JEE Main 2024 Chemistry question
Two reaction are given below : 2Fe(s) + (3/2)O₂(g) → Fe₂O₃(s), ΔH° = –822 kJ/mol; C(s) + ½O₂(g) → CO(g), ΔH° = –110 kJ/mol. Then enthalpy change for following reaction 3C(s) + Fe₂O₃(s) → 2Fe(s) + 3CO(g) is kJ/mol.
Answer: 492
Solution
The required reaction = 3 × (second reaction) − (first reaction): ΔH = 3(−110) − (−822) = −330 + 822 = 492 kJ/mol.