Calculate the lattice energy for MgO(s)using a Born-Haber cycle and the following information: MgO(s)→ Mg²⁺(g)+ O^2-(g)? Mg(s)→ Mg(g)+147.1 kJ/mol Mg(g)→ Mg⁺(g)+ e^- +737.8 kJ/mol Mg⁺(g)→ Mg²⁺(g)+ e^- +1451 kJ/mol 1/2 O₂(g)→ O(g)+249.0 kJ/mol O(g)+ e^- → O^-(g)-141.1 kJ/mol O^-(g)+ e^- → O^2-(g)+798.0 kJ/mol Mg(s)+ 1/2 O₂(g)→ MgO(s)-601.8 kJ/mol A)+1842 kJ/mol B)+2444 kJ/mol C)+3844 kJ/mol D)+4108 kJ/mol

Question

Quizplus · Accepted Answer

Using the Born-Haber cycle, we need to calculate the sum of the enthalpies of formation of each species involved in the formation of MgO(s) from its elements in their standard states.

Mg(s) → Mg(g) + 147.1 kJ/mol (sublimation energy)
Mg(g) → MgS1U1P1 + S1S1P0(g) + eS1U1P1-S1S1P0 + 737.8 kJ/mol (first ionization energy)
MgS1U1P1 + S1S1P0(g) → MgS1U1P12 + S1S1P0(g) + eS1U1P1-S1S1P0 + 1451 kJ/mol (electron affinity of sulfur)
1/2 O2(g) → O(g) + 1/2 OS1U1B12S1U1B0(g) + 124.5 kJ/mol (bond dissociation energy of O2)
O(g) + eS1U1P1-S1S1P0 → OS1U1P1-S1S1P0(g) + 141.1 kJ/mol (electron affinity of oxygen)
OS1U1P1-S1S1P0(g) + eS1U1P1-S1S1P0 → OS1U1P12-S1S1P0(g) + 798.0 kJ/mol (second electron affinity of sulfur)
MgS1U1P12 + OS1U1P12-S1S1P0(g) → MgO(s) + 601.8 kJ/mol (lattice energy)

Adding up these values, we get:

ΔH = 147.1 + 737.8 + 1451 - 124.5 + 141.1 + 798.0 - 601.8 = 2858.7 kJ/mol

Therefore, the lattice energy for MgO(s) is approximately 3844 kJ/mol (sum of all enthalpies minus the enthalpy of formation of MgO(s)), which corresponds to answer choice C.

Calculate the Lattice Energy for MgO(s)using a Born-Haber Cycle and the Following