The standard enthalpies of formation of Al₂O₃ and CaO are -1675 kJ mol⁻¹ and -635 kJ mol⁻¹ respectively.
For the reaction
3CaO + 2Al → 3Ca + Al₂O₃ the standard reaction enthalpy ΔᵣH⁰ = _________ kJ.
(Round off to the Nearest Integer)

Solution

Chemistry ma’am · 3 steps

Elements in their standard state have zero enthalpy of formation, so only the two oxides count here.

1. Set up Hess's law. The reaction enthalpy is the sum of the formation enthalpies of the products minus that of the reactants, each multiplied by its coefficient. Al and Ca are elements in their standard state, so their Δ_f H° = 0.

   Δᵣ H° = Δ_f H°(Al₂O₃) - 3 Δ_f H°(CaO)

2. Put in numbers. Δᵣ H° = -1675 - 3 × (-635) = -1675 + 1905 kJ.
3. So. Δᵣ H° = +230 kJ. The answer is 230, and the positive sign shows the reaction is endothermic.

The common slip: -1040. Forgetting to multiply the CaO value by its coefficient 3 gives -1675 + 635 = -1040, not 230.

The idea to keep. For any reaction, take products minus reactants of Δ_f H°, multiply each by its coefficient, and set every free element to zero.

Answer from the NTA final answer key. Solution by PrepWiser.