Plz explain in detail....

SF4 we break 4 SF bonds. That times 2.

O2 we break 1 O2 bond. That times 1.

OSF4. We form 4 S-F bonds and 1 S=O bond. That times 2.
Then BE reactants - BE products = delta H rxn.

can u plz figure out the answer for me .. its very urgent.. thnks ..

i still didn't get it

thnks a lot Dr bob

Just plug in the numbers and punch your calculator. The idea here is to show you how to do it, not to do it for you.

To estimate the enthalpy change (ΔH) of a reaction using bond enthalpies, you need to consider the bonds that are broken and formed in the reaction. The general approach involves the following steps:

Step 1: Identify the bonds that are broken in the reactants. In this case, the reaction has two SF4 molecules, which means you have to break two S-F bonds.

Step 2: Determine the bond enthalpy for each bond that is broken. The given bond enthalpy for S-F is 327 kJ/mol.

Step 3: Multiply the number of bonds broken by their respective bond enthalpies. In this case, since two S-F bonds are broken, the total energy required would be 2 * 327 kJ/mol = 654 kJ.

Step 4: Identify the bonds that are formed in the products. The reaction produces 2 OSF4 molecules, which means you will form two S=O bonds.

Step 5: Determine the bond enthalpy for each bond that is formed. The given bond enthalpy for S=O is 523 kJ/mol.

Step 6: Multiply the number of bonds formed by their respective bond enthalpies. Since two S=O bonds are formed, the total energy released would be 2 * 523 kJ/mol = 1046 kJ.

Step 7: Calculate the ΔH of the reaction by subtracting the energy released from the energy required: ΔH = Energy required - Energy released = 654 kJ - 1046 kJ = -392 kJ.

Therefore, the estimated enthalpy change of the reaction 2SF4 + O2 -> 2OSF4 is -392 kJ.