Question

Using average bond enthalpies (linked above), estimate the enthalpy change for the following reaction: OF2(g) + H2O(g)O2(g) + 2HF(g)

Answer 1

To estimate the enthalpy change for the reaction 2 H₂(g) + O₂(g) → 2 H₂O(1), we can use the average bond enthalpies from the table. By calculating the energy changes for each bond breaking and formation step, we can estimate the enthalpy change. Using the bond enthalpies for H-H, O=O, and O-H, we can subtract the energy of the bonds broken from the energy of the bonds formed to find the enthalpy change, which is approximately 486 kJ/mol.

Step-by-step explanation:

To estimate the enthalpy change (ΔH) for a chemical reaction, we can use the average bond enthalpies. In this case, we need to calculate the ΔH for the reaction: 2 H₂(g) + O₂(g) → 2 H₂O(1). We can break down the reaction into individual bond breaking and formation steps and calculate the energy changes for each. Using the bond enthalpies from the table, we can estimate the enthalpy change for this reaction by subtracting the energy of the bonds broken from the energy of the bonds formed.

The average bond enthalpy for a H-H bond is 436 kJ/mol, the average bond enthalpy for an O=O bond is 498 kJ/mol, and the average bond enthalpy for an O-H bond is 464 kJ/mol. With this information, we can calculate the ΔH:

ΔH = (4 × O-H bond enthalpy) - (2 × H-H bond enthalpy) - (1 × O=O bond enthalpy)

ΔH = (4 × 464 kJ/mol) - (2 × 436 kJ/mol) - (1 × 498 kJ/mol)

ΔH = 1856 kJ/mol - 872 kJ/mol - 498 kJ/mol

ΔH ≈ 486 kJ/mol

Answer 2

-334

Step-by-step explanation:

Hrxn = BE(bonds broken) - BE(bonds formed)

Moles Bond Bond Energy (kJ/mol)

Bonds broken 2 O-F 184

2 H-O 463

Bonds formed 1 O=O 498

2 H-F 565

H = (2 mol) BE(O-F) + (2 mol) BE(H-O) - [(1 mol) BE(O=O) + (2 mol) BE(H-F)]

H= (2 mol x 184 kJ/mol) + (2 mol x 463 kJ/mol) - (1 mol x 498 kJ/mol) + (2 mol x 565 kJ/mol)

= -334 kJ