Exams › JEE Main › Chemistry

The enthalpies of combustion of carbon and carbon monoxide are −393.5 and −283 kJ mol⁻¹ respectively. The enthalpy of formation of carbon monoxide per mole is

1. −676.5 kJ
2. 676.5 kJ
3. 110.5 kJ
4. −110.5 kJ

Correct answer: −110.5 kJ

Solution

C + 1/2 O2 -> CO equals (C + O2 -> CO2) minus (CO + 1/2 O2 -> CO2): dHf = -393.5 - (-283) = -110.5 kJ/mol.

Related JEE Main Chemistry questions

• Using the following thermochemical equations, determine the enthalpy of vaporisation of water: H₂(g) + 1/2 O₂(g) → H₂O(l), ΔH = -286 kJ H₂(g) + 1/2 O₂(g) → H₂O(g), ΔH = -245.5 kJ
• Given that the standard enthalpy of formation of NH₃ is -46.0 kJ mol⁻¹, and the enthalpy of formation of H₂ from gaseous atoms is -436 kJ mol⁻¹ while that of N₂ from gaseous atoms is -712 kJ mol⁻¹, what is the average N–H bond enthalpy in NH₃?
• For the spontaneous reaction 2C₂H₆(g) + 25O₂(g) → 16CO₂(g) + 18H₂O(g), what are the signs of ΔH, ΔS and ΔG in that order?
• The enthalpy changes for these reactions are given: 1/2 H2(g) + 1/2 O2(g) → OH(g); ΔH = 42.09 kJ mol⁻¹ H2(g) → 2H(g); ΔH = 435.89 kJ mol⁻¹ O2(g) → 2O(g); ΔH = 495.05 kJ mol⁻¹ Using this data, determine the O–H bond energy in the hydroxyl radical.
• The enthalpy of combustion of carbon to form carbon dioxide is −393.5 kJ per mole. What is the heat evolved when 35.2 g of CO₂ is produced from carbon and oxygen gas?
• For a reaction with ΔH = 35.5 kJ mol⁻¹ and ΔS = 83.6 J K⁻¹ mol⁻¹, at what temperature condition will the reaction be spontaneous? (Assume ΔH and ΔS remain constant with temperature.)

⚔️ Practice JEE Main Chemistry free + battle 1v1 →