Hi Mam! I was wondering how to solve this question (with bond enthalpies).

This question is based on bond enthalpies. Remember:

Bond breaking is endothermic (energy is absorbed).

Bond formation is exothermic (energy is released).

So, the enthalpy change of the reaction is found by taking the total energy required to break all bonds in the reactants and subtracting the total energy released when new bonds form in the products.
Part 1: Bond dissociation enthalpy (BDE) method — step by step (no answers)
Step 1: Write the balanced reaction (already given)
Step 2: Count bonds broken (reactants)

In 2 CH₄: each CH₄ has 4 C–H bonds → total C–H bonds broken
= 2 ×4
In 1 O₂: one O=O bond
→ total O=O bonds broken = 1
Step 3: Count bonds formed (products)
Each CH₃OH has: 3 C–H  1 C–O  1 O–H
So in 2 CH₃OH: C–H formed = 2×3
C–O formed = 2 × 1
O–H formed = 2× 1
Use the BDE equation
ΔHrxn�≈∑(BDE of bonds broken)−∑(BDE of bonds formed)
Plug in your counts with the provided BDE table:
 
Broken: (#"C-H" )(413)+(#"O=O" )(498)
Formed: (#"C-H" )(413)+(#"C-O" )(358)+(#"O-H" )(463)
Step 5: Keep track of units and sign
Your result will be in kJ per reaction as written (for producing 2 mol CH₃OH).
Negative value means exothermic.

Part 2: Enthalpies of formation method — step by step (no answers)
Step 1: Use the standard formula


Step 2: Multiply each ΔH_f^∘by its coefficient
Step 3: Substitute and simplify

Step 4: Units and meaning
   Result is in kJ per reaction as written.
   Negative → exothermic, positive → endothermic.