From the following heats of combustion,
CH₂OH(1) + 2O₂(g) → CO₂(g) + 2H₂O(1)
AHxn
= -726.4 kJ/mol
C(graphite) + O₂(g)
- CO₂(g)
→
AHxn
= -393.5 kJ/mol
H₂(g) + O₂(g) →→→ H₂O(1)
AHxn=-285.8 kJ/mol
calculate the enthalpy of formation of methanol
(CH3OH) from its elements:
C(graphite) + 2H₂(g) + O₂(g) → CH₂OH(1)