If 2Mg(s) + O₂(g) $\rarr$ 2MgO(s), $\Delta$ H^$\circ$ = -1203.6 kJ/mol, then for
Mg(s) + (1/2)O₂(g) $\rarr$ MgO(s), the enthalpy change is ^$\circ$H = -601.8 kJ/mol.

Question

If 2Mg(s) + O₂(g) $\rarr$ 2MgO(s), $\Delta$H^$\circ$ = -1203.6 kJ/mol, then for Mg(s) + (1/2)O₂(g) $\rarr$ MgO(s), the enthalpy change is ^$\circ$H = -601.8 kJ/mol.

Quizplus · Accepted Answer

The given equation is a combination of two reactions:
2Mg(s) + O2(g) → 2MgO(s) ΔH = -1203.6 kJ/mol
This reaction gives the enthalpy change for the formation of 2 moles of MgO from Mg and O2.

We need to find the enthalpy change for the formation of 1 mole of MgO from Mg and O2, which can be represented as:
Mg(s) + (1/2)O2(g) → MgO(s) ΔH = ?

Using stoichiometry, we can divide the given reaction by 2:
Mg(s) + (1/4)O2(g) → (1/2)MgO(s) ΔH = -601.8 kJ/mol

Therefore, the statement is true and the enthalpy change for the formation of 1 mole of MgO from Mg and O2 is -601.8 kJ/mol.

If 2Mg(s) + O2(g) →\rarr→ 2MgO(s) Δ\DeltaΔ H ∘\circ∘ = -1203 →\rarr→ MgO(s), the Enthalpy Change Is

If 2Mg(s) + O₂(g) $\rarr$ 2MgO(s) $\Delta$ H^$\circ$ = -1203 $\rarr$ MgO(s), the Enthalpy Change Is