What is the standard enthalpy of reaction (in | vedclass

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(A) The balanced chemical equation for the reaction of $Fe_2O_3$ with $H_2$ is: $Fe_2O_3(s) + 3H_2(g) \rightarrow 2Fe(s) + 3H_2O(l)$.For $2$ moles of

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$-66.58$

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(A) The balanced chemical equation for the reaction of $Fe_2O_3$ with $H_2$ is: $Fe_2O_3(s) + 3H_2(g) ightarrow 2Fe(s) + 3H_2O(l)$.For $2$ moles of $Fe_2O_3$, the equation is: $2Fe_2O_3(s) + 6H_2(g) ightarrow 4Fe(s) + 6H_2O(l)$.The standard enthalpy of reaction is given by: $\Delta H_R^{\circ} = \sum \Delta H_f^{\circ}(	ext{Products}) - \sum \Delta H_f^{\circ}(	ext{Reactants})$.Given $\Delta H_f^{\circ}(Fe_2O_3) = -824.2 \ kJ \ mol^{-1}$, $\Delta H_f^{\circ}(H_2O) = -285.83 \ kJ \ mol^{-1}$, and $\Delta H_f^{\circ}(Fe) = \Delta H_f^{\circ}(H_2) = 0 \ kJ \ mol^{-1}$.Substituting the values: $\Delta H_R^{\circ} = [4(0) + 6(-285.83)] - [2(-824.2) + 6(0)]$.$\Delta H_R^{\circ} = -1714.98 - (-1648.4) = -66.58 \ kJ$.Thus, the correct option is $(A)$.

What is the standard enthalpy of reaction (in $kJ$) when two moles of $Fe_2O_{3(s)}$ react with $H_2$ gas to give $Fe$ metal? $\Delta H_f^{\circ}$ of $Fe_2O_{3(s)}$ and $H_2O_{(l)}$ are $-824.2$ and $-285.83 \ kJ \ mol^{-1}$ respectively.

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