Answer
See the answer below.
Work Step by Step
a) Bonds broken: 3 O=O, 6 C-H, 2 O-H, 2 O-C
Bonds formed: 4 C=O, 8 O-H (net: 6)
Enthalpy of reaction:
$3×498+6×413+2×358-4×803-6×463=-1302\ kJ/mol$
b) Enthalpies of formation: $CH_3OH(g): -201, CO_2(g): -393.509, H_2O(g): -241.83\ kJ/mol$
Enthalpy of reaction: $4×-241.83+2×-393.509-2×-201=-1352.3\ kJ/mol$
Both values are reasonably close to each other.
