The standard enthalpies of formation of Al_{2 | vedclass

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Question

(C) The given reaction is: $3 CaO + 2 Al \rightarrow Al_{2}O_{3} + 3 Ca$The standard enthalpy of reaction is calculated using the formula: $\Delta_{r}

Vedclass · Accepted Answer

$230$

Vedclass · Answer

(C) The given reaction is: $3 CaO + 2 Al ightarrow Al_{2}O_{3} + 3 Ca$The standard enthalpy of reaction is calculated using the formula: $\Delta_{r}H^{\circ} = \Sigma \Delta_{f}H^{\circ}(	ext{products}) - \Sigma \Delta_{f}H^{\circ}(	ext{reactants})$Since the standard enthalpy of formation for elements in their standard state ($Ca$ and $Al$) is $0 \ kJ \ mol^{-1}$,we have:$\Delta_{r}H^{\circ} = [1 	imes \Delta_{f}H^{\circ}(Al_{2}O_{3}) + 3 	imes \Delta_{f}H^{\circ}(Ca)] - [3 	imes \Delta_{f}H^{\circ}(CaO) + 2 	imes \Delta_{f}H^{\circ}(Al)]$$\Delta_{r}H^{\circ} = [1 	imes (-1675) + 3 	imes 0] - [3 	imes (-635) + 2 	imes 0]$$\Delta_{r}H^{\circ} = -1675 + 1905 = +230 \ kJ \ mol^{-1}$

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