If the enthalpies of formation of Al_2O_3 and | vedclass

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(D) The given reaction is: $2Al + Cr_2O_3 	o Al_2O_3 + 2Cr$The enthalpy change of the reaction is calculated as: $\Delta_r H^{\circ} = \Sigma \Delta_

Vedclass · Accepted Answer

$-462$

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(D) The given reaction is: $2Al + Cr_2O_3 	o Al_2O_3 + 2Cr$The enthalpy change of the reaction is calculated as: $\Delta_r H^{\circ} = \Sigma \Delta_f H^{\circ}(	ext{products}) - \Sigma \Delta_f H^{\circ}(	ext{reactants})$$\Delta_r H^{\circ} = [\Delta_f H^{\circ}(Al_2O_3) + 2 	imes \Delta_f H^{\circ}(Cr)] - [2 	imes \Delta_f H^{\circ}(Al) + \Delta_f H^{\circ}(Cr_2O_3)]$Since $Al$ and $Cr$ are in their standard elemental states,their enthalpy of formation is $0$.$\Delta_r H^{\circ} = [-1596 + 2(0)] - [2(0) + (-1134)]$$\Delta_r H^{\circ} = -1596 + 1134 = -462 \, kJ$

If the enthalpies of formation of $Al_2O_3$ and $Cr_2O_3$ are $-1596 \, kJ/mol$ and $-1134 \, kJ/mol$ respectively,then calculate $\Delta H$ for the following reaction in $kJ$:
$2Al + Cr_2O_3 \to Al_2O_3 + 2Cr$

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If the enthalpies of formation of $Al_2O_3$ and $Cr_2O_3$ are $-1596 \, kJ/mol$ and $-1134 \, kJ/mol$ respectively,then calculate $\Delta H$ for the following reaction in $kJ$:$2Al + Cr_2O_3 \to Al_2O_3 + 2Cr$