Fe2O3(s)+32C(s)⟶32CO2(g)+2Fe(s)ΔH∘=+234.1kJC(s)+O2(g)⟶CO2(g) ΔH∘=−393.5kJUse these equations and ∆H° values to calculate ∆H° for this reaction :4Fe(s)+3O2(g)⟶2Fe2O3(s)

3CO2+4Fe⟶2Fe2O3+3C3C+3O2→3CO24Fe+3O2→2Fe2O3ΔH°1=+234.1kJ⇒−234.1×2=−468.2 KJΔH°2=−393.5kJ⇒−393.5×3=−1180.5 KJ−468.2+−1180.5=−1648.7 KJΔH∘=−234.1×2−393.5×3=−1228.7kJ

Fe2O3(s)+32C(s)⟶32CO2(g)+2Fe(s)ΔH∘=+234.1kJC(s)+O2(g)⟶CO2(g) ΔH∘=−393.5kJUse these equations and ∆H° values to calculate ∆H° for this reaction :4Fe(s)+3O2(g)⟶2Fe2O3(s)

3CO2+4Fe⟶2Fe2O3+3C3C+3O2→3CO24Fe+3O2→2Fe2O3ΔH°1=+234.1kJ⇒−234.1×2=−468.2 KJΔH°2=−393.5kJ⇒−393.5×3=−1180.5 KJ−468.2+−1180.5=−1648.7 KJΔH∘=−234.1×2−393.5×3=−1228.7kJ

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$\begin{array}{r} {Fe}_{2} O_{3} (s) + \frac{3}{2} C (s) ⟶ \frac{3}{2} {CO}_{2} (g) + 2 Fe (s) \\ Δ H^{\circ} = + 234.1 kJ \\ C (s) + O_{2} (g) ⟶ {CO}_{2} (g) Δ H^{\circ} = - 393.5 kJ \end{array}$
Use these equations and $∆ H^{°}$ values to calculate $∆ H^{°}$ for this reaction :
$4 Fe (s) + 3 O_{2} (g) ⟶ 2 {Fe}_{2} O_{3} (s)$

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-1021.2 kJ

-1228. 7 kJ

-129.4 kJ

-1255.3 kJ

answer is A.

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Detailed Solution

$3 {CO}_{2} + 4 Fe ⟶ 2 {Fe}_{2} O_{3} + 3 C$

$3 C + 3 O_{2} \to 3 {CO}_{2}$

$4 Fe + 3 O_{2} \to 2 {Fe}_{2} O_{3}$

$\begin{array}{l} Δ {H^{°}}_{1} = + 234.1 kJ \Rightarrow - 234.1 \times 2 = - 468.2 KJ \\ Δ {H^{°}}_{2} = - 393.5 kJ \Rightarrow - 393.5 \times 3 = - 1180.5 KJ \\ - 468.2 + - 1180.5 = - 1648.7 KJ \end{array}$