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JEE Main 30 January 2024, Shift 2, Chemistry Q82

Question 82 of 90 in this shift, Chemistry question 22 of 30, Section B.

Two reactions are given below: 2Fe(s)+32O2(g)→Fe2O3(s), ΔH∘=−822 kJ/mol2\mathrm{Fe_{(s)}} + \frac{3}{2}\mathrm{O_{2(g)}} \to \mathrm{Fe_2O_{3(s)}},\ \Delta H^\circ = -822\ \mathrm{kJ/mol} C(s)+12O2(g)→CO(g), ΔH∘=−110 kJ/mol\mathrm{C_{(s)}} + \frac{1}{2}\mathrm{O_{2(g)}} \to \mathrm{CO_{(g)}},\ \Delta H^\circ = -110\ \mathrm{kJ/mol} Then enthalpy change for following reaction 3C(s)+Fe2O3(s)→2Fe(s)+3CO(g)3\mathrm{C_{(s)}} + \mathrm{Fe_2O_{3(s)}} \to 2\mathrm{Fe_{(s)}} + 3\mathrm{CO_{(g)}} is ________ kJ/mol.

Official answer

492

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