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

Answer:

Ehthalpy change for the reaction is -323 kJ

Explanation:

Enthalpy change for a reaction, $\Delta H_{rxn}$ is given as:

$\Delta H_{rxn}=\sum [n_{i}\times (E_{bond})_{i}]-\sum [n_{j}\times (E_{bond})_{j}]$

Where $(E_{bond})_{i}$ and $(E_{bond})_{j}$ represents average bond energy in breaking "i" th bond and forming "j" th bond respectively. $n_{i}$ and $n_{j}$ are number of moles of bond break and form respectively.

Reaction: $F-O-F+H-O-H\rightarrow O=O+2H-F$

Here, 2 moles of O-F bond and 2 moles of of O-H bond are broken

          1 mol of O=O and 2 moles of H-F bonds are formed.

So, $\Delta H_{rxn}=[2mol\times E_{O-F}]+[2mol\times E_{O-H}]-[1mol\times E_{O=O}]-[2mol\times E_{H-F}]$

So, $\Delta H_{rxn}=[2mol\times 190kJ/mol]+[2mol\times 463kJ/mol]-[1mol\times 495kJ/mol]-[2mol\times 567kJ/mol]=-323kJ$

Answer 2

Answer:

-334

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