Standard enthalpy of combustion of graphite equals formation of CO2(g) Steps:

• Define standard enthalpy of combustion for graphite: C(graphite) + O₂(g) → CO₂(g).
• Define standard enthalpy of formation for CO₂(g): same reaction, C(graphite) + O₂(g) → CO₂(g).
• Compare values: both are -393.5 kJ/mol by definition.
• Check other options using known values to confirm only C matches numerically. Why C is correct:
• By definition, ΔH°_comb(C, graphite) is identical to ΔH°_f(CO₂, g) as both describe the same reaction. Why the others are wrong:
• A: Atomisation of CH₄ requires ~1660 kJ/mol (endothermic); formation is -74.8 kJ/mol.
• B: Combustion of CH₄ is -890 kJ/mol; graphite + H₂ combustion sums to -680 kJ/mol.
• D: Neutralisation of HCl + NaOH is -57.1 kJ/mol; formation of H₂O(l) is -285.8 kJ/mol.

Final answer: C