The standard enthalpies of formation of $$Al_2O_3$$ and CaO are $$-1675$$ kJ mol$$^{-1}$$ and $$-635$$ kJ mol$$^{-1}$$ respectively. For the reaction $$3CaO + 2Al \to 3Ca + Al_2O_3$$ the standard reaction enthalpy $$\Delta_r H^0$$ = ________ kJ. (Round off to the Nearest Integer).

Using Hess's law, the standard reaction enthalpy is calculated from the standard enthalpies of formation: $$\Delta_r H^0 = \sum \Delta_f H^0(\text{products}) - \sum \Delta_f H^0(\text{reactants})$$.

For the reaction $$3CaO + 2Al \to 3Ca + Al_2O_3$$, the enthalpies of formation of elements in their standard states (Al and Ca) are zero. So $$\Delta_r H^0 = [\Delta_f H^0(Al_2O_3) + 3 \times \Delta_f H^0(Ca)] - [3 \times \Delta_f H^0(CaO) + 2 \times \Delta_f H^0(Al)]$$.

Substituting the values: $$\Delta_r H^0 = [(-1675) + 3(0)] - [3(-635) + 2(0)] = -1675 - (-1905) = -1675 + 1905 = 230$$ kJ.

The standard reaction enthalpy is 230 kJ.