Question:

The enthalpy of combustion of carbon and carbon monoxide are \(-393.5\) and \(-283\mathrm{kJ / mol}\) respectively. The enthalpy of formation of carbon monoxide per mole is

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Hess's Law allows calculation of enthalpy of formation from combustion data.
Updated On: Apr 20, 2026
  • \(110.5\mathrm{kJ}\)
  • \(676.5\mathrm{kJ}\)
  • \(-676.5\mathrm{kJ}\)
  • \(-110.5\mathrm{kJ}\)
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
Hess's Law: Enthalpy change is independent of path.

Step 2: Detailed Explanation:
Given:
(1) C(s) + O\(_2\)(g) \(\rightarrow\) CO\(_2\)(g); \(\Delta H = -393.5\) kJ/mol
(2) CO(g) + 1/2 O\(_2\)(g) \(\rightarrow\) CO\(_2\)(g); \(\Delta H = -283.0\) kJ/mol
Subtract (2) from (1):
C(s) + 1/2 O\(_2\)(g) \(\rightarrow\) CO(g); \(\Delta H = -393.5 - (-283.0) = -110.5\) kJ/mol.

Step 3: Final Answer:
\(-110.5\) kJ
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