Exams › JEE Main › Chemistry

In an Ellingham diagram, the line for which reaction has a negative slope with increasing temperature?

1. Mg + 1/2 O2 → MgO
2. 2Ag + 1/2 O2 → Ag2O
3. C + 1/2 O2 → CO
4. CO + 1/2 O2 → CO2

Correct answer: C + 1/2 O2 → CO

Solution

The reaction C + 1/2 O2 → CO has a negative slope in an Ellingham diagram because it represents the formation of carbon monoxide from carbon and oxygen, which becomes more favorable at higher temperatures, indicating that the entropy change is positive and the Gibbs free energy decreases.

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 →