If the standard heat of the reaction Fe_2O_{3 | vedclass

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Question

(B) For the thermochemical equation $Fe_2O_{3(s)} + 3CO_{(g)} = 2Fe_{(s)} + 3CO_{2(g)}$,the standard heat of reaction $\Delta H_{rxn}^o = -6.6 \, kcal

Vedclass · Accepted Answer

$-196.2$

Vedclass · Answer

(B) For the thermochemical equation $Fe_2O_{3(s)} + 3CO_{(g)} = 2Fe_{(s)} + 3CO_{2(g)}$,the standard heat of reaction $\Delta H_{rxn}^o = -6.6 \, kcal$.Using the formula: $\Delta H_{rxn}^o = [\Sigma \Delta H_f^o (products) - \Sigma \Delta H_f^o (reactants)]$Substituting the values:$-6.6 = [2 	imes \Delta H_f^o(Fe_{(s)}) + 3 	imes \Delta H_f^o(CO_{2(g)})] - [\Delta H_f^o(Fe_2O_{3(s)}) + 3 	imes \Delta H_f^o(CO_{(g)})]$Since $\Delta H_f^o$ of an element in its standard state is $0$,$\Delta H_f^o(Fe_{(s)}) = 0$.$-6.6 = [2(0) + 3(-94)] - [\Delta H_f^o(Fe_2O_{3(s)}) + 3(-26.4)]$$-6.6 = [-282] - [\Delta H_f^o(Fe_2O_{3(s)}) - 79.2]$$-6.6 = -282 - \Delta H_f^o(Fe_2O_{3(s)}) + 79.2$$-6.6 = -202.8 - \Delta H_f^o(Fe_2O_{3(s)})$$\Delta H_f^o(Fe_2O_{3(s)}) = -202.8 + 6.6 = -196.2 \, kcal/mol$.

If the standard heat of the reaction $Fe_2O_{3(s)} + 3CO_{(g)} = 2Fe_{(s)} + 3CO_{2(g)}$ is $-6.6 \, kcal$,then $\Delta H_f^o$ for $Fe_2O_{3(s)}$ is $...... \, kcal/mol$. [Given: $\Delta H_f^o$ of $CO_{(g)} = -26.4 \, kcal$ and $\Delta H_f^o$ of $CO_{2(g)} = -94 \, kcal$]

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