Verified Solution • 2025–26 Session moderate QJEEM24Chd55c: JEE Main 2024 Chemistry - Class 11 JEEM24Chd55c Solution: Two reactions are given below: Then enthalpy change for… Two reactions are given below: amp;2Fe(s)+32O2( g)→Fe2O3( s),ΔH∘=−822 kJ/molamp;C(s)+12O2( g)→CO(g),ΔH∘=−110 kJ/mol\begin{aligned}& 2 \mathrm{Fe}_{(\mathrm{s})}+\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})}, \Delta \mathrm{H}^{\circ}=-822 \mathrm{~kJ} / \mathrm{mol} \\& \mathrm{C}_{(\mathrm{s})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{(\mathrm{g})}, \Delta \mathrm{H}^{\circ}=-110 \mathrm{~kJ} / \mathrm{mol}\end{aligned}amp;2Fe(s)+23O2( g)→Fe2O3( s),ΔH∘=−822 kJ/molamp;C(s)+21O2( g)→CO(g),ΔH∘=−110 kJ/mol Then enthalpy change for following reaction 3C(s)+Fe2O3( s)→2Fe(s)+3CO(g)3 \mathrm{C}_{(\mathrm{s})}+\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})} \rightarrow 2 \mathrm{Fe}_{(\mathrm{s})}+3 \mathrm{CO}_{(\mathrm{g})}3C(s)+Fe2O3( s)→2Fe(s)+3CO(g) is _______ kJ/mol\mathrm{kJ} / \mathrm{mol}kJ/mol . Step-by-Step Verified Solution 2Fe(s)+32O2( g)→Fe2O3( s),ΔH∘=−822 kJ/mol2 \mathrm{Fe}_{(\mathrm{s})}+\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})}, \Delta \mathrm{H}^{\circ}=-822 \mathrm{~kJ} / \mathrm{mol}2Fe(s)+23O2( g)→Fe2O3( s),ΔH∘=−822 kJ/mol ........ (1) C(s)+12O2( g)→CO(g),ΔH∘=−110 kJ/mol\mathrm{C}_{(\mathrm{s})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{(\mathrm{g})}, \Delta \mathrm{H}^{\circ}=-110 \mathrm{~kJ} / \mathrm{mol}C(s)+21O2( g)→CO(g),ΔH∘=−110 kJ/mol ........ (2) 3C(s)+Fe2O3( s)→2Fe(s)+3CO(g),ΔH3= ? 3 \mathrm{C}_{(\mathrm{s})}+\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})} \rightarrow 2 \mathrm{Fe}_{(\mathrm{s})}+3 \mathrm{CO}_{(\mathrm{g})}, \Delta \mathrm{H}_3=\text { ? }3C(s)+Fe2O3( s)→2Fe(s)+3CO(g),ΔH3= ? amp;(3)=3×(2)−(1)amp;ΔH3amp;=3×ΔH2−ΔH1amp;=3(−110)+822amp;=492 kJ/mole\begin{aligned}& (3)=3 \times(2)-(1) \\& \begin{aligned}\Delta \mathrm{H}_3 & =3 \times \Delta \mathrm{H}_2-\Delta \mathrm{H}_1 \\& =3(-110)+822 \\& =492 \mathrm{~kJ} / \mathrm{mole}\end{aligned}\end{aligned}amp;(3)=3×(2)−(1)amp;ΔH3amp;=3×ΔH2−ΔH1amp;=3(−110)+822amp;=492 kJ/mole Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring