = 1 × (495 kJ/mol) = 495 kJ/mol

Total bond energy of the reactants = Total bond energy of SF4 + Total bond energy of O2
= 1308 kJ/mol + 495 kJ/mol
= 1803 kJ/mol

Product Bonds:
2 OSF4 - In OSF4 molecule, there are 4 S=O bonds.
Total bond energy of OSF4 = 4 × (523 kJ/mol) = 2092 kJ/mol

Total bond energy of the products = Total bond energy of OSF4
= 2092 kJ/mol

ΔH (rxn) = Total bond energy of the reactants - Total bond energy of the products
= 1803 kJ/mol - 2092 kJ/mol
= -289 kJ/mol

Therefore, the estimated enthalpy change (ΔH) of the reaction is -289 kJ/mol.